Question 15 Marks
Differentiate the following functions with respect to x:
$\text{x}^{(\sin\text{x}-\cos\text{x})}+\frac{\text{x}^2-1}{\text{x}^2+1}$
AnswerLet $\text{y}=\text{x}^{(\sin\text{x}-\cos\text{x})}+\Big(\frac{\text{x}^2-1}{\text{x}^2+1}\Big)$
$\text{y}=\text{e}^{\log\text{x}^{\sin\text{c}-\cos\text{x}}}+\Big(\frac{\text{x}^2-1}{\text{x}^2+1}\Big)$
$\text{y}=\text{e}^{(\sin\text{c}-\cos\text{x})\log\text{x}}+\Big(\frac{\text{x}^2-1}{\text{x}^2+1}\Big)$
$\big[\text{Since},\text{e}^{\log\text{a}}=\text{a},\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating it with respect to x using chain rule and quotient rule,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big[\text{e}^{(\sin\text{x}-\cos\text{x})\log\text{x}}\Big]+\frac{\text{d}}{\text{dx}}\Big[\frac{\text{x}^2-1}{\text{x}^2+1}\Big]$
$=\text{e}^{(\sin\text{x}-\cos\text{x})\log\text{x}}+\frac{\text{d}}{\text{dx}}\big\{(\sin\text{x}-\cos\text{x})\log\text{x}\big\} \\ +\bigg[\frac{(\text{x}^2+1)\frac{\text{d}}{\text{dx}}(\text{x}^2-1)-(\text{x}^2-1)\frac{\text{d}}{\text{dx}}(\text{x}^2+1)}{(\text{x}^2+1)^2}\bigg]$
$=\text{e}^{\log\text{x}^{(\sin\text{x}-\cos\text{x})}}\Big[(\sin\text{x}-\cos\text{x})\frac{\text{d}}{\text{dx}}(\log\text{x})+(\log\text{x})\frac{\text{d}}{\text{dx}}(\sin\text{x}-\cos\text{x})\Big] \\+\Big[\frac{(\text{x}^2+1)(2\text{x})-(\text{x}^2-1)(2\text{x})}{(\text{x}^2+1)^2}\Big]$
$=\text{e}^{(\sin\text{x}-\cos\text{x})}\Big[(\sin\text{x}-\cos\text{x}\big(\frac{1}{\text{x}}\big)+\log\text{x}(\sin\text{x}+\cos\text{x})\Big] \\ +\Big[\frac{2\text{x}^3+2\text{x}-2\text{x}^3+2\text{x}}{(\text{x}^2+1)^2}\Big]$
$\frac{\text{dy}}{\text{dx}}=\text{x}^{\sin\text{x}-\cos\text{x}}\Big[\frac{(\sin\text{x}-\cos\text{x})}{\text{x}}+\log\text{x}(\sin\text{x}+\cos\text{x})\Big]+\frac{4\text{x}}{(\text{x}^2+1)^2}$
View full question & answer→Question 25 Marks
Differentiate $\sin^{-1}\Big(2\text{ax}\sqrt{1-\text{a}^2\text{x}^2}\Big)$ with respect to $\sqrt{1-\text{a}^2\text{x}^2},$ if $-\frac{1}{\sqrt{2}}<\text{ax}<\frac{1}{\sqrt{2}}$.
AnswerLet $\text{u}=\sin^{-1}\Big(2\text{ax}\sqrt{1-\text{a}^2\text{x}^2}\Big)$
Put $\text{ax} =\sin\theta\Rightarrow\theta=\sin^{-1}(\text{ax})$
$\therefore\text{u}=\sin^{-1}\Big(2\sin\theta\sqrt{1-\sin^{2}\theta}\Big)$
$\Rightarrow\text{u}=\sin^{-1}(2\sin\theta\cos\theta)$
$\Rightarrow\text{u}=\sin^{-1}(\sin2\theta)\ .....(\text{i})$
And
Let, $\text{v}=\sqrt{1-\text{a}^2\text{x}^2}$
Differentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{1}{2\sqrt{1-\text{a}^2\text{x}^2}}\times\frac{\text{d}}{\text{dx}}\big(1-\text{a}^2\text{x}^2\big)$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\Big(\frac{0-2\text{a}^2\text{x}}{2\sqrt{1-\text{a}^2\text{x}^2}}\Big)$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{-\text{a}^2\text{x}}{\sqrt{1-\text{a}^2\text{x}^2}}\ .....(\text{ii})$
Here,
$-\frac{1}{\sqrt{2}}<\text{ax}<\frac{1}{\sqrt{2}}$
$\Rightarrow-\frac{1}{\sqrt{2}}<\sin\theta<\frac{1}{\sqrt{2}}$
$\Rightarrow-\frac{\pi}{4}<\theta<\frac{\pi}{4}$
So, from equation (i),
$\text{u}=2\theta\Big[\text{Since},\sin^{-1}(\sin\theta)=\theta,\text{if }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\Rightarrow\text{u}=2\sin^{-1}\text{x}$
Differentiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=2\times\frac{1}{\sqrt{1-(\text{ax})^2}}\frac{\text{d}}{\text{dx}}(\text{ax})$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\frac{2}{1-\text{a}^2\text{x}^2}(\text{a})$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\frac{2\text{a}}{1-\text{a}^2\text{x}^2}\ .....(\text{iii})$
Dividing equation (iii) by (ii),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\Big(\frac{2\text{a}}{\sqrt{1-\text{a}^2\text{x}^2}}\Big)\Big(\frac{\sqrt{1-\text{a}^2\text{x}^2}}{\text{-a}^2\text{x}}\Big)$
$\therefore\frac{\text{du}}{\text{dv}}=-\frac{2}{\text{ax}}$
View full question & answer→Question 35 Marks
Differentiate the following functions from first principles:
$\text{e}^{\sqrt{2\text{x}}}$
AnswerLet $\text{f(x)}=\text{e}^{\sqrt{2\text{x}}}$
$\Rightarrow\text{f}(\text{x}+\text{h})=\text{e}^{\sqrt{2(\text{x}+\text{h})}}$
$\therefore\frac{\text{d}}{\text{dx}}(\text{f(x)})=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0 }\frac{\text{e}^{\sqrt{2(\text{x}+\text{h})}}-\text{e}^{\sqrt{2\text{x}}}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\text{e}^{\sqrt{2\text{x}}}\frac{\text{e}^{\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}}}-1}{\text{h}}$
$=\text{e}^{\sqrt{2\text{x}}}\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{\big(\text{e}^{2(\text{x}+\text{h})-\sqrt{2\text{x}}}-1\big)}{\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}}}\Bigg)\bigg(\frac{\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}}}{\text{h}}\bigg)$
$=\text{e}^{\sqrt{2\text{x}}}\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{2}(\text{x}+\text{h})-\sqrt{2\text{x}}}{\text{h}}\ \Big[\text{Since},\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^\text{h}-1}{\text{h}}=1\Big]$
$=\text{e}^{\sqrt{2\text{x}}}\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}}}{\text{h}}\times\frac{\sqrt{2(\text{x}+\text{h})}+\sqrt{2\text{x}}}{\sqrt{2(\text{x}+\text{x})}+\sqrt{2\text{x}}}$
$[\text{Rationalizing the numerator]}$
$=\text{de}^{\sqrt{2\text{x}}}\lim\limits_{\text{h}\rightarrow0}\frac{2(\text{x}+\text{h})-2\text{x}}{\text{h}\big(\sqrt{2(\text{x}+\text{h})}+\sqrt{2\text{x}}\big)}$
$=\text{e}^{\sqrt{2\text{x}}}\lim\limits_{\text{h}\rightarrow0}\frac{2\text{x}+2\text{h}-2\text{x}}{\text{h}\big(\sqrt{2}(\text{x}+\text{h})+\sqrt{2\text{x}}\big)}$
$=\text{e}^{\sqrt{2\text{x}}}\lim\limits_{\text{h}\rightarrow0}\frac{2\text{h}}{\text{h}\big(\sqrt{2(\text{x}+\text{h})}+\sqrt{2\text{x}}\big)}$
$=\text{e}^{\sqrt{2\text{x}}}\lim\limits_{\text{h}\rightarrow0}\frac{2\text{h}}{\big(\sqrt{2(\text{x}+\text{h})}+\sqrt{2\text{x}}\big)}$
$=\frac{\text{e}^{\sqrt{2\text{x}}}}{\sqrt{2\text{x}}}$
So,
$\frac{\text{d}}{\text{dx}}\big(\text{e}^{\sqrt{2\text{x}}}\big)=\frac{\text{e}^{2\text{x}}}{\sqrt{2\text{x}}}$
View full question & answer→Question 45 Marks
Differentiate the following functions from first principles:
$\text{e}^{\sqrt{\cot\text{x}}}$
AnswerLet $\text{f(x)} =\text{e}^{\sqrt{\cot\text{x}}}$
$\Rightarrow\ \text{f}(\text{x}+\text{h})=\text{e}^{\sqrt{\cot\text{x}}}$
$\therefore \frac{\text{d}}{\text{dx}}\{\text{f(x)}\}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})=\text{f(x)}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{\sqrt{\cot(\text{x}+\text{h})}}-\text{e}^{\sqrt{\cot\text{x}}}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{\sqrt{\cot\text{x}}}\Big(\text{e}^{\sqrt{\cot(\text{x}+\text{h})}-\sqrt{\cot\text{x}}}-1\Big)}{\text{h}}$
$=\text{e}^{\sqrt{\cot\text{x}}}\lim\limits_{\text{h}\rightarrow0}\bigg(\frac{\text{e}^{\sqrt{\cot(\text{x}+\text{h})}-\sqrt{\cot\text{x}}}-1}{{\sqrt{\cot(\text{x}+\text{h})}}-\sqrt{\cot\text{x}}}\bigg)\times\bigg(\frac{\sqrt{\cot(\text{x}+\text{h})}-\sqrt{\cot\text{x}}}{\text{h}}\bigg)$
$=\text{e}^{\sqrt{\cot\text{x}}}\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{\cot(\text{x}+\text{h})}-\sqrt{\cot\text{x}}}{\text{h}}\times\frac{\sqrt{\cot(\text{x}+\text{h})}+\sqrt{\cot\text{x}}}{\sqrt{\cot(\text{x}+\text{h})}+\sqrt{\cot\text{x}}}$
$\Big[\text{Since, } \lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^\text{x}-1}{\text{x}}=1\text{ and rationalizing numerator}\Big]$
$\text{e}^{\sqrt{\cot\text{x}}}\lim\limits_{\text{h}\rightarrow0}\frac{\cot(\text{x}+\text{h})-\cot\text{x}}{\text{h}\big(\sqrt{\cot(\text{x}+\text{h})}+\sqrt{\cot\text{x}}\big)}$
$=\text{e}^{\sqrt{\cot\text{x}}}\lim\limits_{\text{h}\rightarrow0}\frac{\frac{\cot(\text{x}+\text{h})\cot\text{x}+1}{\cot(\text{x}+\text{h}-\text{x})}}{\text{x}\big(\sqrt{\cot(\text{x}+\text{h})}+\sqrt{\cot\text{x}}\big)}$
$\Big[\text{Since,} \cot(\text{A}-\text{B})=\frac{\cot\text{A}\cot\text{B}+1}{\cot\text{A}-\cot\text{B}}\Big]$
$=\text{e}^{\sqrt{\cot\text{x}}}\lim\limits_{\text{h}\rightarrow0}\frac{\cot(\text{x}+\text{h})\cot\text{x}+1}{\cot(-\text{h})\times\text{h}\big(\sqrt{\cot(\text{x}+\text{h})}+\sqrt{\cot\text{x}}\big)}$
$=\text{e}^{\sqrt{\cot\text{x}}}\lim\limits_{\text{h}\rightarrow0}\frac{\cot(\text{x}+\text{h})\cot\text{x}+1}{\Big(\frac{\text{h}}{\cot\text{h}}\Big)\big(\sqrt{\cot(\text{x}+\text{h})}+\sqrt{\cot\text{x}}\big)}$
$=\frac{\text{e}^\sqrt{\cot\text{x}}\times(\cot^2\text{x}+1)}{2\sqrt{\cot\text{x}}}\ \Big[\because\ \lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}=1\Big]$
$=-\frac{\text{e}^\sqrt{\cot\text{x}}\times\text{cosec}^2\text{x}}{2\sqrt{\cot\text{x}}}\ \big[\because(1+\cot^2\text{x})=\text{cosec}^2\text{x}\big]$
$\therefore\ \frac{\text{d}}{\text{dx}}\Big(\text{e}^\sqrt{\cot\text{x}}\Big)=-\frac{\text{e}^\sqrt{\cot\text{x}}\times\text{cosec}^2\text{x}}{2\sqrt{\cot}\text{x}}$
View full question & answer→Question 55 Marks
If $\text{y}=\frac{\text{e}^\text{x}-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}},$ prove that $\frac{\text{dy}}{\text{dx}}=1-\text{y}^2$
AnswerGivne, $\text{y}=\frac{\text{e}^{\text{x}}-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$
Differentiate with respect to x,
$\frac{\text{d}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{e}^{\text{x}}-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}\Big)$
$=\Bigg[\frac{\big(\text{e}^{\text{x}}+\text{e}^{-\text{x}}\big)\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}}-\text{e}^{-\text{x}}\big)\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}}+\text{e}^{-\text{x}}\big)}{\big(\text{e}^{\text{x}}+\text{e}^{-\text{x}}\big)^2}\Bigg]$
[Using quotient rule and chain rule]
$=\begin{bmatrix} \frac{\big(\text{e}^{\text{x}}+\text{e}^{-\text{x}}\big)\Big[\text{e}^{\text{x}}-\text{e}^{-\text{x}}\frac{\text{d}}{\text{dx}}(-\text{x})-\big(\text{e}^{\text{x}}-\text{e}^{-\text{x}}\big)\Big(\text{e}^{\text{x}}+\text{e}^{-\text{x}}\frac{\text{d}}{\text{dx}}(-\text{x})\Big)\Big]}{\big(\text{e}^{\text{x}}+\text{e}^{-\text{x}}\big)^2} \end{bmatrix}$
$=\begin{bmatrix} \frac{\big(\text{e}^{\text{x}}+\text{e}^{-\text{x}}\big)\big(\text{e}^{\text{x}}+\text{e}^{-\text{x}}\big)-\big(\text{e}^{\text{x}}-\text{e}^{-\text{x}}\big)\big(\text{e}^{\text{x}}-\text{e}^{-\text{x}}\big)}{\big(\text{e}^{\text{x}}+\text{e}^{-\text{x}}\big)^2} \end{bmatrix}$
$=\bigg[\frac{\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}+2\text{e}^{\text{x}}\times\text{e}^{-\text{x}}-\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}+2\text{e}^{\text{x}}\text{e}^{-\text{x}}}{\big(\text{e}^{\text{x}}+\text{e}^{-\text{x}}\big)^2}\bigg]$
$\frac{\text{dy}}{\text{dx}}\bigg[\frac{4}{(\text{e}^{\text{x}}+\text{e}^{-\text{x}})^2}\bigg]\ .....(\text{i})$
Now,
$1-\text{y}^2=1-\Big(\frac{\text{e}^{\text{x}}-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}\Big)^2$
$=1-\frac{(\text{e}^{\text{x}}-\text{e}^{-\text{x}})^2}{(\text{e}^{\text{x}}+\text{e}^{-\text{x}})^2}$
$=\frac{(\text{e}^{\text{x}}+\text{e}^{-\text{x}})^2-(\text{e}^{\text{x}}-\text{e}^{-\text{x}})^2}{(\text{e}^{\text{x}}+\text{e}^{-\text{x}})^2}$
$=\frac{4}{(\text{e}^{\text{x}}+\text{e}^{-\text{x}})^2}$
View full question & answer→Question 65 Marks
If $\sqrt{1-\text{x}^2}+\sqrt{1-\text{y}^2}=\text{a}(\text{x}-\text{y}),$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\sqrt{1-\text{y}^2}}{1-\text{x}^2}$
AnswerWe have, $\sqrt{1-\text{x}^2}+\sqrt{1-\text{y}^2}=\text{a}\big(\text{x}-\text{y}\big)$
Let $\text{x}=\sin\text{A},\text{y}=\sin\text{B}$
$\Rightarrow\sqrt{1-\sin^2\text{A}}+\sqrt{1-\sin^2\text{B}}=\text{a}\big(\sin\text{A}-\sin\text{B}\big)$
$\Rightarrow\cos\text{A}+\cos\text{B}=\text{a}\big(\sin\text{A}-\sin\text{B}\big)$
$\Rightarrow\text{a}=\frac{\cos\text{A}+\cos\text{B}}{\sin\text{A}-\sin\text{B}}$
$\Rightarrow\text{a}=\frac{2\cos\frac{\text{A}+\cos\text{B}}{2}\cos\frac{\text{A}-\text{B}}{2}}{2\cos\frac{\text{A}+\text{B}}{2}\sin\frac{\text{A}-\text{B}}{2}}$
$\begin{bmatrix}\because\sin\text{A}-\sin\text{B}=2\cos\frac{\text{A}+\text{B}}{2}\sin\frac{\text{A}-\text{B}}{2} \\ \because\cos \text{A}+\cos\text{B}=2\cos\frac{\text{A}+\text{B}}{2}\cos\frac{\text{A}-\text{B}}{2}\end{bmatrix}$
$\Rightarrow\text{a}=\cot\Big(\frac{\text{A}-\text{B}}{2}\Big)$
$\Rightarrow\cot^{-1}\text{a}=\frac{\text{A}-\text{B}}{2}$
$\Rightarrow2\cot^{-1}\text{a}={\text{A}-\text{B}}$
$\Rightarrow2\cot^{-1}\text{a}=\sin^{-1}\text{x}-\sin^{-1}\text{y}$
$\Big[\because \text{x}=\sin\text{A},\text{y}=\sin\text{B}\Big]$
Differentiating with respect to x, we get
$\frac{\text{d}}{\text{dx}}\big(2\cot^{-1}\text{a}\big)=\frac{\text{d}}{\text{dx}}\big(\sin^{-1}\text{x}\big)-\frac{\text{d}}{\text{dx}}\big(\sin^{-1}\text{y}\big)$
$\Rightarrow0=\frac{1}{\sqrt{1-\text{x}^2}}-\frac{1}{\sqrt{1-\text{y}^2}}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{1}{\sqrt{1-\text{y}^2}}\frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sqrt{1-\text{y}^2}}{\sqrt{1-\text{x}^2}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\sqrt\frac{1-\text{y}^2}{1-\text{x}^2}$
View full question & answer→Question 75 Marks
Differentiate the following functions with respect to x:
$\text{x}^{\text{x}^2-3}+(\text{x}-3)^{\text{x}^2}$
AnswerLet $\text{y}=\text{x}^{\text{x}^2-3}+(\text{x}-3)^{\text{x}^2}$
Also, let $\text{u}=\text{x}^{\text{x}^2-3}\text{ and v}=(\text{x}-3)^{\text{x}^2}$
$\therefore \text{y}=\text{u}+\text{v}$
Differentiating both sides with respect to x, we obtain
$\frac{\text{dy}}{\text{dx}}=\frac{\text{du}}{\text{dx}}+\frac{\text{dv}}{\text{dx}}\ .....(\text{i})$
$\text{u}=\text{x}^{\text{x}^2-3}$
$\log\text{u}=(\text{x}^2-3)\log\text{x}$
Differentiating with respect to x, we obtain
$\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\log\text{x}\times\frac{\text{d}}{\text{dx}}\big(\text{x}^2-3\big)+\big(\text{x}^2-3\big)\times\frac{\text{d}}{\text{dx}}(\log\text{x})$
$\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\log\text{x}\times2\text{x}+(\text{x}^2-3)\times\frac{1}{\text{x}}$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{x}^{\text{x}^2-3}\times\Big[\frac{\text{x}^2-3}{\text{x}}+2\text{x}\log\text{x}\Big]$
Also,
$\text{v}=(\text{x}-3)^{\text{x}^2}$
$\therefore\log\text{v}=\log(\text{x}-3)^{\text{x}^2}$
$\Rightarrow\log\text{v}=\text{x}^2\log(\text{x}-3)$
Differentaiting both sides with respect to x, we obtain
$\frac{1}{\text{v}}\times\frac{\text{dv}}{\text{dx}}=\log(\text{x}-3)\times\frac{\text{d}}{\text{dx}}(\text{x}^2)+\text{x}^2\times\frac{\text{d}}{\text{dx}}[\log(\text{x}-3)]$
$\Rightarrow\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\log(\text{x}-3)\times2\text{x}+\text{x}^2\times\frac{1}{\text{x}-3}\times\frac{\text{d}}{\text{dx}}(\text{x}-3)$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\text{v}\Big[2\text{x}\log(\text{x}-3)+\frac{\text{x}^2}{\text{x}-3}\times1\Big]$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=(\text{x}-3)^{\text{x}^2}\Big[\frac{\text{x}^2}{\text{x}-3}+2\text{x}\log(\text{x}-3)\Big]$
Substituting the expressions of $\frac{\text{du}}{\text{dx}}$ and $\frac{\text{dv}}{\text{dx}}$ in equation (1), we obtain
$\frac{\text{du}}{\text{dx}}=\text{x}^{\text{x}^2-3}\Big[\frac{\text{x}^2-3}{\text{x}}+2\text{x}\log\text{x}\Big] \\ +(\text{x}-3)^{\text{x}^2}\Big[\frac{\text{x}^2}{\text{x}-3}+2\text{x}\log(\text{x}-3)\Big]$
View full question & answer→Question 85 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{2\text{a}^{\text{x}}}{1-\text{a}^{2\text{x}}}\Big),\text{a}>1, -\infty<\text{x}<0$
AnswerLet $\text{y}=\tan^{-1}\Big(\frac{2\text{a}^{\text{x}}}{1-\text{a}^{2\text{x}}}\Big)$
Put $\text{a}^{\text{x}}=\tan\theta$
$\Rightarrow\text{y}=\tan^{-1}\Big\{\frac{2\times\text{a}^\text{x}}{1-(\text{a}^{\text{x}})^2}\Big\}$
$\Rightarrow\text{y}=\tan^{-1}\Big(\frac{2\tan\theta}{1-\tan^2\theta}\Big)$
$\Rightarrow\text{y}=\tan^{-1}(\tan2\theta)\ .....(\text{i})$
Here, $-\infty<\text{x}<0$
$\Rightarrow\text{a}^{-\infty}<\text{a}^{\text{x}}<2^{0}$
$\Rightarrow 0<\tan\theta<1$
$\Rightarrow 0<\theta<\frac{\pi}{4}$
$\Rightarrow 0<2\theta<\frac{\pi}{2}$
So, from equation (i),
$\text{y}=2\theta\Big[\text{Since},\tan^{-1}(\tan\theta)=\theta,\text{if }\theta\in\big(-\frac{\pi}{2},\frac{\pi}{2}\big)\Big]$
$\Rightarrow\text{y}=2\tan^{-1}(\text{a}^{\text{x}})$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{2}{1+(\text{a}^{\text{x}})^2}\frac{\text{d}}{\text{dx}}(\text{a}^{\text{x}})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2\times\text{a}^{\text{x}}\log_\text{e}\text{a}}{1+\text{a}^{2\text{x}}}$
$\therefore \frac{\text{dy}}{\text{dx}}=\frac{2\text{a}^{\text{x}}\log_\text{e}\text{a}}{1+\text{a}^{2\text{x}}}$
View full question & answer→Question 95 Marks
Differentiate $\cos^{-1}(4\text{x}^3-3\text{x})$ with respect to $\tan^{-1}\Big(\frac{\sqrt{1-\text{x}^2}}{\text{x}}\Big),$ if $\frac{1}{2}<\text{x}<1$
AnswerLet, $\text{u}=\cos^{-1}(4\text{x}^3-3\text{x})$
Put, $\text{x}=\cos\theta$
$\Rightarrow\theta=\cos^{-1}\text{x}$
Now, $\text{u}=\cos^{-1}(4\cos^3\theta-3\cos\theta)$
$\Rightarrow \text{u}=\cos^{}-1(\cos3\theta)\ .....(\text{i})$
Let, $\text{v}=\tan^{-1}\Big(\frac{\sqrt{1-\text{x}^2}}{\text{x}}\Big)$
$\Rightarrow\text{v}=\tan^{-1}\bigg(\frac{\sqrt{-1\cos^2\theta}}{\cos\theta}\bigg)$
$\Rightarrow\text{v}=\tan^{-1}\Big(\frac{\sin\theta}{\cos\theta}\Big)$
$\Rightarrow\text{v}=\tan^{-1}(\tan\theta)\ .....(\text{ii})$
Here,
$\frac{1}{2}<\text{x}<1$
$\Rightarrow\frac{1}{2}<\cos<1$
$\Rightarrow0<\theta<\frac{\pi}{3}$
So, from equation (i),
$\text{u}=3\theta\big[\text{Since,} \cos^{-1}(\cos\theta)=\theta,\text{if }\theta\in[0,\pi]\big]$
$\Rightarrow\text{u}=3\cos^{-1}\text{x}$
Differenting it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{-3}{\sqrt{1-\text{x}^2}}\ .....(\text{iii})$
From equation (ii),
$\text{v}=\theta \Big[\text{since,}\tan^{-1}(\tan\theta)=\theta,\text{if }\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\Big]$
$\Rightarrow\text{v}=\cos^{-1}\text{x}$
Differentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{-1}{\sqrt{1-\text{x}^2}}\ .....(\text{iv})$
Dividing equation (iii) by (iv),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\Big(\frac{-3}{\sqrt{1-\text{x}^2}}\Big)\Big(\frac{-\sqrt{1-\text{x}^2}}{1}\Big)$
$\therefore\frac{\text{du}}{\text{dv}}=3$
View full question & answer→Question 105 Marks
Differentiate $\tan^{-1}\Big(\frac{1+\text{ax}}{1-\text{ax}}\Big)$ with respect to $\sqrt{1+\text{a}^2\text{x}^2}$
AnswerLet, $\text{u}=\tan^{-1}\Big(\frac{1+\text{ax}}{1-\text{ax}}\Big)$
Put $\text{ax}= \tan\theta$
$\Rightarrow\text{u}=\tan^{-1}\Big(\frac{1+\tan\theta}{1-\tan\theta}\Big)$
$\Rightarrow\text{u}=\tan^{-1}\bigg(\frac{\tan\frac{\pi}{4}+\tan\theta}{1-\tan\frac{\pi}{4}\tan\theta}\bigg)$
$\Rightarrow\text{u}=\tan^{-1}\Big[\tan\Big(\frac{\pi}{4}+\theta\Big)\Big]$
$\Rightarrow\text{u}=\frac{\pi}{4}+\theta$
$\Rightarrow\text{u}=\frac{\pi}{4}+\tan^{-1}(\text{ax}) [\text{since,}\tan\theta=\text{ax}]$
Differentiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=0+\frac{1}{1+(\text{ax}^2)}\frac{\text{d}}{\text{dx}}(\text{ax})$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\frac{\text{a}}{1+\text{a}^2\text{x}^2}\ .....(\text{i})$
Now,
Let, $\text{v}=\sqrt{1+\text{a}^2\text{x}^2}$
Differentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{1}{2\sqrt{1+\text{a}^2\text{x}^2}}\frac{\text{d}}{\text{dx}}(1+\text{a}^2\text{x}^2)$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{1}{2\sqrt{1+\text{a}^2\text{x}^2}}(2\text{a}^2\text{x})$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{\text{a}^2\text{x}}{\sqrt{1+\text{a}^2\text{x}^2}}\ .....(\text{ii})$
Dividing equation (i) by (ii),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{\text{a}}{1+\text{a}^2\text{x}^2}\times\frac{\sqrt{1+\text{a}^2\text{x}^2}}{\text{a}^2\text{x}}$
$\frac{\text{du}}{\text{dv}}=\frac{1}{\text{ax}\sqrt{1+\text{a}^2\text{x}^2}}$
View full question & answer→Question 115 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=\frac{(\text{x}^2-1)^3(2\text{x}-1)}{\sqrt{(\text{x}-3)(4\text{x}-1)}}$
AnswerWe have, $\text{y}=\frac{(\text{x}^2-1)^3(2\text{x}-1)}{\sqrt{(\text{x}-3)(4\text{x}-1)}}.....(\text{i})$$\text{y}=\frac{(\text{x}^2-1)^3(2\text{x}-1)}{(\text{x}-3)^{\frac{1}{2}}(4\text{x}-1)^{\frac{1}{2}}}$
Taking log on both sides, $\log\text{y}=\log\Bigg[\frac{(\text{x}^2-1)^3(2\text{x}-1)}{(\text{x}-3)^{\frac{1}{2}}(4\text{x}-1)^{\frac{1}{2}}}\Bigg]$$\Rightarrow\log\text{y}=\log(\text{x}^2-1)^3+\log(2\text{x}-1)-\log(\text{x}-3)^{\frac{1}{2}}-\log(4\text{x}-1)^{\frac{1}{2}}$
$\Rightarrow\log\text{y}=\log(\text{x}^2-1)^3+\log(2\text{x}-1)-\frac{1}{2}\log(\text{x}-3)-\frac{1}{2}\log(4\text{x}-1)$
Differentiating with respect to x using chain rule, $\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=3\frac{\text{d}}{\text{dx}}\Big\{\log(\text{x}^2-1)\Big\}+\frac{\text{d}}{\text{dx}}\Big\{\log(2\text{x}-1)\Big\}\\-\frac{1}{2}\frac{\text{d}}{\text{dx}}\Big\{\log(\text{x}-3)\Big\}-\frac{1}{2}\Big\{\log(4\text{x}-1)\Big\}$$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=3\Big(\frac{1}{\text{x}^2-1}\Big)\frac{\text{d}}{\text{dx}}(\text{x}^2-1)+\frac{1}{(2\text{x}-1)}\frac{\text{d}}{\text{dx}}(2\text{x}-1)\\-\frac{1}{2}\Big(\frac{1}{\text{x}-3}\Big)\frac{\text{d}}{\text{dx}}(\text{x}-3)-\frac{1}{2}\frac{1}{(4\text{x}-1)}\frac{\text{d}}{\text{dx}}(4\text{x}-1)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=3\Big(\frac{1}{\text{x}^2-1}\Big)(2\text{x})+\frac{1}{2\text{x}-1}(2)-\frac{1}{2}\Big(\frac{1}{\text{x}-3}\Big)(1)-\frac{1}{2}\Big(\frac{1}{4\text{x}-1}\Big)(4)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\Big[\frac{6\text{x}}{\text{x}^2-1}+\frac{2}{2\text{x}-1}-\frac{1}{2(\text{x}-3)}-\frac{2}{4\text{x}-1}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{6\text{x}}{\text{x}^2-1}+\frac{2}{2\text{x}-1}-\frac{1}{2(\text{x}-3)}-\frac{2}{4\text{x}-1}\Big]$
$\Rightarrow\frac{(\text{x}^2-1)^3(2\text{x}-1)}{\sqrt{(\text{x}-3)(4\text{x}-1)}}\Big[\frac{6\text{x}}{\text{x}^2-1}+\frac{2}{2\text{x}-1}-\frac{1}{2(\text{x}-3)}-\frac{2}{4\text{x}-1}\Big]$
[Using equation (i)]
View full question & answer→Question 125 Marks
If $\text{x}=\sin^{-1}\Big(\frac{2\text{t}}{1+\text{t}^2}\Big)$ and $\text{y}=\tan^{-1}\Big(\frac{2\text{t}}{1-\text{t}^2}\Big),-1<\text{t}<1,$ prove that $\frac{\text{dy}}{\text{dx}}=1$
AnswerWe have, $\text{x}=\sin^{-1}\Big(\frac{2\text{t}}{1+\text{t}^{2}}\Big)$
Put $\text{t}=\tan\theta$
$ \Rightarrow-1<\tan\theta<1$
$\Rightarrow-\frac{\pi}{4}<\theta<\frac{\pi}{4}$
$\Rightarrow-\frac{\pi}{2}<2\theta<\frac{\pi}{2}$
$\therefore\text{x}=\sin^{-1}\Big(\frac{2\tan\theta}{1+\tan^{2}\theta}\Big)$
$\Rightarrow\text{x}=\sin^{-1}(\sin2\theta)$
$\Rightarrow\text{x}=2\theta$
$\Big[\therefore-\frac{\pi}{2}<2\theta<\frac{\pi}{2}\Big]$
$\Rightarrow\text{x}=2(\tan^{-1}\text{t})$
$\big[\therefore \text{t}=\sin\theta\big]$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{2}{1+\text{t}^{2}}\ .....(\text{i})$
Now, $\text{y}=\tan^{-1}\Big(\frac{2\text{t}}{1-\text{t}^{2}}\Big)$
Put $\text{t}=\tan\theta$
$\Rightarrow\text{y}=\tan^{-1}\Big(\frac{2\text{t}}{1-\text{t}^{2}}\Big)$
$\Rightarrow\text{y}=\tan^{-1}\Big(\frac{2\tan\theta}{1-\tan\theta}\Big)$
$\Rightarrow\text{y}=\tan^{-1}(\tan2\theta)$
$\Rightarrow\text{y}=2\theta$
$\Big[\therefore-\frac{\pi}{2}<2\theta<\frac{\pi}{2}\Big]$
$\Rightarrow\text{y}= 2 \tan^{-1}\text{t}$
$\big[\therefore\text{t}=\tan\theta\big]$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{2}{1+\text{t}^{2}}\ .....(\text{ii})$
Dividing equation (ii) by (i),
$\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{2}{1+\text{t}^{2}}\times\frac{1+\text{t}^{2}}{2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=1$
View full question & answer→Question 135 Marks
Differentiate $\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$ with respect to $\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big),$ if -1 < x < 1.
AnswerLet $\text{u}=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$
put $\text{x}=\tan\theta\Rightarrow \theta=\tan^{-1}\text{x},\text{so}$
$\text{u}=\sin^{-1}\Big(\frac{2\tan\theta}{1+\tan^2\theta}\Big)$
$\text{u}=\sin^{-1}(\sin2\theta)\ .....(\text{i})$
Let $\text{v}=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)$
$=\tan^{-1}\Big(\frac{2\tan\theta}{1-\tan^2\theta}\Big)$
$\text{v}=\tan^{-1}(\tan2\theta)\ .....(\text{ii})$
Here, $-1<\text{x}<1$
$\Rightarrow-1<\tan\theta<1$
$\Rightarrow -\frac{\pi}{4}<\theta<\frac{\pi}{4}$
So, from equation (i),
$\text{u}=2\theta$
$\Big[\text{Since,}\sin^{-1}(\sin\theta)=\theta,\text{if }\theta\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\Big]$
$\text{u}=2\tan^{-1}\text{x}$
Diffrerentiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=\frac{2}{(1+\text{x}^2)}\ .....\text{(iii)}$
From equation (ii),
$\text{v}=2\theta$
$\Big[\text{Since,}\tan^{-1}(\tan\theta)=\theta,\text{if }\theta\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\Big]$
$\text{v}=2\tan^{-1}\text{x}$
Differentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{2}{1+\text{x}^2}\ .....\text{(iv)}$
Dividing equation (iii) by (iv),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{2}{1+\text{x}^2}\times\frac{1+\text{x}^2}{2}$
$\frac{\text{du}}{\text{dv}}=1$
View full question & answer→Question 145 Marks
Find $\frac{\text{dy}}{\text{dx}},$ when
$\text{x}=\frac{1-\text{t}^2}{1+\text{t}^2}\text{ and y}=\frac{2\text{t}}{1+\text{t}^2}$
AnswerWe have, $\text{y}=\frac{2\text{t}}{1+\text{t}^{2}}$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\bigg[\frac{(1+\text{t}^{2})\frac{\text{d}}{\text{dt}}(2\text{t})-2\text{t}\frac{\text{d}}{\text{dt}}(1+\text{t}^{2})}{(1+\text{t}^{2})^{2}}\bigg]$
[using quotient rule]
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\bigg[\frac{(1+\text{t}^{2})(2)-2\text{t}(2\text{t})}{(1+\text{t}^{2})^{2}}\bigg]$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\bigg[\frac{2+2\text{t}^{2}-4\text{t}^{2}}{(1+\text{t}^{2})}\bigg]$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\bigg[\frac{2-2\text{t}^{2}}{(1+\text{t}^{2})^{2}}\bigg]...(\text{i})$
and, $\text{x}=\frac{1-\text{t}^{2}}{1+\text{t}^{2}}$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\bigg[\frac{(1+\text{t}^{2})\frac{\text{d}}{\text{dt}}(1-\text{t}^{2})-(1-\text{t}^{2})\frac{\text{d}}{\text{dt}}(1+\text{t}^{2})}{(1+\text{t}^{2})^{2}}\bigg]$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\bigg[\frac{(1+\text{t}^{2})(-2\text{t})-(1-\text{t}^{2})(2\text{t})}{(1+\text{t}^{2})^{2}}\bigg]$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\bigg[\frac{-4\text{t}}{(1+\text{t}^{2})^{2}}\bigg].....(\text{ii})$
Dividing equation (i) by (ii), We get,
$\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{2(1-\text{t}^{2})}{(1+\text{t}^{2})^{2}}\times\frac{(1+\text{t}^{2})^{2}}{-4\text{t}} $
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2(1-\text{t}^{2})}{-4\text{t}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{t}^{2}-1}{2\text{t}}$
View full question & answer→Question 155 Marks
If $\text{x}=2\cos\theta-\cot2\theta$ and $\text{y}=2\sin\theta-\sin2\theta,$ prove that $\frac{\text{dy}}{\text{dx}}=\tan\big(\frac{3\theta}{2}\big)$
AnswerHere, $\text{x}=2\cos\theta-\cos2\theta$
Diffierentiating it with respect to $\theta$ using chain rule,
$\frac{\text{dx}}{\text{d}\theta}=2(-\sin\theta)-(-\sin2\theta)\frac{\text{d}}{\text{d}\theta}(2\theta)$
$=-2\sin\theta+2\sin2\theta$
$\frac{\text{dx}}{\text{d}\theta}=2(\sin2\theta-\sin\theta)...(\text{i})$
And, $\text{y}=2\sin\theta-\sin\theta$
Differentiating it with respect to $\theta$ using chain rule,
$\frac{\text{dt}}{\text{d}\theta}=2\cos\theta-\cos2\theta\frac{\text{d}}{\text{d}\theta}(2\theta)$
$=2\cos\theta-\cos2\theta(2)$
$=2\cos\theta-\cos2\theta(2)$
$\frac{\text{dy}}{\text{d}\theta}=2(\cos\theta-\cos2\theta)...(\text{ii})$
Dividing equation (ii) by equation (i),
$\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}=\frac{2(\cos\theta-\cos2\theta)}{2(\sin2\theta-\sin\theta)}$
$=\frac{(\cos\theta-\cos2\theta)}{(\sin2\theta-\sin\theta)}$
$\frac{\text{dy}}{\text{dx}}=\frac{-2\sin\big(\frac{\theta+2\theta}{2}\big)\sin\big(\frac{\theta-2\theta}{2}\big)}{2\cos\big(\frac{2\theta+\theta}{2}\big)\sin\big(2\theta-\theta\big)}$
$\Big[\sin\text{A}-\sin\text{B}=2\cos\Big(\frac{\text{A+B}}{2}\Big)\sin\Big(\frac{\text{A+B}}{2}\Big)\Big]$
$\Rightarrow-\frac{\sin\big(\frac{3\theta}{2}\big)\big(\sin\big(\frac{-\theta}{2}\big)\big)}{\cos\big(\frac{3\theta}{2}\big)\sin\big(\frac{\theta}{2}\big)}$
$\Rightarrow-\frac{\sin\big(\frac{3\theta}{2}\big)\big(-\sin\frac{-\theta}{2}\big)}{\cos\big(\frac{3\theta}{2}\big)\sin\big(\frac{\theta}{2}\big)}$
$\Rightarrow\frac{\sin\big(\frac{3\theta}{2}\big)}{\cos\big(\frac{3\theta}{2}\big)}$
$\frac{\text{dy}}{\text{dx}}=\tan\big(\frac{3\theta}{2}\big)$
View full question & answer→Question 165 Marks
Find $\frac{\text{dy}}{\text{dx}}$ in the following cases:
$\sin\text{ xy}+\cos(\text{x}+\text{y})=1$
AnswerWe have, $\sin\text{ xy}+\cos(\text{x}+\text{y})=1$Differentiating with respect to x,
$\frac{\text{d}}{\text{dx}}\sin\text{xy}+\frac{\text{d}}{\text{dx}}\cos(\text{x}+\text{y})=\frac{\text{d}}{\text{dx}}(1)$
$\Rightarrow\cos \text{xy}\frac{\text{d}}{\text{dx}}(\text{xy})-\sin(\text{x}+\text{y})\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})=0$
[Using chain rule]
$\Rightarrow\cos(\text{xy})\Big[\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\frac{\text{d}}{\text{dx}}(\text{x})\Big]-\sin(\text{x}+\text{y})\Big[1+\frac{\text{dy}}{\text{dx}}\Big]=0$
$\Rightarrow\cos(\text{xy})\Big[\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}(1)\Big]-\sin(\text{x}+\text{y})+\sin(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\text{x}\cos(\text{xy})\frac{\text{dy}}{\text{dx}}+\text{y}\cos(\text{xy})-\sin(\text{x}+\text{y})+\sin(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\big[\text{x}\cos(\text{xy})+\sin(\text{x}+\text{y})\big]\frac{\text{dy}}{\text{dx}} \\ =\big[\sin(\text{x}+\text{y})-\text{y}\cos(\text{xy})\big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\Big[\frac{\sin(\text{x}+\text{y})-\text{y}\cos\text{xy}}{\text{x}\cos\text{xy}+\sin(\text{x}+\text{y})}\Big]$
View full question & answer→Question 175 Marks
Differentiate the following functions with respect to x:
$\sqrt{\frac{1+\sin\text{x}}{1-\sin\text{x}}}$
AnswerLet $\text{y}=\sqrt{\frac{1+\sin\text{x}}{1-\sin\text{x}}}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\frac{1+\sin\text{x}}{1-\sin\text{x}}\Big)^{\frac{1}{2}}$
$=\frac{\text{1}}{\text{2}}\Big(\frac{1+\sin\text{x}}{1-\sin\text{x}}\Big)^{\frac{1}{2}-1}\frac{\text{d}}{\text{dx}}\Big(\frac{1+\sin\text{x}}{1-\sin\text{x}}\Big)$
$=\frac{\text{1}}{\text{2}}\Big(\frac{1+\sin\text{x}}{1-\sin\text{x}}\Big)^{\frac{1}{2}}\Big[\frac{(1-\sin\text{x})(\cos\text{x})-(1+\sin\text{x})(-\cos\text{x})}{(1-\sin\text{x})^2}\Big]$
$=\frac{\text{1}}{\text{2}}\frac{(1+\sin\text{x})^\frac{1}{2}}{(1-\sin\text{x})^\frac{1}{2}}\Big[\frac{\cos\text{x}-\cos\text{x}+\cos\text{x}\sin\text{x}+\sin\text{x}\cos\text{x}}{(1-\sin\text{x})^2}\Big]$
$=\frac{1}{2}\times\frac{2\cos\text{x}}{\sqrt{1+\sin\text{x}}(1-\sin\text{x})^\frac{3}{2}}$
$=\frac{\cos\text{x}}{\sqrt{1+\sin\text{x}}(1-\sin\text{x})^\frac{3}{2}}$
$=\frac{\cos\text{x}}{\sqrt{1+\sin\text{x}}\sqrt{1-\sin\text{x}}(1-\sin\text{x})}$
$=\frac{\cos\text{x}}{\sqrt{1-\sin^2\text{x}}(1-\sin\text{x})}$
$=\frac{\cos\text{x}}{\cos\text{x}(1-\sin\text{x})}$ $\big[\text{Using }1-\sin^2\text{x}=\cos^2\text{x}\big] $
$=\frac{1}{(1-\sin\text{x})}\times\frac{(1+\sin\text{x})}{(1+\sin\text{x})}$
Thus, $ \frac{\text{dy}}{\text{dx}}=\frac{1}{\cos^2\text{x}}+\frac{\sin\text{x}}{\cos^2\text{x}}$
$\frac{\text{dy}}{\text{dx}}=\sec^2\text{x}+\tan\text{x}\sec\text{x}$
$\frac{\text{dy}}{\text{dx}}=\sec\text{x}[\tan\text{x}+\sec\text{x}] $
View full question & answer→Question 185 Marks
Find $\frac{\text{dy}}{\text{dx}},$ when
$\text{x}=\cos^{-1}\frac{1}{\sqrt{1+\text{t}^2}}\text{ and y}=\sin^{-1}\frac{\text{t}}{\sqrt{1+\text{t}^2}},\text{t}\in\text{R}$
AnswerWe have, $\text{x}=\cos^{-1}\Big(\frac{1}{\sqrt{1+\text{t}^{2}}}\Big)$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{-1}{\sqrt{1-\Big(\frac{1}{\sqrt{1+\text{t}^{2}}}\Big)^{2}}}\frac{\text{d}}{\text{dt}}\Big(\frac{1}{\sqrt{1+\text{t}}^{2}}\Big)$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{-1}{\sqrt{1-\frac{1}{(1+\text{t}^{2})}}}\left\{\frac{-1}{2(1+\text{t}^{2})^\frac{3}{2}}\right\}\frac{\text{d}}{\text{dt}}(1+\text{t}^{2})$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{(1+\text{t}^{2})^{\frac{1}{2}}}{\sqrt{1+\text{t}^{2}-1}}\times\frac{1}{2(1+\text{t}^{2})^\frac{3}{2}}(2\text{t})$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{\text{t}}{\sqrt{\text{t}^{2}}\times(1+\text{t}^{2})}$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{1}{1+\text{t}^{2}}...(\text{i})$
Now, $\text{y}=\sin^{-1}\Big(\frac{1}{\sqrt{1+\text{t}^{2}}}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{1}{\sqrt{1-\Big(\frac{1}{\sqrt{1+\text{t}^{2}}}\Big)^{2}}}\frac{\text{d}}{\text{dt}}\Big(\frac{1}{\sqrt{1+\text{t}^{2}}}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{1}{\sqrt{1-\frac{1}{(1+\text{t}^{2})}}}\left\{\frac{-1}{2(1+\text{t}^{2})\frac{3}{2}}\right\}\frac{\text{d}}{\text{dt}}(1+\text{t}^{2})$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{(1+\text{t}^{2})^{\frac{1}{2}}}{\sqrt{1+\text{t}^{2}-1}}\times\frac{-1}{2(1+\text{t}^{2})^{\frac{3}{2}}}(2\text{t})$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{-1}{2\sqrt{\text{t}^{2}}\times(1+\text{t}^{2})}(2\text{t})$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{-1}{1+\text{t}^{2}}....(\text{ii})$
Dividing equation (ii) by (i),
$\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{1}{(1+\text{t}^{2})}\times\frac{(1+\text{t}^{2})}{-1}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-1$
View full question & answer→Question 195 Marks
Differentiate the following functions with respect to x:
$\cos^{-1}\Big(\frac{1-\text{x}^{2\text{n}}}{1+\text{x}^{2\text{n}}}\Big), <\text{x}<\infty$
AnswerLet $\text{y}=\cos^{-1}\Big(\frac{1-\text{x}^{2\text{n}}}{1+\text{x}^{2\text{n}}}\Big)$
Put $\text{x}=\tan\theta,\text{ so}$
$\text{y}=\cos^{-1}\bigg(\frac{1-(\text{x}^{\text{n}})^2}{1+(\text{x}^{\text{n}})^2}\bigg)$
$=\cos^{-1}\Big(\frac{1-\tan^2\theta}{1+\tan^2\theta}\Big)$
$\text{y}=\cos^{-1}(\cos2\theta)\ .....(\text{i})$
Here, $0<\text{x}<\infty$
$\Rightarrow 0<\text{x}^{\text{n}}<\infty$
$\Rightarrow 0<\theta<\frac{\pi}{2}$
$\Rightarrow 0<(2\theta)<\pi$
So, from equation (i),
$\text{y}=2\theta\big[\text{Since}, \cos^{-1}(\cos\theta)=\theta,\text{ if }\theta\in[0,\pi]\big]$
$\text{y}=2\tan^{-1}\big(\text{x}^{\text{n}}\big)$
Differantiating it with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=2\Big(\frac{1}{1+(\text{x}^{\text{n}})^2}\Big)\frac{\text{d}}{\text{dx}}(\text{x}^{\text{n}})$
$=\frac{2}{1+\text{x}^{2\text{n}}}\times(\text{nx}^{\text{n}-1})$
$\frac{\text{dy}}{\text{dx}}=\frac{2\text{nx}^{\text{n}-1}}{1+\text{x}^{2\text{n}}}$
View full question & answer→Question 205 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{2^{\text{x}+1}}{1-4^{\text{x}}}\Big),-\infty<\text{x}<0$
AnswerLet $\text{y}=\tan^{-1}\Big\{\frac{2^{\text{x}+1}}{1-4^{\text{x}}}\Big\}$
Put $2\text{x}=\tan\theta,\text{ so}$
$\text{y}=\tan^{-1}\Big\{\frac{2^{\text{x}}\times2}{1-(2^\text{x})^2}\Big\}$
$=\tan^{-1}\Big\{\frac{2\tan\theta}{1-\tan^2\theta}\Big\}$
$\text{y}=\tan^{-1}\big\{\tan(2\theta)\big\}\ .....(\text{i})$
Here, $-\infty<\text{x}<0$
$\Rightarrow 2^{-\infty}<2^{\text{x}}<2^{0}$
$\Rightarrow 0<2^{\text{x}}<1$
$\Rightarrow 0< \theta< \frac{\pi}{4}$
$\Rightarrow 0 < (2\theta) <\frac{\pi}{2}$
From equation (i),
$\text{y}=2\theta \Big[\text{Since}, \tan^{-1}(\tan\theta)=\theta, \text{ if }\theta\in \Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\Big]$
$\text{y}=2\tan^{-1}(2^\text{x})$
Differentiate it with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=\frac{2}{1+(2^\text{x})^2}\frac{\text{d}}{\text{dx}}(2^\text{x})$
$=\frac{2\times2^\text{x}\log2}{1+4^{\text{x}}}$
$\frac{\text{dy}}{\text{dx}}=\frac{2^{\text{x}+1}\log2}{1+4^{\text{x}}}$
View full question & answer→Question 215 Marks
If $\text{y}=\frac{1}{2}\log\Big(\frac{1-\cos2\text{x}}{1+\cos2\text{x}}\Big),$ Prvoe that $\frac{\text{dy}}{\text{dx}}=2\text{ cosec }2\text{x}$
AnswerGiven, $\text{y}=\frac{1}{2}\log\Big(\frac{1-\cos2\text{x}}{1+\cos2\text{x}}\Big)$
$\Rightarrow\text{y}=\frac{1}{2}\log\Big(\frac{2\sin^2\text{x}}{2\cos^2\text{x}}\Big)$
$\begin{bmatrix} \text{Since}, 1-\cos2\text{x}=2\sin^2\text{x}, \\ 1+\cos2\text{x}=2\cos^2\text{x} \end{bmatrix}$
$\Rightarrow\text{y}=\frac{1}{2}\log\big(\tan^2\text{x}\big)$
$\Rightarrow\text{y}=\frac{2}{2}\log\tan\text{x}$
$\big[\text{Since}, \log\text{a}^{\text{b}}=\text{b}\log\text{a}\big]$
$\Rightarrow\text{y}=\log\tan\text{x}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=(\log\tan\text{x})$
$=\frac{1}{\tan\text{x}}\times\frac{\text{d}}{\text{dx}}(\tan\text{x})$
[Using chain rule]
$=\frac{\sec^2\text{x}}{\tan\text{x}}$
$=\frac{1}{\cos^2\text{x}\times\frac{\sin\text{x}}{\cos\text{x}}}$
$=\frac{1}{\sin\text{x}\cos\text{x}}$
$=\frac{2}{2\sin\text{x}\cos\text{x}}$
$=\frac{2}{\sin2\text{x}}\Big[\text{Since},\frac{1}{\sin\text{x}=\text{cosec x}}\Big]$
So,
$\frac{\text{dy}}{\text{dx}}=2\text{ cosec }2\text{x}$
View full question & answer→Question 225 Marks
If $\sin(\text{xy})+\frac{\text{y}}{\text{x}}=\text{x}^2-\text{y}^2,$ find $\frac{\text{dy}}{\text{dx}}$
AnswerWe have, $\sin(\text{xy})+\frac{\text{y}}{\text{x}}=\text{x}^2-\text{y}^2$
Differentiating with respect to x, we get
$\Rightarrow\frac{\text{d}}{\text{dx}}(\sin\text{ xy})+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{y}}{\text{x}}\Big)=\frac{\text{d}}{\text{dx}}(\text{x}^2)-\frac{\text{d}}{\text{dx}}(\text{y}^2)$
$\Rightarrow \cos(\text{xy})\frac{\text{d}}{\text{dx}}(\text{xy})+\Bigg\{\frac{\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}\frac{\text{d}}{\text{dx}}(\text{x})}{\text{x}^2}\Bigg\}=2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow \cos(\text{xy})\Big\{\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\frac{\text{d}}{\text{dx}}(\text{x})\Big\}+\Bigg\{\frac{\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}(1)}{\text{x}^2}\Bigg\}=2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow \cos(\text{xy})\Big\{\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}(1)\Big\}+\frac{1}{\text{x}^2}\Big(\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}\Big)=2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow \text{x}\cos(\text{xy})\frac{\text{dy}}{\text{dx}}+\text{y}\cos(\text{xy})+\frac{1}{\text{x}}\frac{\text{dy}}{\text{dx}}-\frac{\text{y}}{\text{x}^2}=2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}\Big\{\text{x}\cos(\text{xy})+\frac{1}{\text{x}}+2\text{y}\Big\}=\frac{\text{y}}{\text{x}^2}-\text{y}\cos(\text{xy})+2\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\Big\{\frac{\text{x}^2\cos(\text{xy})+1+2\text{xy}}{\text{x}}\Big\}=\frac{1}{\text{x}^2}\big(\text{y}-\text{x}^2\text{y}\cos(\text{xy})+2\text{x}^2\big)$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\frac{2\text{x}^3+\text{y}-\text{x}^2\text{y}\cos(\text{xy})}{\text{x}\big(\text{x}^2\cos(\text{xy})+1+2\text{xy}\big)}$
View full question & answer→Question 235 Marks
If $\text{y}=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)+\sec^{-1}\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big), 0<\text{x}<1$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{4}{1+\text{x}^2}$
AnswerLet $\text{y}=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)+\sec^{-1}\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)$
$\text{y}=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)+\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$
Put $\text{x}=\tan\theta$
$\text{y}=\sin^{-1}\Big(\frac{2\tan\theta}{1+\tan^2\theta}\Big)+\cos^{-1}\Big(\frac{1-\tan^2\theta}{1+\tan^2\theta}\Big)$
$\text{y}=\sin^{-1}(\sin2\theta)+\cos^{-1}(\cos2\theta)\ .....(1)$
Here, $0<\text{x}<1$
$\Rightarrow 0<\tan\theta<1$
$\Rightarrow 0<\theta<\frac{\pi}{4}$
$\Rightarrow0<(2\theta)<\frac{\pi}{2}$
So, from equation (i),
$\text{y}=2\theta+2\theta$
$\begin{bmatrix} \text{Since}, \sin^{-1}(\sin\theta)=\theta,\text{ if }\theta\in \Big[-\frac{\pi}{2},\frac{\pi}{2}\Big] \\ \cos^{-1}(\cos\theta)=\theta,\text{ if }\theta\in[0,\pi] \end{bmatrix}$
$\text{y}=4\tan^{-1}\text{x}\ [\text{Since, x}=\tan\theta]$
Diffrentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{4}{1+\text{x}^2}$
View full question & answer→Question 245 Marks
Differentiate the following functions with respect to x:
$\sin^{-1}\big(1-2\text{x}^2\big),0<\text{x}<1$
AnswerLet $\text{y}=\sin^{-1}\big\{1-2\text{x}^2\big\}$
Let $\text{x}=\sin\theta,\text{ So},$
$\text{y}=\sin^{-1}\big(1-2\sin^2\theta\big)$
$=\sin^{-1}(\cos2\theta)$
$\text{y}=\sin^{-1}\Big\{\sin\Big(\frac{\pi}{2}-2\theta\Big)\Big\}\ .....(\text{i})$
Here, $0<\text{x}<1$
$\Rightarrow 0<\sin\theta<1$
$\Rightarrow 0<\theta<\frac{\pi}{2}$
$\Rightarrow 0<2\theta<\pi$
$\Rightarrow 0> -2\theta>-\pi$
$\Rightarrow\ \frac{\pi}{2}>\Big(\frac{\pi}{2}-2\theta\Big)>\frac{\pi}{2}-\pi$
$\Rightarrow\ \frac{\pi}{2}>\Big(\frac{\pi}{2}-2\theta\Big)>\Big(-\frac{\pi}{2}\Big)$
So, from equatuion (i),
$\text{y}=\frac{\pi}{2}-2\theta$
$\Big[\text{Since}, \sin^{-1}(\sin\theta)=\theta,\text{ if }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\text{y}\frac{\pi}{2}-2\sin^{-1}\text{x}\big[\text{Since x}=\sin\theta\big]$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=0-2\Big(\frac{1}{\sqrt{1-\text{x}^2}}\Big)$
$\frac{\text{dy}}{\text{dx}}=-\frac{2}{\sqrt{1-\text{x}^2}}$
View full question & answer→Question 255 Marks
If $\text{y}=\sin^{-1}\Big(\frac{\text{x}}{1+\text{x}^2}\Big)+\cos^{-1}\Big(\frac{1}{\sqrt{1+\text{x}^2}}\Big), 0<\text{x}<\infty$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{4}{1+\text{x}^2}$
AnswerLet $\text{y}=\sin^{-1}\Big(\frac{\text{x}}{\sqrt{1+\text{x}^2}}\Big)+\cos^{-1}\Big(\frac{1}{\sqrt{1+\text{x}^2}}\Big)$
Put $\text{x}=\tan\theta$
$\therefore\text{y}=\sin^{-1}\Big(\frac{\tan\theta}{\sqrt{1+\tan^2\theta}}\Big)+\cos^{-1}\Big(\frac{1}{\sqrt{1+\tan^2\theta}}\Big)$
$\Rightarrow \text{y}=\sin^{-1}\bigg(\frac{\frac{\sin\theta}{\cos\theta}}{\sec\theta}\bigg)+\cos^{-1}\Big(\frac{1}{\sec\theta}\Big)$
$\Rightarrow\text{y}=\sin^{-1}\bigg(\frac{\frac{\sin\theta}{\cos\theta}}{\frac{1}{\cos\theta}}\bigg)+\cos^{-1}(\cos\theta)$
$\Rightarrow\text{y}=\sin^{-1}(\sin\theta)+\cos^{-1}(\cos\theta)\ .....(\text{i})$
Here, $0<\text{x}<\infty$
$\Rightarrow 0<\tan\theta<\infty$
$\Rightarrow 0 <\theta<\frac{\pi}{2}$
So, from equation (i),
$\text{y}=\theta+\theta$
$\begin{bmatrix} \text{Since}, \sin^{-1}(\sin\theta)=\theta,\text{ if }\theta\in \Big[-\frac{\pi}{2},\frac{\pi}{2}\Big] \\ \cos^{-1}(\cos\theta)=\theta,\text{ if }\theta\in[0,\pi] \end{bmatrix}$
$\Rightarrow\text{y}=2\theta$
$\Rightarrow\text{y}=2\tan^{-1}\text{x}\ \big[\text{Since, x}=\tan\theta\big]$
Differentiate it with respect to x,
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{2}{1+\text{x}^2}$
View full question & answer→Question 265 Marks
If $\text{x}\sin(\text{a}+\text{y})+\sin\text{a}\cos(\text{a}+\text{y})=0,$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\sin^2(\text{a}+\text{y})}{\sin\text{a}}$
AnswerWe have,
$\text{x}\sin(\text{a}+\text{y})+\sin\text{a}\cos(\text{a}+\text{y})=0$
Differentiating with respect to x using chain rule,
$\frac{\text{d}}{\text{dx}}\big[\text{x}\sin(\text{a}+\text{y})+\sin\text{a}\cos(\text{a}+\text{y})\big]=0$
$\Rightarrow\text{x}\frac{\text{d}}{\text{dx}}\sin(\text{a}+\text{y})+\sin(\text{a}+\text{y})\frac{\text{d}}{\text{dx}}(\text{x}) \\ +\sin\text{a}\frac{\text{d}}{\text{dx}}\cos(\text{a}+\text{y})+\cos(\text{a}+\text{y})\frac{\text{d}}{\text{dx}}\sin\text{a}=0$
$\Rightarrow\text{x}\cos(\text{a}+\text{y})\Big(0+\frac{\text{dy}}{\text{dx}}\Big)+\sin(\text{a}+\text{y}) \\ +\sin\text{a}\Big\{-\sin(\text{a}+\text{y})\frac{\text{dy}}{\text{dx}}\Big\}+0=0$
$\Rightarrow\big[\text{x}\cos(\text{a}+\text{y})-\sin\text{a}\sin(\text{a}+\text{y})\big]\frac{\text{dy}}{\text{dx}}+\sin(\text{a}+\text{y})=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\sin(\text{a}+\text{y})}{\text{x}\cos(\text{a}+\text{y})-\sin\text{a}\sin(\text{a}+\text{y})}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-\sin(\text{a}+\text{y})}{\Big\{-\frac{\sin\text{a}\cos(\text{a}+\text{y})}{\sin(\text{a}+\text{y})}\Big\}\cos(\text{a}+\text{y})-\sin\text{a}\sin(\text{a}+\text{y})}$
$\Big[\because\text{x}=-\frac{\sin\text{a}\cos(\text{a}+\text{y})}{\sin(\text{a}+\text{y})}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sin^2(\text{a}+\text{y})}{\sin\text{a}\cos^2(\text{a}+\text{y})+\sin\text{a}\sin^2(\text{a}+\text{y})}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sin^2(\text{a}+\text{y})}{\sin\text{a}\big[\cos^2(\text{a}+\text{y})+\sin^2(\text{a}+\text{y})\big]}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sin^2(\text{a}+\text{y})}{\sin\text{a}} \\ \big[\because\cos^2(\text{a}+\text{y})+\sin^2(\text{a}+\text{y})=1\big]$
View full question & answer→Question 275 Marks
If $x^y + y^x = (x + y)^{x+y}$, find $\frac{\text{dy}}{\text{dx}}$
AnswerHere,
$x^y + y^x = (x + y)^{x+y}$
$\text{e}^{\log\text{x}^\text{y}}+\text{e}^{\log\text{y}^\text{x}}=\text{e}^{\log(\text{x}+\text{y})^{(\text{x}+\text{y})}}$
$\text{e}^{\text{y}\log\text{x}}+\text{e}^{\text{x}\log\text{y}}=\text{e}^{{(\text{x}+\text{y})}\log(\text{x}+\text{y})}$
$\big[\text{Since},\text{e}^{\log\text{a}}=\text{a},\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating it with respect to x using chain rule, product rule,
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{y}\log\text{x}}\big)+\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}\log\text{y}}\big)=\frac{\text{d}}{\text{dx}}^{(\text{x}+\text{y})\log(\text{x}+\text{y})}$
$\Rightarrow\text{e}^{\text{y}\log\text{x}}\Big[\text{y}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{dy}}{\text{dx}}\Big] \\ +\text{e}^{\text{x}\log\text{y}}\Big[\text{x}\frac{\text{d}}{\text{dx}}\log\text{y}+\log\text{y}\frac{\text{d}}{\text{dx}}(\text{x})\Big] \\ =\text{e}^{(\text{x}+\text{y})\log(\text{x}+\text{y})}\frac{\text{d}}{\text{dx}}\big[(\text{x}+\text{y})\log(\text{x}+\text{y})\big]$
$\Rightarrow\text{e}^{\text{y}\log\text{x}}\Big[\text{y}\big(\frac{1}{\text{x}}\big)+\log\text{x}\frac{\text{dy}}{\text{dx}}\Big]+\text{e}^{\log\text{x}}\Big[\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}+\log\text{y}(1)\Big] \\ =\text{e}^{\log(\text{x}+\text{y})^{(\text{x}+\text{y})}}\Big[(\text{x}+\text{y})\frac{\text{d}}{\text{dx}}\log(\text{x}+\text{y})+\log(\text{x}+\text{y})\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})\Big]$
$\Rightarrow\text{x}^\text{y}\Big[\frac{\text{y}}{\text{x}}+\log\text{x}\frac{\text{dy}}{\text{dx}}\Big]+\text{y}^\text{x}\Big[\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}+\log\text{y}\Big] \\ =(\text{x}+\text{y})^{(\text{x}+\text{y})}\Big[(\text{x}+\text{y})\frac{1}{(\text{x}+\text{y})}\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})+\log(\text{x}+\text{y})\Big(1+\frac{\text{dy}}{\text{dx}}\Big)\Big]$
$\Rightarrow\text{x}^\text{y}\times\frac{\text{y}}{\text{x}}+\text{x}^{\text{y}}\log\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}^\text{x}\times\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}+\text{y}^\text{x}\log\text{y} \\ =(\text{x}+\text{y})^{(\text{x}+\text{y})}\Big[1\times\Big(1+\frac{\text{dy}}{\text{dx}}\Big)+\log(\text{x}+\text{y})\Big(1+\frac{\text{dy}}{\text{dx}}\Big)\Big]$
$\Rightarrow\text{x}^{\text{y}-1}\times\text{y}+\text{x}^\text{y}\log\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}^{\text{x}-1}\times\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}^{\text{x}}\log\text{y} \\ =(\text{x}+\text{y})^{(\text{x}+\text{y})}+(\text{x}+\text{y})^{(\text{x}+\text{y})}\frac{\text{dy}}{\text{dx}} \\ +(\text{x}+\text{y})^{(\text{x}+\text{y})}\log(\text{x}+\text{y})+(\text{x}+\text{y})^{(\text{x}+\text{y})}\log(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\Big[\text{x}^{\text{y}}\log\text{x}+\text{xy}^{\text{x}-1}-(\text{x}+\text{y})^{\text{x}+\text{y}}(1+\log(\text{x}+\text{y}))\Big] \\ =(\text{x}+\text{y})^{\text{x}+\text{y}}(1+\log(\text{x}+\text{y}))-\text{x}^{\text{y}-1}\times\text{y}-\text{y}^{\text{x}}\log\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\bigg[\frac{(\text{x}+\text{y})^{\text{x}+\text{y}}(1+\log(\text{x}+\text{y}))-\text{x}^{\text{y}-1}\times\text{y}-\text{y}^\text{x}\log\text{y}}{\text{x}^\text{y}\log\text{x}+\text{xy}^{\text{x}-1}-(\text{x}+\text{y})^{\text{x}+\text{y}}(1+\log(\text{x}+\text{y}))}\bigg]$
View full question & answer→Question 285 Marks
Differentiate the following functions from first principles:
$\text{e}^{\cos\text{x}}$
AnswerLet $\text{f(x)}=\text{e}^{\cos\text{x}}$
$\Rightarrow\text{f}(\text{x}+\text{h})=\text{e}^{\cos(\text{x}+\text{h})}$
$\therefore\frac{\text{d}}{\text{dx}}(\text{f(x)})=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{\cos(\text{x}+\text{h})}-\text{e}^{\cos\text{x}}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\text{e}^{\cos\text{x}}\Big[\frac{\text{e}^{\cos(\text{x}+\text{h})-\cos\text{x}}-1}{\text{h}}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{ e}^{\cos\text{x}}\Big[\frac{\text{e}^{\cos(\text{x}+\text{h})-\cos\text{x}}-1}{\cos(\text{x}+\text{h})-\cos\text{x}}\Big]\times\frac{\cos(\text{x}+\text{h})-\cos\text{x}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\text{ e}^{\cos\text{x}}\times\Big(\frac{\cos(\text{x}+\text{h})-\cos\text{x}}{\text{h}}\Big)$
$\Big[\text{Since},\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^\text{x}-1}{\text{x}}=1\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\ \text{e}^{\cos\text{x}}\times\bigg(\frac{-2\sin\frac{\text{x}+\text{h}+\text{x}}{2}\times\sin\frac{\text{x}+\text{h}-\text{x}}{2}}{\text{h}}\bigg)$
$\begin{bmatrix} \cos\text{A}-\cos\text{B}=-2\sin\frac{\text{A}+\text{B}}{2}\sin\frac{\text{A}-\text{B}}{2} \end{bmatrix}$
$=\text{e}^{\cos\text{x}}\lim\limits_{\text{h}\rightarrow0}\frac{-\sin\Big(\frac{2\text{x}+\text{h}}{2}\Big)}{2}\times\frac{\sin\Big(\frac{\text{h}}{2}\Big)}{\frac{\text{h}}{2}}$
$=\text{e}^{\cos\text{x}}\lim\limits-2\sin\Big(\frac{2\text{x}+\text{h}}{2}\Big)\times\frac{1}{2}$
$\Big[\text{Since},\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\Big]$
$=\text{e}^{\cos\text{x}}(-\sin\text{x})$
$=-\sin\text{xe}^{\cos\text{x}}$
Hence,
$\frac{\text{d}}{\text{dx}}\big(\text{e}^{\cos\text{x}}\big)=-\sin\text{xe}^{\cos\text{x}}$
View full question & answer→Question 295 Marks
Differentiate $\sin^{-1}\sqrt{1-\text{x}^2}$ with respect to $\cot^2\Big(\frac{\text{x}}{\sqrt{1-\text{x}^2}}\Big),$ if 0 < x < 1.
AnswerLet $\text{u}=\sin^{-1}\Big(\sqrt{1-\text{x}^2}\Big)$
Put $\text{x}=\cos\theta\Rightarrow\theta=\cos^{-1}\text{x},\text{ so}$
$\text{u}=\sin^{-1}(\sin\theta)\ .....(\text{i})$
And,
Let $\text{v}=\cot^{-1}\Big(\frac{\text{x}}{\sqrt{1-\text{x}^2}}\Big)$
$=\cot^{-1}\Big(\frac{\cot\theta}{\sqrt{1-\cos^2\theta}}\Big)$
$=\cot^{-1}\Big(\frac{\cos\theta}{\sin\theta}\Big)$
$\text{v}=\cot^{-1}(\cot\theta)\ .....(\text{ii})$
Here, $0<\text{x}<1$
$\Rightarrow0<\cos\theta<1$
$\Rightarrow0<\theta<\frac{\pi}{2}$
So, from equation (i),
$\text{u}=\theta\Big[\text{Since,}\sin^{-1}(\sin\theta)=\theta,\text{if }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\text{u}=\cos^{-1}\text{x}$
Differentiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=\frac{-1}{\sqrt{1-\text{x}^2}}\ .....(\text{iii})$
From equation (ii),
$\text{v}=\theta\big[\text{Since,}\cot^{-1}(\cot\theta)=\theta,\text{if }\theta\in(0,\pi) \big]$
$\text{v}=\cos^{-1}\text{x}$
Differentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{-1}{\sqrt{1-\text{x}^2}}\ .....(\text{iv})$
Dividing equation (iii) by (iv),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{-1}{\sqrt{1-\text{x}^2}}\times\frac{\sqrt{1-\text{x}^2}}{-1}$
$\frac{\text{du}}{\text{dv}}=1$
View full question & answer→Question 305 Marks
Differentiate the following functions with respect to x:
$(\text{x}\cos\text{x})^\text{x}+(\text{x}\sin\text{x})^\frac{1}{\text{x}}$
AnswerLet $\text{y}=(\text{x}\cos\text{x})^\text{x}+(\text{x}\sin\text{x})^\frac{1}{\text{x}}$
Also, $\text{u}=(\text{x}\cos\text{x})^\text{x}\text{ and }\text{v}(\text{x}\sin\text{x})^\frac{1}{\text{x}}$
$\therefore\ \text{y}=\text{u}+\text{v}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{du}}{\text{dx}}+\frac{\text{dv}}{\text{dx}}\ .....(\text{i})$
Now, $\text{u}=(\text{x}\cos\text{x})^\text{x}$
$\Rightarrow\log\text{u}=\log(\text{x}\cos\text{x})^\text{x}$
$\Rightarrow\log\text{u}=\text{x}\log(\text{x}\cos\text{x})$
$\Rightarrow\log\text{u}=\text{x}\big[\log\text{x}+\log\cos\text{x}\big]$
$\Rightarrow\log\text{u}=\text{x}\log\text{x}+\text{x}\log\cos\text{x}$
Differentiate both sides with respect to x,
$\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x}\log\text{x})+\frac{\text{d}}{\text{dx}}(\text{x}\log\cos\text{x})$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{u}\Big[\Big\{\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x})+\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})\Big\} \\ +\Big\{\log\cos\text{x}\frac{\text{d}}{\text{dx}}(\text{x})+\text{x}\frac{\text{d}}{\text{dx}}(\log\cos\text{x})\Big\}\Big]$
$\Rightarrow\frac{\text{du}}{\text{dx}}=(\text{x}\cos\text{x})^\text{x}\Big[\Big(\log\text{x}(1)+\text{x}\big(\frac{1}{\text{x}}\big)\Big) \\ +\Big\{\log\cos\text{x}(1)+\text{x}\frac{1}{\cos\text{x}}\frac{\text{d}}{\text{dx}}(\cos\text{x})\Big\}\Big]$
$\Rightarrow\frac{\text{du}}{\text{dx}}=(\text{x}\cos\text{x})^\text{x}\Big[(\log\text{x}+1)+\Big\{\log\cos\text{x}\frac{\text{x}}{\cos\text{x}}(-\sin\text{x})\Big\}\Big]$
$\Rightarrow\frac{\text{du}}{\text{dx}}=(\text{x}\cos\text{x})^\text{x}\big[(1+\log\text{x})+(\log\cos\text{x}-\text{x}\tan\text{x})\big]$
$\Rightarrow\frac{\text{du}}{\text{dx}}=(\text{x}\cos\text{x})^\text{x}\big[1-\text{x}\tan\text{x}+(\log\text{x}+\log\cos\text{x})\big]$
$\Rightarrow\frac{\text{du}}{\text{dx}}=(\text{x}\cos\text{x})^\text{x}\big[1-\text{x}\tan\text{x}+\log(\text{x}\cos\text{x})\big]\ .....(\text{ii})$
Again, $\text{v}=(\text{x}\sin\text{x})^\frac{1}{\text{x}}$
$\Rightarrow\log\text{v}=\log(\text{x}\sin\text{x})^\frac{1}{\text{x}}$
$\Rightarrow\log\text{v}=\frac{1}{\text{x}}\log(\text{x}\sin\text{x})$
$\Rightarrow\log\text{v}=\frac{1}{\text{x}}(\log\text{x}+\log\sin\text{x})$
$\Rightarrow\log\text{v}=\frac{1}{\text{x}}\log\text{x}+\frac{1}{\text{x}}\log\sin\text{x}$
Differentiating both sides with respect to x,
$\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\frac{1}{\text{x}}\log\text{x}\Big)+\frac{\text{d}}{\text{dx}}\Big[\frac{1}{\text{x}}\log(\sin\text{x})\Big]$
$\Rightarrow\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\Big[\log\text{x}\frac{\text{d}}{\text{dx}}\Big(\frac{1}{\text{x}}\Big)+\frac{1}{\text{x}}\frac{\text{d}}{\text{dx}}(\log\text{x})\Big] \\ +\Big[\log(\sin\text{x})\frac{\text{d}}{\text{dx}}\Big(\frac{1}{\text{x}}\Big)+\frac{1}{\text{x}}\frac{\text{d}}{\text{dx}}\big\{\log(\sin\text{x})\big\}\Big]$
$\Rightarrow\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\Big[\log\text{x}\Big(-\frac{1}{\text{x}^2}\Big)+\Big(\frac{1}{\text{x}}\Big)\Big(\frac{1}{\text{x}}\Big)\Big] \\ +\Big[\log(\sin\text{x})\Big(-\frac{1}{\text{x}^2}\Big)+\frac{1}{\text{x}}\Big(\frac{1}{\sin\text{x}}\Big)\frac{\text{d}}{\text{dx}}(\sin\text{x})\Big]$
$\Rightarrow\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\frac{1}{\text{x}^2}(1-\log\text{x})+\Big[-\frac{\log(\sin\text{x})}{\text{x}^2}+\frac{1}{\text{x}\sin\text{x}}(\cos\text{x})\Big]$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=(\text{x}\sin\text{x})^{\frac{1}{\text{x}}}\Big[\frac{1-\log\text{x}}{\text{x}^2}+\frac{\log(\sin\text{x}+\text{x}\cos\text{x})}{\text{x}^2}\Big]$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=(\text{x}\sin\text{x})^\frac{1}{\text{x}}\Big[\frac{1-\log\text{x}-\log(\sin\text{x})+\text{x}\cot\text{x}}{\text{x}^2}\Big]$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=(\text{x}\sin\text{x})^\frac{1}{\text{x}}\Big[\frac{1-\log(\text{x}\sin\text{x})+\text{x}\cot\text{x}}{\text{x}^2}\Big]\ .....(\text{iii})$
From (i), (ii) and (iii), we obtain
$\frac{\text{dy}}{\text{dx}}=(\text{x}\cos\text{x})^\text{x}\big[1-\text{x}\tan\text{x}+\log(\text{x}\cos\text{x})\big]\\+(\text{x}\sin\text{x})^\frac{1}{\text{x}} \Big[\frac{\text{x}\cot\text{x}+1-\log(\text{x}\sin\text{x})}{\text{x}^2}\Big]$
View full question & answer→Question 315 Marks
If $x^x + y^x = 1$, prove that $\frac{\text{dy}}{\text{dx}}=-\Big\{\frac{\text{x}^\text{x}(1+\log\text{x})+\text{y}^\text{x}\times\log\text{y}}{\text{x}\times\text{y}^{\text{x}-1}}\Big\}$
AnswerHere,
$x^x + y^x = 1$
$\text{e}^{\log\text{x}^\text{x}}+\text{e}^{\log\text{y}^\text{x}}=1$
$\text{e}^{\text{x}\log\text{x}}+\text{e}^{\text{x}\log\text{y}}=1$
$\big[\text{Since},\text{e}^{\log\text{a}}=\text{a}.\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating it with respect to x using product rule and chain rule,
$\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}\log\text{x}}\big)+\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}\log\text{y}}\big)=\frac{\text{d}}{\text{dx}}(1)$
$\text{e}^{\text{x}\log\text{x}}\frac{\text{d}}{\text{dx}}(\text{x}\log\text{x})+\text{e}^{\text{x}\log\text{y}}\frac{\text{d}}{\text{dx}}(\text{x}\log\text{y})=0$
$\text{e}^{\text{x}\log\text{x}}\Big[\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big] \\ +\text{e}^{\log\text{y}^\text{x}}\Big[\text{x}\frac{\text{d}}{\text{dx}}(\log\text{y})+\log\text{y}\frac{\text{d}}{\text{dx}}(\text{x})\Big]=0$
$\text{x}^\text{x}\Big[\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}(1)\Big]+\text{y}^\text{x}\Big[\text{x}\Big(\frac{1}{\text{y}}\Big)\frac{\text{dy}}{\text{dx}}+\log\text{y}(1)\Big]=0$
$\text{x}^\text{x}[1+\log\text{x}]+\text{y}^\text{x}\Big(\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}+\log\text{y}\Big)=0$
$\text{y}^\text{x}\times\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}=-\big[\text{x}^\text{x}(1+\log\text{x})+\text{y}^\text{x}\log\text{y}\big]$
$\big(\text{xy}^{\text{x}-1}\big)\frac{\text{dy}}{\text{dx}}=-\big[\text{x}^\text{x}(1+\log\text{x})+\text{y}^\text{x}\log\text{y}\big]$
$\frac{\text{dy}}{\text{dx}}=-\Big[\frac{\text{x}^\text{x}(1+\log\text{x})+\text{y}^\text{x}\log\text{y}}{\text{xy}^{\text{x}-1}}\Big]$
View full question & answer→Question 325 Marks
Differentiate the following functions from first principles:
$\log\cos\text{x}$
AnswerLet $\text{f(x)} = \log \cos \text{x}$
$\Rightarrow\ \text{f}(\text{x}+\text{h})=\log\cos(\text{x}+\text{h})$
$\therefore \frac{\text{d}}{\text{dx}}\{\text{f(x)}\}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})=\text{f(x)}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log\cos(\text{x}+\text{h})-\log\cos\text{x}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log^{\log(\text{x}+\text{h})}_{\cos\text{x}}}{\text{h}}\ \Big[\because\ \log\text{A}-\log\text{B}=\log\Big(\frac{\text{A}}{\text{B}}\Big)\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log\Big[1+\left\{\frac{\cos(\text{z}+\text{h})}{\cos\text{z}}-1\right\}\Big]}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log\left\{1+\frac{\cos(\text{x}+\text{h})-\cos\text{z}}{\cos\text{z}}\right\}}{\left\{\frac{\cos(\text{x}+\text{h})-\cos\text{x}}{\cos\text{x}}\right\}}\times\lim\limits_{\text{h}\rightarrow0}\left\{\frac{\cos(\text{x}+\text{h})-\cos\text{x}}{\cos\text{x}}\right\}$
$=1\times\lim\limits_{\text{h}\rightarrow0}\frac{\cos(\text{x}+\text{h})-\cos\text{x}}{\cos\text{x}\times\text{h}}\ \Big[\because\lim\limits_{\text{h}\rightarrow0}\frac{\log(1+\text{x})}{\text{x}}=1\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{-2\sin\Big(\frac{\text{x}+\text{h}+\text{x}}{2}\Big)\sin\Big(\frac{\text{x}+\text{h}+\text{x}}{2}\Big)}{\cos\text{x}\times\text{h}}$
$=-2\lim\limits_{\text{h}\rightarrow0}\frac{\sin\Big(\frac{2\text{x}+\text{h}}{2}\Big)\times\sin\big(\frac{\text{h}}{2}\big)}{2\cos\text{x}\times\big(\frac{\text{x}}{2}\big)}$
$=\frac{-2\sin\text{x}}{2\cos\text{x}}\Big[\because \lim\limits\frac{\sin\text{x}}{\text{x}}=1\Big]$
$=-\tan\text{x}$
So,
$\frac{\text{d}}{\text{dx}}(\log\cos\text{x})=-\tan\text{x}$
View full question & answer→Question 335 Marks
Differentiate the following functions with respect to x:
$\cos^{-1}\Big(\frac{\text{x}+\sqrt{1-\text{x}^2}}{\sqrt{2}}\Big),-1<\text{x}<1$
AnswerLet $\text{y}=\cos^{-1}\Big(\frac{\text{x}+\sqrt{1-\text{x}^2}}{\sqrt{2}}\Big)$
Put $\text{x}=\cos\theta$
$\text{y}=\sin^{-1}\Big\{\frac{\cos\theta+\sqrt{1-\cos^2\theta}}{\sqrt{2}}\Big\}$
$\text{y}=\cos^{-1}\Big\{\frac{\cos\theta+\sin\theta}{\sqrt{2}}\Big\}$
$\text{y}=\cos^{-1}\Big\{\cos\theta\Big(\frac{1}{\sqrt{2}}\Big)+\sin\theta\Big(\frac{1}{\sqrt{2}}\Big)\Big\}$
$\text{y}=\cos^{-1}\Big\{\cos\theta\cos\frac{\pi}{2}+\sin\theta\sin\frac{\pi}{2}\Big\}$
$\text{y}=\cos^{-1}\Big\{\cos\Big(\theta-\frac{\pi}{4}\Big)\Big\}\ .....(\text{i})$
Here, $-1<\text{x}<1$
$\Rightarrow -1<\cos\theta<1$
$\Rightarrow\frac{3\pi}{4}<\theta<\frac{5\pi}{4}$
$\Rightarrow\Big(\frac{3\pi}{4}-\frac{\pi}{4}\Big)<\Big(\theta-\frac{\pi}{4}\Big)<\frac{5\pi}{4}-\frac{\pi}{4}$
$\Rightarrow\Big(\frac{\pi}{4}\Big)<\Big(\theta-\frac{\pi}{4}\Big)<\pi$
So, from equation (i),
$\text{y}=\Big(\theta-\frac{\pi}{4}\Big)\ \big[\text{Since}, \cos^{-1}(\cos\theta)=\theta,\text{ if }\theta\in[0,\pi]\big]$
$\text{y}=\cos^{-1}\text{x}-\frac{\pi}{4}\big[\text{Since, x}=\sin\theta\big]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=-\frac{1}{\sqrt{1-\text{x}^2}}+0$
$\frac{\text{dy}}{\text{dx}}=-\frac{1}{\sqrt{1-\text{x}^2}}$
View full question & answer→Question 345 Marks
Differentiate $(\cos\text{x})^{\sin\text{x}}$ with respect to $(\sin\text{x})^{\cos\text{x}}$
AnswerLet, $\text{u} = (\cos)^{\sin\text{x}}$
Taking log on both sides,
$\log\text{u} = \log(\cos\text{x})^{\sin \text{x}}$
$\Rightarrow \log\text{u} = \sin \text{x}\log(\cos\text{x})$
Differentiating it with respect to x using rule,
$\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\sin \text{x}\frac{\text{d}}{\text{dx}}(\log\cos\text{x})+\log \cos \text{x}\frac{\text{d}}{\text{dx}}(\sin\text{x})$
[using product rule]
$\Rightarrow\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\sin \text{x}\big(\frac{1}{\cos\text{x}}\big)\frac{\text{d}}{\text{dx}}(\cos\text{x})+\log\cos\text{x}(\cos\text{x})$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{u}[(\tan\text{x})\times(-\sin\text{x})+\log\log\text{x}(\cos\text{x})]$
$\Rightarrow\frac{\text{du}}{\text{dx}}=(\cos\text{x})^{\sin\text{x}}[\cos\text{x}\log\cos\text{x}-\sin\text{x}\tan\text{x}]\ .....\text{(i)} $
Let, $\text{v = }(\sin\text{x})^{\cos\text{x}}$
Taking log on both sides,
$\log\text{v}=\log(\sin\text{x})^{\cos\text{x}}$
$\Rightarrow\log\text{v}=\cos\text{x}\log(\sin\text{x})$
Differentiating it with respect to x using chain rule,
$\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\cos\text{x}\frac{\text{d}}{\text{dx}}(\log\sin\text{x})+\log\sin\text{x}\frac{\text{d}}{\text{dx}}(\cos\text{x}) $
[using product rule]
$\Rightarrow\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\cos\text{x}\big(\frac{1}{\sin\text{x}}\big)\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}(-\sin\text{x})$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\text{v}[\cot\text{x}(\cos \text{x})-\sin\text{x}\log\sin\text{x}]$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=(\sin\text{x})^{\cos\text{x}}[\cot \text{x} (\cos \text{x})-\sin \text{x}\log\sin\text{x}]\ ...(\text{ii})$
Dividing equation (i) by (ii)
$\therefore\frac{\text{du}}{\text{dv}}=\frac{(\cos\text{x})^{\sin\text{x}}[\cos \text{x}\log\cos\text{x}-\sin\text{x}\tan\text{x ]}}{(\sin\text{x})^{\cot\text{x}}[\cot \text{x}(\cos\text{x})-\sin\text{x}\log\sin\text{x}]}$
View full question & answer→Question 355 Marks
Differentiate the following functions with respect to x:
$\log\sqrt{\frac{1-\cos\text{x}}{1+\cos\text{x}}}$
AnswerLet, $\text{y}=\log\sqrt{\frac{1-\cos\text{x}}{1+\cos\text{x}}}$
$\Rightarrow\ \text{y}=\log\Big(\frac{1-\cos\text{x}}{1+\cos\text{x}}\Big)^\frac{1}{2}$
$\Rightarrow\ \text{y}=\frac{1}{2}\log\Big(\frac{1-\cos\text{x}}{1+\cos\text{x}}\Big) \big[\text{Using }\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiate with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big\{\frac{1}{2}\log\Big(\frac{1-\cos\text{x}}{1+\cos\text{x}}\Big)\Big\}$
$=\frac{1}{2}\times\frac{1}{\Big(\frac{1-\cos\text{x}}{1+\cos\text{x}}\Big)}\times\frac{\text{d}}{\text{dx}}\Big(\frac{1-\cos\text{x}}{1+\cos\text{x}}\Big)$
[Using chain rule]
$=\frac{1}{2}\Big(\frac{1+\cos\text{x}}{1-\cos\text{x}}\Big)\bigg[\frac{(1+\cos\text{x})\frac{\text{d}}{\text{dx}}(1-\cos\text{x})-(1-\cos\text{x})\frac{\text{d}}{\text{dx}}(1+\cos\text{x})}{(1+\cos\text{x})^2}\bigg]$
$=\frac{1}{2}\Big(\frac{1+\cos\text{x}}{1-\cos\text{x}}\Big)\bigg[\frac{(1+\cos\text{x})(\sin\text{x})-(1-\cos\text{x})(-\sin\text{x})}{(1+\cos\text{x})^2}\bigg]$
$=\frac{1}{2}\Big(\frac{1+\cos\text{x}}{1-\cos\text{x}}\Big)\Big[\frac{\sin\text{x}+\sin\text{x}\cos\text{x}+\sin\text{x}-\sin\text{x}\cos\text{x}}{(1+\cos\text{x})^2}\Big]$
$=\frac{1}{2}\Big(\frac{1+\cos\text{x}}{1-\cos\text{x}}\Big)\Big[\frac{2\sin\text{x}}{(1+\cos\text{x})^2}\Big]$
$=\frac{\sin\text{x}}{(1-\cos\text{x})(1+\cos\text{x})}$
$=\frac{\sin\text{x}}{1-\cos^2\text{x}}$
$=\frac{\sin\text{x}}{\sin^2\text{x}}$
$=\frac{1}{\sin\text{x}}$
$=\text{cosec x}$
So,
$\frac{\text{d}}{\text{dx}}\Big(\log\sqrt{\frac{1-\cos\text{x}}{1+\cos\text{x}}}\Big)=\text{cosec x}$
View full question & answer→Question 365 Marks
Differentiate the following functions with respect to x:
$\frac{\sqrt{\text{x}^2+1}+\sqrt{\text{x}^2-1}}{\sqrt{\text{x}^2+1}-\sqrt{\text{x}^2-1}}$
AnswerWe have, $\frac{\sqrt{\text{x}^2+1}+\sqrt{\text{x}^2-1}}{\sqrt{\text{x}^2+1}-\sqrt{\text{x}^2-1}}$
By rationalising we get,
$\frac{\sqrt{\text{x}^2+1}+\sqrt{\text{x}^2-1}}{\sqrt{\text{x}^2+1}-\sqrt{\text{x}^2-1}}\times\frac{\sqrt{\text{x}^2+1}+\sqrt{\text{x}^2-1}}{\sqrt{\text{x}^2+1}+\sqrt{\text{x}^2-1}}$
$=\frac{\big(\sqrt{\text{x}^2+1}+\sqrt{\text{x}^2-1}\big)^2}{\big(\sqrt{\text{x}^2+1}\big)^2-\big(\sqrt{\text{x}^2-1}\big)^2}$
$=\frac{\big(\sqrt{\text{x}^2+1}\big)^2+\big(\sqrt{\text{x}^2-1}\big)^2+2\big(\sqrt{\text{x}^2+1}\big)\big(\sqrt{\text{x}^2-1}\big)}{\text{x}^2+1-\text{x}^2+1}$
$=\frac{\text{x}^2+1+\text{x}^2-1+2\sqrt{\text{x}^4-1}}{2}$
$=\frac{2\text{x}^2+2\sqrt{\text{x}^4-1}}{2}$
$=\text{x}^2+\sqrt{\text{x}^4-1}$
Now, Let $\text{y}=\text{x}^2+\sqrt{\text{x}^4-1}$
Differentiate it with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{x}^2+\sqrt{\text{x}^4-1}\big)$
$=2\text{x}+\frac{1}{2\sqrt{\text{x}^4-1}}\times\frac{\text{d}}{\text{dx}}(\text{x}^4-1)$
$=2\text{x}+\frac{1}{2\sqrt{\text{x}^4-1}}\times(4\text{x}^3)$
$=2\text{x}+\frac{2\text{x}^3}{\sqrt{\text{x}^4-1}}$
View full question & answer→Question 375 Marks
Find $\frac{\text{dy}}{\text{dx}},$ when
$\text{x}=\frac{3\text{at}}{1+\text{t}^2}\text{ and y}=\frac{3\text{at}^2}{1+\text{t}^2}$
AnswerHere, $\text{x}=\frac{3\text{at}}{1+\text{t}^{2}}$
Differentiating it with respect to t using quotiont rule,
$\frac{\text{dx}}{\text{dt}}= \bigg[\frac{(1+\text{t}^{2})\frac{\text{d}}{\text{dt}}(3\text{at})-3\text{at}\frac{\text{d}}{\text{dt}}(1+\text{t}^{2})}{(1+\text{t}^{2})}\bigg]$
$\frac{\text{dx}}{\text{dt}}=\bigg[\frac{(1+\text{t}^{2})(3\text{a})-3\text{at}(2\text{t})}{(1+\text{t}^{2})}\bigg]$
$\frac{\text{dx}}{\text{dt}}=\bigg[\frac{(1+\text{t}^{2})(3\text{a})-3\text{at}(2\text{t})}{(1+\text{t}^{2})}\bigg]$
$\frac{\text{dx}}{\text{dt}}=\Big[\frac{3\text{a}-3\text{at}^{2}}{(1-\text{t}^{2})^{2}}\Big]$ And,
$\frac{\text{dx}}{\text{dt}}=\frac{3\text{a}(1-\text{t}^{2})}{(1+\text{t}^{2})^{2}}.....(\text{i})$
$\text{y}=\frac{3\text{at}^{2}}{(1+\text{t}^{2})}$
Differentiating it with respect to t using quotient rule,
$\frac{\text{dy}}{\text{dt}}=\bigg[\frac{(1+\text{t}^{2})\frac{\text{d}}{\text{dt}}(3\text{at}^{2})-3\text{at}\frac{\text{d}}{\text{dt}}(1+\text{t}^{2})}{(1+\text{t}^{2})^{2}}\bigg]$
$\frac{\text{dy}}{\text{dt}}=\Big[\frac{(1+\text{t}^{2})(6\text{at})-(3\text{at}^{2})(2\text{t})}{(1+\text{t}^{2})^{2}}\Big]$
$\frac{\text{dy}}{\text{dt}}=\Big[\frac{6\text{at}+6\text{at}^{3}-6\text{at}^{3}}{(1+\text{t}^{2})^{2}}\Big]$
$\frac{\text{dy}}{\text{dt}}=\frac{6\text{at}}{(1+\text{t}^{2})^{2}}....(\text{ii})$
Dividing equation (ii) by (i),
$\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{6\text{at}}{(1+\text{t}^{2})^{2}}\times\frac{(1+\text{t}^{2})^{2}}{3\text{a}(1-\text{t}^{2})}$
$\frac{\text{dy}}{\text{dx}}=\frac{2\text{t}}{1-\text{t}^{2}}$
View full question & answer→Question 385 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=\text{x}^{\sin\text{x}}+\big(\sin\text{x}\big)^\text{x}$
AnswerLet $\text{y}=\text{x}^{\sin\text{x}}+(\sin\text{x})^\text{x}$
Also, let $\text{u}=\text{x}^{\sin\text{x}}\text{ and v}=(\sin\text{x})^\text{x}$
$\therefore\text{y}=\text{u}+\text{v}$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{\text{du}}{\text{dx}}+\frac{\text{dv}}{\text{dx}}\ .....(\text{i})$
$\text{u}=\text{x}^{\sin\text{x}}$
$\Rightarrow\log\text{u}=\log\big(\text{x}^{\sin\text{x}}\big)$
$\Rightarrow\log\text{u}=\sin\text{x}\log\text{x}$
Differentiating both sides with respect to x, we obtain
$\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\sin\text{x})\times\log\text{x}+\sin\text{x}\times\frac{\text{d}}{\text{dx}}(\log\text{x})$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{u}\Big[\cot\text{x}\log\text{x}+\sin\text{x}\times\frac{1}{\text{x}}\Big]$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{x}^{\sin\text{x}}\Big[\cos\text{x}\log\text{x}+\frac{\sin\text{x}}{\text{x}}\Big]\ .....(\text{ii})$
$\text{v}=(\sin\text{x})^\text{x}$
$\Rightarrow\log\text{v}=\log(\sin\text{x})^\text{x}$
$\Rightarrow\log\text{v}=\text{x}\log(\sin\text{x})$
Differentiating both sides with respect to x, we obtain
$\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x})\times\log(\sin\text{x})+\text{x}\times\frac{\text{d}}{\text{dx}}\big[\log(\sin\text{x})\big]$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\text{v}\Big[\log(\sin\text{x})+\text{x}\times\frac{1}{\sin\text{x}}\times\frac{\text{d}}{\text{dx}}(\sin\text{x})\Big]$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=(\sin\text{x})^\text{x}\Big[\log\sin\text{x}+\frac{\text{x}}{\sin\text{x}}\cos\text{x}\Big]$
$\Rightarrow\frac{\text{dv}}{\text{dx}}(\sin\text{x})^\text{x}\big[\log\sin\text{x}+\text{x}\cot\text{x}\big]$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=(\sin\text{x})^\text{x}\big[\log\sin\text{x}+\text{x}\cot\text{x}\big]\ .....(\text{iii})$
From (i), (ii) and (iii), we obtain
$\frac{\text{dy}}{\text{dx}}=\text{x}^{\sin\text{x}}\Big(\cos\text{x}\log\text{x}+\frac{\sin\text{x}}{\text{x}}\big)+(\sin\text{x})^\text{x}\big[\log\sin\text{x}+\text{x}\cot\text{x}\big]$
View full question & answer→Question 395 Marks
If $y^x + x^y + x^x = a^b,$ find $\frac{\text{dy}}{\text{dx}}$
AnswerGiven that $y^x + x^y + x^x = a^b$
Putting $u = y^x, v = x^y$ and $w = x^x$, we get $u + v + w = a^b$
Therefore $\frac{\text{du}}{\text{dx}}+\frac{\text{dv}}{\text{dx}}+\frac{\text{dw}}{\text{dx}}=0\ .....(\text{i})$
Now, $u = y^x$. Taking logrithm on both sides, we have
$\log\text{u}=\text{x}\log\text{y}$
Differentiating both sides w.r.t. x, we have
$\frac{1}{\text{u}}\times\frac{\text{du}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}(\log\text{y})+\log\text{y}\frac{\text{d}}{\text{dx}}(\text{x})$
$=\text{x}\frac{1}{\text{y}}\times\frac{\text{dy}}{\text{dx}}+\log\text{y}\times1$
So, $\frac{\text{du}}{\text{dx}}=\text{u}\Big(\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}+\log\text{y}\Big)=\text{y}^\text{x}\Big[\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}+\log\text{y}\Big]\ .....(\text{ii})$
Also $v = x^y$
Taking logarithm on both sides, we have
$\log\text{v}=\text{y}\log\text{x}$
Differentiating both sides w.r.t. x, we have
$\frac{1}{\text{v}}\times\frac{\text{dv}}{\text{dx}}=\text{y}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{dy}}{\text{dx}}$
$=\text{y}\times\frac{1}{\text{x}}+\log\text{x}\times\frac{\text{dy}}{\text{dx}}$
So, $\frac{\text{dv}}{\text{dx}}=\text{v}\Big[\frac{\text{y}}{\text{x}}+\log\text{x}\times\frac{\text{dy}}{\text{dx}}\Big]$
$=\text{x}^\text{y}\Big[\frac{\text{y}}{\text{x}}+\log\text{x}\times\frac{\text{dy}}{\text{dx}}\Big]$
Again $w = x^x$
Taking logarithm on both sides, we have
$\log\text{w}=\text{x}\log\text{x}$
Differentiating both sides w.r.t x, we have
$\frac{1}{\text{w}}\times\frac{\text{dw}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{x}}{\text{dx}}(\text{x})$
$=\text{x}.\frac{1}{\text{x}}+\log\text{x}\times1$
i.e. $\frac{\text{dw}}{\text{dx}}=\text{w}(1+\log\text{x})$
$=\text{x}^\text{x}(1+\log\text{x})\ .....(\text{iv})$
From (i), (ii), (iii), (iv), we have
$\text{y}^\text{x}\Big(\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}+\log\text{y}\Big)+\text{x}^\text{y}\Big(\frac{\text{y}}{\text{x}}+\log\text{x}\frac{\text{dy}}{\text{dx}}\Big) \\ +\text{x}^\text{x}(1+\log\text{x})=0$
$\big(\text{x}\times\text{y}^{\text{x}-1}+\text{x}^\text{y}\times\log\text{x}\big) \\ \frac{\text{dy}}{\text{dx}}=-\text{x}^\text{x}(1+\log\text{x})-\text{y}\times\text{x}^{\text{y}-1}-\text{y}^\text{x}\log\text{y}=0$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{-\big[\text{y}^\text{x}\log\text{y}+\text{y}\times\text{x}^{\text{y}-1}+\text{x}^\text{x}(1+\log\text{x})\big]}{\text{x}\times\text{y}^{\text{x}-1}+\text{x}^\text{y}\log\text{y}}$
View full question & answer→Question 405 Marks
Find $\frac{\text{dy}}{\text{dx}},$ when
$\text{x}=\text{e}^{\theta}\Big(\theta+\frac{1}{\theta}\Big)\text{ and y}=\text{e}^{-\theta}\Big(\theta-\frac{1}{\theta}\Big)$
AnswerWe have, $\text{x}=\text{e}^\theta\Big(\theta+\frac{1}{\theta}\Big)$
$\frac{\text{dx}}{\text{d}\theta}=\text{e}^\theta\frac{\text{d}}{\text{d}\theta}\Big(\theta+\frac{1}{\theta}\Big)+\Big(\theta+\frac{1}{\theta}\Big)\frac{\text{d}}{\text{d}\theta}(\text{e}^\theta)$
[Using product rule]
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{e}^\theta\Big(1-\frac{1}{\theta^{2}}\Big)+\Big(\frac{\theta^{2}+1}{\theta}\Big)\text{e}^{\theta}$
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{e}^{\theta}\Big(1-\frac{1}{\theta^{2}}+\frac{\theta^{2}+1}{\theta}\Big)$
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{e}^\theta\Big(\frac{\theta^{2}-1+\theta^{3}+\theta}{\theta{2}}\Big)$
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=\frac{\text{e}^\theta(\theta^{3}+\theta^{2}+\theta-1)}{\theta^{2}}\ .....(\text{i})$
And, $\text{y}=\text{e}^\theta\Big(\theta-\frac{1}{\theta}\Big)$
View full question & answer→Question 415 Marks
Differentiate the following functions with respect to x:
$\frac{3\text{x}^2\sin\text{x}}{\sqrt{7-\text{x}^2}}$
AnswerLet $\text{y}=\frac{3\text{x}^2\sin\text{x}}{\sqrt{7-\text{x}^2}}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\bigg\{\frac{3\text{x}^2\sin\text{x}}{(\sqrt{7-\text{x}^2})^\frac{1}{2}}\bigg\}$
$=\frac{(7-\text{x}^2)^\frac{1}{2}\times\frac{\text{d}}{\text{dx}}(3\text{x}^3\sin\text{x})-(3\text{x}^2\sin\text{x})\frac{\text{d}}{\text{dx}}(7-\text{x}^2)^\frac{1}{2}}{\Big[(7-\text{x}^2)^\frac{1}{2}\Big]^2}$
[Using quotient rule, chain rule and product rule]
$\Bigg[\frac{(7-\text{x}^2)^\frac{1}{2}\times3\Big[\text{x}^2\frac{\text{d}}{\text{dx}}(\sin\text{x})+\sin\text{x}\frac{\text{d}}{\text{dx}}(\text{x}^2)\Big]-3\text{x}^2\sin\text{x}\times\frac{1}{2}(7-\text{x}^2)\times\frac{\text{d}}{\text{dx}}(7-\text{x}^2)}{(7-\text{x}^2)}\Bigg]$
$\Bigg[\frac{(7-\text{x}^2)^\frac{1}{2}3(\text{x}^2\cos\text{x}+2\text{x}\sin\text{x})-3\text{x}^2\sin\text{x}\times\frac{1}{2}(7-\text{x}^2)^\frac{-1}{2}(-2\text{x})}{(7-\text{x}^2)}\Bigg]$
$=\Bigg[\frac{(7-\text{x}^2)^\frac{1}{2}\times3(\text{x}^2\cos+2\text{x}\sin\text{x})}{(7-\text{x}^2)}+\frac{3\text{x}^2\sin\text{x}(7-\text{x}^2)^\frac{-1}{2}}{(7-\text{x}^2)}\Bigg]$
$\bigg[\frac{6\text{x}\sin\text{x}+3\text{x}^2\cos\text{x}}{\sqrt{(7-\text{x}^2)}}+\frac{3\text{x}^3\sin\text{x}}{(7-\text{x}^2)^\frac{3}{2}}\bigg]$
So,
$\frac{\text{d}}{\text{dx}}\Big(\frac{3\text{x}^2\sin\text{x}}{\sqrt{7-\text{x}^2}}\Big)\bigg[\frac{6\text{x}\sin\text{x}+3\text{x}^2\cos\text{x}}{\sqrt{(7-\text{x}^2)}}+\frac{3\text{x}^3\sin\text{x}}{(7-\text{x}^2)^\frac{3}{2}}\bigg]$
View full question & answer→Question 425 Marks
If $\text{x}\sqrt{1+\text{y}}+\text{y}\sqrt{1+\text{x}}=0,$ prove that $(1+\text{x})^2\frac{\text{dx}}{\text{dx}}+1=0$
AnswerWe have $\text{x}\sqrt{1+\text{y}}+\text{y}\sqrt{1+\text{x}}=0$
$\Rightarrow\text{x}\sqrt{1+\text{y}}=-\text{y}\sqrt{1+\text{x}}$
Squaring both sides, we get,
$\Rightarrow\big(\text{x}\sqrt{1+\text{y}}\big)^2=(-\text{y}\sqrt{1+\text{x}}\big)^2$
$\Rightarrow\text{x}^2\big(1+\text{y}\big)=\text{y}^2(1+\text{x}\big)$
$\Rightarrow\text{x}^2+\text{x}^2\text{y}=\text{y}^2+\text{y}^2\text{x}$
$\Rightarrow\text{x}^2-\text{y}^2=\text{y}^2\text{x}-\text{x}^2\text{y}$
$\Rightarrow(\text{x}-\text{y})(\text{x}+\text{y})=\text{x}\text{y}(\text{y}-\text{x})$
$\Rightarrow(\text{x}+\text{y})=-\text{x}\text{y}$
$\Rightarrow\text{y}+\text{x}\text{y}=-\text{x}$
$\Rightarrow\text{y}(1+\text{x})=-\text{x}$
$\Rightarrow\text{y}=\frac{-\text{x}}{(1+\text{x})}$
Differentiating with respect to x, we get
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\bigg[\frac{-(1+\text{x})\frac{\text{d}}{\text{dx}}(\text{x})-(-\text{x})\frac{\text{d}}{\text{dx}}(\text{x}+1)}{(1+\text{x})^2}\bigg]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\Big[\frac{-(1+\text{x})(1)+\text{x}(1)}{(1+\text{x})^2}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\Big[\frac{-1-\text{x}+\text{x}}{(1+\text{x})^2}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-1}{(1+\text{x})^2}$
$\Rightarrow(1+\text{x})^2\frac{\text{dy}}{\text{dx}}=-1$
$\Rightarrow(1+\text{x})^2\frac{\text{dy}}{\text{dx}}+1=0$
View full question & answer→Question 435 Marks
If $\text{y}=\big\{\log_{\cos\text{x}}\sin\text{x}\big\}\big\{\log_{\sin\text{x}}\cos\text{x}\big\}^{-1}+\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big),$ find $\frac{\text{dy}}{\text{dx}}$ at $\text{x}=\frac{\pi}{4}$
AnswerWe have, $\text{y}=\big\{\log_{\cos\text{x}}\sin\text{x}\big\}\big\{\log_{\sin\text{x}}\cos\text{x}\big\}^{-1}+\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$
$\Rightarrow \text{y}\big\{\log_{\cos\text{x}}\sin\text{x}\big\}\big\{\log_{\cos\text{x}}\sin\text{x}\big\}+\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big) \\ \big[\because\log_{\text{a}}\text{b}=(\log_\text{b}\text{a})^{-1}\big]$
$\Rightarrow \text{y}=\Big[\frac{\log\sin\text{x}}{\log\cos\text{x}}\Big]^2+\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)\ \Big[\because \log_\text{a}\text{b}=\frac{\log\text{b}}{\log\text{a}}\Big]$
Differentiating with respect to x,
$\frac{\text{d}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big[\frac{\log\sin\text{x}}{\log\sin\text{x}}\Big]^2+\frac{\text{d}}{\text{dx}}\Big\{\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)\Big\}$
$\Rightarrow\frac{\text{d}}{\text{dx}}=2\Big[\frac{\log\sin\text{x}}{\log\sin\text{x}}\Big]\frac{\text{d}}{\text{dx}}\Big(\frac{\log\sin\text{x}}{\log\sin\text{x}}\Big)+\frac{1}{\sqrt{-1\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)^2}}\times\frac{\text{d}}{\text{dx}}\Big[\frac{2\text{x}}{1+\text{x}^2}\Big]$
$\Rightarrow\frac{\text{d}}{\text{dx}}=2\Big[\frac{\log\sin\text{x}}{\log\sin\text{x}}\Big]\bigg[\frac{(\log\cos\text{x})\frac{\text{d}}{\text{dx}}(\log\sin\text{x})-\log\sin\text{x}\frac{\text{d}}{\text{dx}}(\log\cos\text{x})}{(\log\cos\text{x})^3}\bigg] \\ +\Big[\frac{(1+\text{x}^2)}{\sqrt{1+\text{z}^2-2\text{x}^2}}\Big]\Big[\frac{(1+\text{x}^2)(2)-(2\text{x})(2\text{x})}{(1+\text{x}^2)^3}\Big]$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=2\Big[\frac{\log\sin\text{x}}{\log\cos\text{x}}\Big]\bigg[\frac{\log\cos\times\frac{1}{\sin\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x})-\log\sin\text{x}\times\frac{1}{\cos\text{x}}\frac{\text{d}}{\text{dx}}(\cos\text{x})}{(\log\cos\text{x})^3}\bigg] \\ +\Big[\frac{(1+\text{x}^2)}{\sqrt{1+\text{x}^4-2\text{x}^2}}\Big]\Big[\frac{(1+\text{x}^2)(2)-(2\text{x})(2\text{x})}{(1+\text{x}^2)^2}\Big]$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=2\Big[\frac{\log\sin\text{x}}{\log\cos\text{x}}\Big]\bigg[\frac{\log\cos\text{x}\Big(\frac{\cos\text{x}}{\sin\text{x}}\Big)+\log\sin\text{x}\times\Big(\frac{\sin\text{x}}{\cos\text{x}}\Big)}{(\log\cos\text{x})^2}\bigg] \\ +\Big[\frac{1+\text{x}^2}{\sqrt{(1-\text{x}^2)^3}}\Big]\Big[\frac{2+2\text{x}^2-4\text{x}^2}{(1+\text{x}^2)^2}\Big]$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=2\frac{\log\sin\text{x}}{(\log\cos\text{x})^3}(\cos\text{x}\log\cos\text{x}+\tan\text{x}\log\sin\text{x})+\frac{2}{1+\text{x}^2}$
Put $\text{x}=\frac{\pi}{4}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=2\Bigg\{\frac{\log\sin\frac{\pi}{4}}{\big(\log\cos\frac{\pi}{4}\big)^3}\Bigg\} \\ \Big(\cot\frac{\pi}{4}\log\cos\frac{\pi}{4}+\tan\frac{\pi}{4}\log\sin\frac{\pi}{4}\Big)+2\bigg\{\frac{1}{1+\big(\frac{\pi}{4}\big)^2}\bigg\}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=2\Bigg\{\frac{1}{\big(\log\frac{1}{\sqrt{2}}\big)^2}\Bigg\} \\ \Big(1\times\log\frac{1}{\sqrt{2}}+1\times\log\frac{1}{\sqrt{2}}\Big)+2\Big(\frac{16}{16+\pi^2}\Big)$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=2\frac{2\log\Big(\frac{1}{\sqrt{2}}\Big)}{\Big\{\log\Big(\frac{1}{\sqrt{2}}\Big)^2\Big\}}+\frac{32}{16+\pi^2}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=4\frac{1}{\log\Big(\frac{1}{\sqrt{2}}\Big)}+\frac{32}{16+\pi^2}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=4\frac{1}{-\frac{1}{2}\log2}+\frac{32}{16+\pi^2}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=-\frac{8}{\log2}+\frac{32}{16+\pi^2}$
So, $\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{a}=\frac{\pi}{4}}=8\Big[\frac{4}{16+\pi^2}-\frac{1}{\log2}\Big]$
View full question & answer→Question 445 Marks
Differentiate $\cos^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$ with respect to $\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big),$ if $0<\text{x}<1$
AnswerLet $\text{u}=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$
Put $\text{x}=\tan\theta,$
$\text{u}=\sin^{-1}\Big(\frac{2\tan\theta}{1+\tan^2\theta}\Big)$
$\text{u}= \sin^{-1}(\sin2\theta)\ ..... (\text{i})$
Let $\text{v}= \cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$
$=\cos^{-1}\Big(\frac{1-\tan^2\theta}{1+\tan^2\theta}\Big)$
$\text{v}=\cos^{-1}(\cos2\theta)\ .....(\text{ii})$
Here, 0 < x < 1
$0<\tan\theta<1$
$\Rightarrow 0<\theta<\frac{\pi}{4}$
So, from equation (i),
$\text{u}=2 \theta\ \Big [\text{Since,} \sin^-1(\sin\theta)=\theta, \text{if } \theta \in\Big[\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\text{u}=2\tan^{-1}\text{x}\ \big[\text{Since, x}=\tan\frac{\pi}{2}\big]$
Differentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{2}{1+\text{x}^2}\ .....(\text{iii})$
From equation (ii),
$\text{v}=2 \theta \ \big[ \text{since,}\cos^{-1}(\cos\theta)=\theta,\text{if }\theta\in[0,\pi]\big]$
$\text{v}=2\tan^{-1}\text{x}[{\text{since,x}=\tan\theta}]$
Differentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{2}{1+\text{x}^2}\ .....(\text{iv})$
Dividing equation (iii) by (iv),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{2}{(1+\text{x}^2)}\times\frac{(1+\text{x}^2)}{2}$
$\frac{\text{du}}{\text{dv}}=1$
View full question & answer→Question 455 Marks
Differentiate the following functions with respect to x:
$\log\big\{\text{x}+2+\sqrt{\text{x}^2+4\text{x}+1}\big\}$
AnswerLet $\text{y}=\log\big\{\text{x}+2+\sqrt{\text{x}^2+4\text{x}+1}\big\}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big[\log\big\{\text{x}+2+\sqrt{\text{x}^2+4\text{x}+1}\big\}\Big]$
$=\frac{1}{\big[\text{x}+2+\sqrt{\text{x}^2+4\text{x}+1}\big]}\frac{\text{d}}{\text{dx}}\Big[\text{x}+2+\big(\text{x}^2+4\text{x}+1\big)^\frac{1}{2}\Big]$
[Using chain rule]
$=\frac{1}{\big[\text{x}+2+\sqrt{\text{x}^4+4\text{x}+1}\big]}\times\Big[1+0+\frac{1}{2}\big(\text{x}^2+4\text{x}+1\big)^\frac{1}{2}\frac{\text{d}}{\text{dx}}(\text{x}^2+4\text{x}+1)\Big]$
$=\frac{1+\frac{(2\text{x}+4)}{2\big(\sqrt{\text{x}^2+4\text{x}+1}\big)}}{\big[\text{x}+2+\sqrt{\text{x}^4+4\text{x}+1}\big]}$
$=\frac{\sqrt{\text{x}^2+4\text{x}+1}+\text{x}+2}{\big[\text{x}+2+\sqrt{\text{x}^2+4\text{x}+1}\big]\times\sqrt{\text{x}^2+4\text{x}+1}}$
$=\frac{1}{\sqrt{\text{x}^2+4\text{x}+1}}$
So,
$\frac{\text{d}}{\text{dx}}\Big[\log\big\{\text{x}+2+\sqrt{\text{x}^2+4\text{x}+1}\big\}\Big]=\frac{1}{\sqrt{\text{x}^2+4\text{x}+1}}$
View full question & answer→Question 465 Marks
If $\text{y}\sqrt{1-\text{x}^2}+\text{x}\sqrt{1-\text{y}^2}=1,$ prove that $\frac{\text{dy}}{\text{dx}}=-\sqrt{\frac{1-\text{y}^2}{1-\text{x}^2}}$
AnswerWe have, $\text{y}\sqrt{1-\text{x}^2}+\text{x}\sqrt{1-\text{y}^2}=1$
Let $\text{x}=\sin\text{A},\text{y}=\sin\text{B}$
$\Rightarrow\sin\text{B}\sqrt{1-\sin^2\text{A}}+\sin\text{A}\sqrt{1-\sin^2\text{B}}=1$
$\Rightarrow\sin\text{B}\cos\text{A}+\sin\text{A}\cos\text{B}=1$
$\big[\because\sin(\text{x}+\text{y})=\sin\text{x}\cos\text{y}+\cos\text{x}\sin\text{y}\big]$
$\Rightarrow\sin\big(\text{A}+\text{B}\big)=1$
$\Rightarrow\text{A}+\text{B}=\sin^{-1}(1)$
$\Rightarrow\sin^{-1}\text{x}+\sin^{-1}\text{y}=\frac{\pi}{2} \big[\because\text{x}=\sin\text{A},\text{y}=\sin\text{B}\big]$
Differentiating with respect to x, we get
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\sin^{-1}\text{x}\big)+\frac{\text{d}}{\text{dx}}\big(\sin^{-1}\text{y}\big)=\frac{\text{d}}{\text{dx}}\Big(\frac{\pi}{2}\Big)$
$\Rightarrow\frac{1}{\sqrt{1-\text{x}^2}}+\frac{1}{\sqrt{1-\text{y}^2}}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\sqrt\frac{1-\text{y}^2}{1-\text{x}^2}$
View full question & answer→Question 475 Marks
Differentiate $\sin^{-1}\Big(2\text{x}\sqrt{1-\text{x}^2}\Big)$ with respect to $\sec^{-1}\Big(\frac{1}{\sqrt{1+\text{x}^2}}\Big),$ if:
$\text{x}\in\Big(\frac{1}{\sqrt{2}},1\Big)$
AnswerLet $\text{u}=\sin^{-1}\Big(2\text{x}\sqrt{1-\text{x}^2}\Big)$
Put $\text{x}=\sin\theta$
$\Rightarrow\text{u}=\sin^{-1}\Big(2\sin\theta\sqrt{1-\sin^2\theta}\Big)$
$\Rightarrow\text{u}=\sin^{-1}(2\sin\theta\cos\theta)$
$\Rightarrow\text{u}=\sin^{-1}(\sin2\theta)\ .....(\text{i})$
And,
Let $\text{v}=\sec^{-1}\Big(\frac{1}{\sqrt{1-\text{x}^2}}\Big)$
$\Rightarrow\text{v}=\sec^{-1}\Big(\frac{1}{\sqrt{1-\sin^2\theta}}\Big)$
$\Rightarrow\text{v}=\sec^{-1}\Big(\frac{1}{\cos\theta}\Big)$
$\Rightarrow\text{v}=\sec^{-1}(\sec\theta)$
$\Rightarrow\text{v}=\cos^{-1}\bigg(\frac{1}{\frac{1}{\cos\theta}}\bigg)\ \Big[\text{Since},\sec^{-1}\text{x}=\cos^{-1}\big(\frac{1}{\text{x}}\big)\Big]$
$\Rightarrow\text{v}=\cos^{-1}(\cos\theta)\ .....(\text{ii})$
Here,
$\text{x}\in\Big(\frac{1}{\sqrt{2}},1\Big)$
$\Rightarrow\sin\theta\in\Big(\frac{1}{\sqrt{2}},1\Big)$
$\Rightarrow\theta\in\Big(\frac{\pi}{4},\frac{\pi}{2}\Big)$
So, from equation (i),
$\text{u}=2\theta\Big[\text{Since},\sin^{-1}(\sin\theta)=\theta,\text{ if }\theta\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\Big]$
Let, $\text{u}=2\sin^{-1}\text{x}\big[\text{Since},\text{x}=\sin\theta\big]$
$\frac{\text{du}}{\text{dx}}=2\Big(\frac{1}{\sqrt{1-\text{x}^2}}\Big)$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\frac{2}{\sqrt{1-\text{x}^2}}\ .....(\text{iii})$
And, from equation (ii),
$\text{v}=\theta\ \big[\text{Since},\cos^{-1}(\cos\theta)=\theta,\text{if }\theta\in[0,\pi]\big]$
$\Rightarrow\text{v}=\sin^{-1}\text{x}\big[\text{Since},\text{x}=\sin\theta\big]$
$\frac{\text{dv}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}\ .....(\text{iv})$
Dividing equation (iii) by (iv),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{2}{\sqrt{1-\text{x}^2}}\times\frac{\sqrt{1-\text{x}^2}}{1}$
$\therefore\frac{\text{du}}{\text{dv}}=2$
View full question & answer→Question 485 Marks
Differentiate the following functions with respect to x:
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^\text{x}+\text{x}^{\Big(1+\frac{1}{\text{x}}\Big)}$
AnswerLet $\text{y}=\Big(\text{x}+\frac{1}{\text{x}}\Big)^\text{x}+\text{x}^{\Big(1+\frac{1}{\text{x}}\Big)}$
Also, let $\Big(\text{x}+\frac{1}{\text{x}}\Big)^\text{x}$ and $\text{v}=\text{x}^{\Big(1+\frac{1}{\text{x}}\Big)}$
$\therefore\text{y}=\text{u}+\text{v}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{du}}{\text{dx}}+\frac{\text{dv}}{\text{dx}}....(1)$
Then, $\text{u}=\Big(\text{x}+\frac{1}{\text{x}}\Big)^\text{x}$
$\Rightarrow\log\text{u}=\log\Big(\text{x}+\frac{1}{\text{x}}\Big)^\text{x}$
$\Rightarrow\log\text{u}=\text{x}\log\Big(\text{x}+\frac{1}{\text{x}}\Big)$
Differentiating both sides with respect to x, we obtain,
$\frac{1}{\text{u}}\cdot\frac{\text{du}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x})\times\log\Big(\text{x}+\frac{1}{\text{x}}\Big)+\text{x}\times\frac{\text{d}}{\text{dx}}\Big[\log\Big(\text{x}+\frac{1}{\text{x}}\Big)\Big]$
$\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=1\times\log\Big(\text{x}+\frac{1}{\text{x}}\Big)+\text{x}\times\frac{1}{\Big(\text{x}+\frac{1}{\text{x}}\Big)}\cdot\frac{\text{d}}{\text{dx}}\Big(\text{x}+\frac{1}{\text{x}}\Big)$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\Big(\text{x}+\frac{1}{\text{x}}\Big)^\text{x}\Big[\log\Big(\text{x}+\frac{1}{\text{x}}\Big)+\frac{\text{x}^2-1}{\text{x}^2+1}\Big]$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\Big(\text{x}+\frac{1}{\text{x}}\Big)^\text{x}\Big[\frac{\text{x}^2-1}{\text{x}^2+1}+\log\Big(\text{x}+\frac{1}{\text{x}}\Big)\Big]$
$\text{v}=\text{x}^{\Big(1+\frac{1}{\text{x}}\Big)}$
$\Rightarrow\log\text{v}=\log\Bigg[\text{x}^{\Big(1+\frac{1}{\text{x}}\Big)}\Bigg]$
$\Rightarrow\log\text{v}=\Big(1+\frac{1}{\text{x}}\Big)\log\text{x}....(2)$
Differentiating both sides with respect to x, we obtain
$\frac{1}{\text{v}}\cdot\frac{\text{dv}}{\text{dx}}=\Big[\frac{\text{d}}{\text{dx}}\Big(1+\frac{1}{\text{x}}\Big)\Big]\times\log\text{x}+\Big(1+\frac{1}{\text{x}}\Big)\cdot\frac{\text{d}}{\text{dx}}\log\text{x}$
$\Rightarrow\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\Big(-\frac{1}{\text{x}^2}\Big)\log\text{x}+\Big(1+\frac{1}{\text{x}}\Big)\cdot\frac{1}{\text{x}}$
$\Rightarrow\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=-\frac{\log\text{x}}{\text{x}^2}+\frac{1}{\text{x}}+\frac{1}{\text{x}^2}$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\text{v}\Big[\frac{-\log\text{x}+\text{x}+1}{\text{x}^2}\Big]$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\text{x}^{\big(1+\frac{1}{\text{x}}\big)}\Big(\frac{\text{x}+1-\log\text{x}}{\text{x}^2}\Big) .....(3)$
Therefore, from (i), (ii) and (iii), we obtain
$\frac{\text{dy}}{\text{dx}}=\Big(\text{x}+\frac{1}{\text{x}}\Big)^\text{x}\Bigg[\frac{\text{x}^2-1}{\text{x}^2+1}+\log\Big(\text{x}+\frac{1}{\text{x}}\Big)\Bigg]+\text{x}^{\big(1+\frac{1}{\text{x}}\big)}\Big(\frac{\text{x+1}-\log\text{x}}{\text{x}^2}\Big)$
View full question & answer→Question 495 Marks
If $\text{y}\sin(\text{x}^\text{x}),$ prove that $\frac{\text{dy}}{\text{dx}}=\cos(\text{x}^\text{x})\times\text{x}^\text{x}(1+\log\text{x})$
AnswerLet $\text{y}=\sin(\text{x}^\text{x})\ .....(\text{i})$
Also, Let $\text{u}=\text{x}^\text{x}\ .....(\text{ii})$
Taking log on both sides,
$\Rightarrow\log\text{u}=\log\text{x}^\text{x}$
$\Rightarrow\log\text{u}=\text{x}\log\text{x}$
Differentiating both sides with respect to x,
$\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x}\log\text{x})$
$\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}\text{x}$
$\Rightarrow\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\text{x}\big(\frac{1}{\text{x}}\big)+\log\text{x}(1)$
$\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=1+\log\text{x}$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{u}(1+\log\text{x})$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{x}^\text{x}(1+\log\text{x})\ .....(\text{iii})$
[Using equation (ii)]
Now, using equation (ii) in equation (i),
$\text{y}=\sin\text{u}$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\sin\text{u})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\cos\text{u}\frac{\text{du}}{\text{dx}}$
Using equation (ii) and (iii),
$\frac{\text{dy}}{\text{dx}}=\cos(\text{x}^\text{x})\times\text{x}^\text{x}(1+\log\text{x})$
View full question & answer→Question 505 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=(\tan\text{x})^{\cot\text{x}}+(\cot\text{x})^{\tan\text{x}}$
AnswerHere,
$\text{y}=(\tan\text{x})^{\cot\text{x}}+(\cot\text{x})^{\tan\text{x}}$
$\text{y}=\text{e}^{\log(\tan\text{x})^{\cot\text{x}}}+\text{e}^{\log(\cot\text{x})^{\tan\text{x}}}$
$\big[\text{Since},\log_\text{e}\text{e}=1,\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
$\text{y}=\text{e}^{\cot\text{x}\log\tan\text{x}}+\text{e}^{\tan\text{x}\log(\cot\text{x})}$
Differentiating it with respect to x using rule and product rule,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\cot\text{x}\log\tan\text{x}}\big)+\frac{\text{d}}{\text{dx}}\big(\text{e}^{\tan\text{x}\log\cot\text{x}}\big)$
$=\text{e}^{\cot\text{x}\log\tan\text{x}}\frac{\text{d}}{\text{dx}}(\cot\text{x}\log\tan\text{x})+\text{e}^{\tan\text{x}\log\cot\text{x}}\frac{\text{d}}{\text{dx}}(\tan\text{x}\log\cot\text{x})$
$=\text{e}^{\log(\tan\text{x})^{\cot\text{x}}}\Big[\cot\text{x}\frac{\text{d}}{\text{dx}}\log\tan\text{x}+\log\tan\text{x}\frac{\text{d}}{\text{dx}}\cot\text{x}\Big] \\ +\text{e}^{\log(\cot\text{x})^{\tan\text{x}}}\Big[ \tan\text{x}\frac{\text{d}}{\text{dx}} \log\cot\text{x}+\log\cot\text{x}\frac{\text{d}}{\text{dx}}(\tan\text{x})\Big]$
$=(\tan\text{x})^{\cot\text{x}}\Big[\cot\text{x}\times\Big(\frac{1}{\tan\text{x}}\Big)\frac{\text{d}}{\text{dx}}(\tan\text{x})+\log\tan\text{x}(-\text{cosec}^2\text{x})\Big] \\ +(\cot\text{x})^{\tan\text{x}}\Big[\tan\text{x}\big(\frac{1}{\cot\text{x}}\big)\frac{\text{d}}{\text{dx}}(\cot\text{x})+\log\cot\text{x}\big(\sec^2\text{x}\big)\Big]$
$=(\tan\text{x})^{\cot\text{x}}\Big[(1)\big(\sec^2\text{x}\big)-\text{cosec}^2\text{x}\log\tan\text{x}\Big] \\ +(\cot)^{\tan\text{x}}\Big[(1)\big(-\text{cosec}^2\text{x}\big)+\sec^2\text{x}\log\cot\text{x}\Big]$
$\frac{\text{dy}}{\text{dx}}=(\tan\text{x})^{\cot\text{x}}\Big[\sec^{2\text{x}}-\text{cosec}^2\text{x}\log\tan\text{x}\Big] \\ +(\cot)^{\tan\text{x}}\Big[\sec^2\text{x}\log\cot\text{x}-\text{cosec}^2\text{x}\Big]$
View full question & answer→Question 515 Marks
Differentiate the following functions with respect to x:
$\cos^{-1}\Big\{\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}}\Big\}$
AnswerLet $\text{y}=\cos^{-1}\Big\{\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}}\Big\}$
Let $\text{x}=\text{a}\cot\theta$
$\Rightarrow\ \text{y}=\cos^{-1}\Big\{\frac{\text{a}\cot\theta}{\sqrt{\text{a}^2\cot^2\theta+\text{a}^2}}\Big\}$
$\Rightarrow\text{y}=\cos^{-1}\Big\{\frac{\text{a}\cot\theta}{\sqrt{\text{a}^2(\cot^2\theta+1)}}\Big\}$
$\Rightarrow\ \text{y}=\sin^{-1}\Big(\frac{\text{a}\cot\theta}{\text{a cosec}\theta}\Big)$
$\Rightarrow\ \text{y}=\cos^{-1}\Bigg(\frac{\frac{\cos\theta}{\sin\theta}}{\frac{1}{\sin\theta}}\Bigg)$
$\Rightarrow\ \text{y}=\cos^{-1}(\cos\theta)$
$\Rightarrow\ \text{y}=\theta$
$\Rightarrow\ \text{y}=\cot^{-1}\big(\frac{\text{x}}{\text{a}}\big)\ \big[\text{Since, x}=\text{a}\cot\theta\big]$
Differentiating it with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{1+\big(\frac{\text{x}}{\text{a}}\big)^2}\frac{\text{d}}{\text{dx}}\big(\frac{\text{x}}{\text{a}}\big)$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{-\text{a}^2}{\text{a}^2+\text{x}^2}\times\big(\frac{1}{\text{a}}\big)$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{-\text{a}}{\text{a}^2+\text{x}^2}$
View full question & answer→Question 525 Marks
If $\sqrt{\text{y}+\text{x}}+\sqrt{\text{y}-\text{x}}=\text{c},$ show that $\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}-\sqrt{\frac{\text{y}^2}{\text{x}^2}-1}$
AnswerHere,
$\sqrt{\text{y}+\text{x}}+\sqrt{\text{y}-\text{x}}=\text{c}$
Differentiating with respect to x,
$\Rightarrow\frac{\text{d}}{\text{dx}}(\sqrt{\text{y}+\text{x}})+\frac{\text{d}}{\text{dx}}\sqrt{\text{y}-\text{x}}=\frac{\text{d}}{\text{dx}}(\text{c})$
$\Rightarrow\frac{1}{2\sqrt{\text{y}+\text{x}}}\frac{\text{d}}{\text{dx}}(\text{y}+\text{x})+\frac{1}{2\sqrt{\text{y}-\text{x}}}\frac{\text{d}}{\text{dx}}(\text{y}-\text{z})=0$
$\Rightarrow \frac{1}{2\sqrt{\text{y}+\text{x}}}\Big(\frac{\text{dy}}{\text{dx}}+1\Big)+\frac{1}{2\sqrt{\text{y}-\text{x}}}\Big(\frac{\text{dy}}{\text{dx}}-1\Big)=0$
$\Rightarrow \frac{\text{dy}}{\text{dx}}\Big(\frac{1}{2\sqrt{\text{y}+\text{x}}}\Big)+\frac{\text{dy}}{\text{dx}}\Big(\frac{1}{2\sqrt{\text{y}-\text{x}}}\Big)=\frac{1}{2\sqrt{\text{y}-\text{x}}}-\frac{1}{2\sqrt{\text{y}+\text{x}}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}\times\Big[\frac{1}{\sqrt{\text{y}+\text{x}}}+\frac{1}{\sqrt{\text{y}-\text{x}}}\Big]=\frac{1}{2}\Big[\frac{\sqrt{\text{y}+\text{x}}-\sqrt{\text{y}-\text{x}}}{\sqrt{\text{y}-\text{x}}\sqrt{\text{y}+\text{x}}}\Big]$
$\Rightarrow \frac{\text{dy}}{\text{dx}}\Big[\frac{\sqrt{\text{y}-\text{x}}-\sqrt{\text{y}+\text{x}}}{\sqrt{\text{y}+\text{x}}\sqrt{\text{y}-\text{x}}}\Big]=\Big[\frac{\sqrt{\text{y}+\text{x}}-\sqrt{\text{y}-\text{x}}}{\sqrt{\text{y}-\text{x}}\sqrt{\text{y}+\text{x}}}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sqrt{\text{y}+\text{x}}-\sqrt{\text{y}-\text{x}}}{\sqrt{\text{y}-\text{x}}+\sqrt{\text{y}-\text{x}}}\times\frac{\sqrt{\text{y}+\text{x}}-\sqrt{\text{y}-\text{x}}}{\sqrt{\text{y}+\text{x}}-\sqrt{\text{y}+\text{x}}}$
[Rationalizing the denominator]
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\frac{(\text{y}+\text{x})+(\text{y}-\text{x})-2\sqrt{\text{y}+\text{x}}\sqrt{\text{y}-\text{x}}}{\text{y}+\text{x}-\text{y}+\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2\text{y}-2\sqrt{\text{y}^2-\text{x}^2}}{2\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2\text{y}}{2\text{x}}-\frac{2\sqrt{\text{y}^2-\text{x}^2}}{2\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}-\sqrt{\frac{\text{y}^2-\text{x}^2}{\text{x}^2}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}-\sqrt{\frac{\text{y}^2}{\text{x}^2}-1}$
View full question & answer→Question 535 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big\{\frac{\text{x}}{\text{a}+\sqrt{\text{a}^2-\text{x}^2}}\Big\},-\text{a}<\text{x}<\text{a}$
AnswerLet $\tan^{-1}\Big\{\frac{\text{x}}{\text{a}+\sqrt{\text{a}^2-\text{x}^2}}\Big\}$
Put $\text{x}=\text{a}\sin\theta,\text{ So}$
$\text{y}=\tan^{-1}\Big\{\frac{\text{a}\sin\theta}{\text{a}+\sqrt{\text{a}^2+\text{a}^2\sin^2\theta}}\Big\}$
$\tan^{-1}\bigg\{\frac{\text{a}\sin\theta}{\text{a}+\sqrt{\text{a}^2(1-\sin^2\theta)}}\bigg\}$
$=\tan^{-1}\Big\{\frac{\text{a}\sin\theta}{\text{a}+\text{a}\cos\theta}\Big\}$
$=\tan^{-1}\Big\{\frac{\text{a}\sin\theta}{\text{a}(1+\cos\theta)}\Big\}$
$=\tan^{-1}\Big(\frac{\sin\theta}{1+\cos\theta}\Big)$
$=\tan^{-1}\bigg(\frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\cos^2\times\frac{\theta}{2}}\bigg)$
$\text{y}=\tan^{-1}\Big(\tan\frac{\theta}{2}\Big)\ .....\text{(i)}$
Here, $-\text{a}<\text{x}<\text{a}$
$\Rightarrow -1<\frac{\text{x}}{\text{a}}<1$
$\Rightarrow -\frac{\pi}{2}<\theta<\frac{\pi}{2}$
$\Rightarrow-\frac{\pi}{4}<\frac{\theta}{2}<\frac{\pi}{4}$
So, from equation (i),
$\text{y}=\frac{\theta}{2}\ \Big[\text{Since}, \tan^{-1}(\tan\theta)=\theta,\text{ if }\theta\in \Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\text{y}=\frac{1}{2}+\sin^{-1}\Big(\frac{\text{x}}{\text{a}}\Big) \big[\text{Since, x}=\text{a}\sin\theta\big]$
Differentiating it with respect to x using chian rule,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2}\times\frac{1}{\sqrt{1-\big(\frac{\text{x}}{\text{a}}\big)^2}}\frac{\text{d}}{\text{dx}}\big(\frac{\text{x}}{\text{a}}\big)$
$=\frac{\text{a}}{2\sqrt{\text{a}^2-\text{x}^2}}\Big(\frac{1}{\text{a}}\Big)$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2\sqrt{\text{a}^2-\text{x}^2}}$
View full question & answer→Question 545 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=\text{x}^{\sin\text{x}}+\big(\sin\text{x}\big)^\text{x}$
AnswerLet $\text{y}=\text{x}^{\sin\text{x}}+(\sin\text{x})^\text{x}$
Also, let $\text{u}=\text{x}^{\sin\text{x}}\text{ and v}=(\sin\text{x})^\text{x}$
$\therefore\text{y}=\text{u}+\text{v}$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{\text{du}}{\text{dx}}+\frac{\text{dv}}{\text{dx}}\ .....(\text{i})$
$\text{u}=\text{x}^{\sin\text{x}}$
$\Rightarrow\log\text{u}=\log\big(\text{x}^{\sin\text{x}}\big)$
$\Rightarrow\log\text{u}=\sin\text{x}\log\text{x}$
Differentiating both sides with respect to x, we obtain
$\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\sin\text{x})\times\log\text{x}+\sin\text{x}\times\frac{\text{d}}{\text{dx}}(\log\text{x})$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{u}\Big[\cot\text{x}\log\text{x}+\sin\text{x}\times\frac{1}{\text{x}}\Big]$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{x}^{\sin\text{x}}\Big[\cos\text{x}\log\text{x}+\frac{\sin\text{x}}{\text{x}}\Big]\ .....(\text{ii})$
$\text{v}=(\sin\text{x})^\text{x}$
$\Rightarrow\log\text{v}=\log(\sin\text{x})^\text{x}$
$\Rightarrow\log\text{v}=\text{x}\log(\sin\text{x})$
Differentiating both sides with respect to x, we obtain
$\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x})\times\log(\sin\text{x})+\text{x}\times\frac{\text{d}}{\text{dx}}\big[\log(\sin\text{x})\big]$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\text{v}\Big[\log(\sin\text{x})+\text{x}\times\frac{1}{\sin\text{x}}\times\frac{\text{d}}{\text{dx}}(\sin\text{x})\Big]$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=(\sin\text{x})^\text{x}\Big[\log\sin\text{x}+\frac{\text{x}}{\sin\text{x}}\cos\text{x}\Big]$
$\Rightarrow\frac{\text{dv}}{\text{dx}}(\sin\text{x})^\text{x}\big[\log\sin\text{x}+\text{x}\cot\text{x}\big]$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=(\sin\text{x})^\text{x}\big[\log\sin\text{x}+\text{x}\cot\text{x}\big]\ .....(\text{iii})$
From (i), (ii) and (iii), we obtain
$\frac{\text{dy}}{\text{dx}}=\text{x}^{\sin\text{x}}\Big(\cos\text{x}\log\text{x}+\frac{\sin\text{x}}{\text{x}}\big)+(\sin\text{x})^\text{x}\big[\log\sin\text{x}+\text{x}\cot\text{x}\big]$
View full question & answer→Question 555 Marks
Differentiate the following functions with respect to x:
$\sin^{-1}\Big(\frac{\text{x}+\sqrt{1-\text{x}^3}}{\sqrt{2}}\Big),-1<\text{x}<1$
AnswerLet $\text{y}=\sin^{-1}\Big(\frac{\text{x}+\sqrt{1-\text{x}^3}}{\sqrt{2}}\Big)$
Put $\text{x}=\text{a}\sin\theta,\text{ So}$
$=\sin^{-1}\Big\{\frac{\sin\theta+\sqrt{1-\sin^2\theta}}{\sqrt{2}}\Big\}$
$=\sin^{-1}\Big\{\frac{\sin\theta+\cos\theta}{\sqrt{2}}\Big\}$
$=\sin^{-1}\Big\{\sin\theta\Big(\frac{1}{\sqrt{2}}\Big)+\cos\theta\Big(\frac{1}{\sqrt{2}}\Big)\Big\}$
$=\sin^{-1}\Big\{\sin\theta\cos\frac{\pi}{4}+\cos\theta\sin\frac{\pi}{4}\Big\}$
$\text{y}=\sin^{-1}\Big\{\sin\Big(\theta+\frac{\pi}{4}\Big)\Big\}\ .....(\text{i})$
Here, $-1<\text{x}<1$
$\Rightarrow\ -1<\sin\theta<1$
$\Rightarrow -\frac{\pi}{2}<\theta<\frac{\pi}{2}$
$\Rightarrow\Big(-\frac{\pi}{2}+\frac{\pi}{4}\Big)<\Big(\frac{\pi}{4}+\theta\Big)<\frac{3\pi}{4}$
So, from equation (i),
$\text{y}=\theta+\frac{\pi}{4}\Big[\text{Since},\sin^{-1}(\sin\theta)=\theta,\text{ as }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\text{y}=\sin^{-1}\text{x}+\frac{\pi}{4} \big[\text{Since},\sin\theta=\text{x}\big]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}+0$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}$
View full question & answer→Question 565 Marks
If $\text{xy}\log(\text{x}+\text{y})=1,$ prove that $\frac{\text{dx}}{\text{dx}}=-\frac{\text{y}(\text{x}^2\text{y}+\text{x}+\text{y})}{\text{x}(\text{xy}^2+\text{x}+\text{y})}$
AnswerHere,
$\text{xy }\log(\text{x}+\text{y})=1$
Differentiating with respect to x, we get
$\Rightarrow\frac{\text{d}}{\text{dx}}\big[\text{xy}\log(\text{x}+\text{y})\big]=\frac{\text{d}}{\text{dx}}(1)$
$\Rightarrow\text{xy}\frac{\text{d}}{\text{dx}}\log(\text{x}+\text{y})+\text{x}\log(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}+\text{y}\log(\text{x}+\text{y})\frac{\text{d}}{\text{dx}}(\text{x})=0$
[Using chain rule and product rule]
$\Rightarrow\text{xy}\times\Big(\frac{1}{\text{x}+\text{y}}\Big)\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})+\text{x}\log(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}+\text{y}\log(\text{x}+\text{y})(1)=0$
$\Rightarrow\Big(\frac{\text{xy}}{\text{x}+\text{y}}\Big)\Big(1+\frac{\text{dy}}{\text{dx}}\Big)+\text{x}\log(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}+\text{y}\log(\text{x}+\text{y})(1)=0$
$\Rightarrow\Big(\frac{\text{xy}}{\text{x}+\text{y}}\Big)\frac{\text{dy}}{\text{dx}}+\Big(\frac{\text{xy}}{\text{x}+\text{y}}\Big)+\text{x}\Big(\frac{1}{\text{xy}}\Big)\frac{\text{dy}}{\text{dx}}+\text{y}\Big(\frac{1}{\text{xy}}\Big)=0$
$\Big[\text{Since from equation (i)}\log(\text{x}+\text{y})=\frac{1}{\text{xy}}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\Big[\frac{\text{xy}}{\text{x}+\text{y}}+\frac{1}{\text{y}}\Big]=-\Big[\frac{1}{\text{x}}+\frac{\text{xy}}{\text{x}+\text{y}}\Big]$
$\frac{\text{dy}}{\text{dx}}\Big[\frac{\text{xy}^2+\text{x}+\text{y}}{(\text{x}+\text{y})\text{y}}\Big]=-\Big[\frac{\text{x}+\text{y}+\text{x}^2\text{y}}{\text{x}(\text{x}+\text{y})}\Big]$
$\frac{\text{dy}}{\text{dx}}=-\Big(\frac{\text{x}+\text{y}+\text{x}^2\text{y}}{\text{x}(\text{x}+\text{y})}\Big)\Big(\frac{(\text{x}+\text{y})\text{y}}{\text{xy}^2+\text{x}+\text{y}}\Big)$
$=-\frac{\text{y}}{\text{x}}\Big(\frac{\text{x}+\text{y}+\text{x}^2\text{y}}{\text{x}+\text{y}+\text{xy}^2}\Big)$
So,
$\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}}{\text{x}}\Big(\frac{\text{x}^2\text{y}+\text{x}+\text{y}}{\text{xy}^2+\text{x}+\text{y}}\Big)$
View full question & answer→Question 575 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{4\text{x}}{1-4\text{x}^2}\Big),-\frac{1}{2}<\text{x}<\frac{1}{2}$
Answer Let $\text{y}=\tan^{-1}\Big\{\frac{4\text{x}}{1-4\text{x}^2}\Big\}$
Put $2\text{x}=\tan\theta,\text{ so}$
$\text{y}=\tan^{-1}\Big\{\frac{2\tan\theta}{1-\tan^2\theta}\Big\}$
$\text{y}=\tan^{-1}\{\tan2\theta\}\ .....(\text{i})$
Here, $-\frac{1}{2}<\text{x}<\frac{1}{2}$
$\Rightarrow -1<2\text{x}<1$
$\Rightarrow -1<\tan\theta<1$
$\Rightarrow -\frac{\pi}{4}<\theta<\frac{\pi}{4}$
$\Rightarrow -\frac{\pi}{2}<(2\theta)<\frac{\pi}{2}$
So, from equation (i),
$\text{y}=\theta\Big[\text{Since}, \tan^{-1}(\tan\theta)=\theta,\text{ if }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\text{y}=2\tan^{-1}(2\text{x})\ \big[\text{Since}, 2\text{x}=\tan\theta\big]$
Differentiating ti with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=2\Big(\frac{1}{1+(2\text{x})^2}\Big)\frac{\text{d}}{\text{dx}}(2\text{x})$
$\frac{\text{dy}}{\text{dx}}=\frac{4}{1+4\text{x}^2}$
View full question & answer→Question 585 Marks
If $\text{x}\sin(\text{a}+\text{y})+\sin\text{a}\cos(\text{a}+\text{y})=0,$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\sin^2(\text{a}+\text{y})}{\sin\text{a}}$
AnswerWe have, $\text{x}\sin(\text{a}+\text{y})+\sin\text{a}\cos(\text{a}+\text{y})=0$
Differentiate with respect to x, we get
$\Rightarrow\frac{\text{d}}{\text{dx}}\big[\text{x}\sin(\text{a}+\text{y})\big]+\frac{\text{d}}{\text{dx}}\big[\sin\text{a}\cos(\text{a}+\text{y})\big]=0$
$\Rightarrow\Big[\text{x}\frac{\text{d}}{\text{dx}}\sin(\text{a}+\text{y})+\sin(\text{a}+\text{y})\frac{\text{d}}{\text{dx}}(\text{x})\Big]+\sin\text{a}\frac{\text{d}}{\text{dx}}\cos(\text{a}+\text{y})=0$
$\Rightarrow\Big[\text{x}\cos(\text{a}+\text{y})\frac{\text{d}}{\text{dx}}(\text{a}+\text{y})+\sin(\text{a}+\text{y})(1)\Big] \\ +\sin\text{a}\Big[-\sin(\text{a}+\text{y})\frac{\text{d}}{\text{dx}}(\text{a}+\text{y})\Big]=0$
$\Rightarrow\text{x}\cos(\text{a}+\text{y})\frac{\text{dy}}{\text{dx}}(\text{a}+\text{y})+\sin(\text{a}+\text{y})-\sin\text{a}\sin(\text{a}+\text{y})\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\big[\text{x}\cos(\text{a}+\text{y})-\sin\text{a}\sin(\text{a}+\text{y})\big] \\ =-\sin(\text{a}+\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\Big[-\sin\text{a}\frac{\cos^2(\text{a}+\text{y})}{\sin(\text{a}+\text{y})}-\sin\text{a}\sin(\text{a}+\text{y})\Big] \\ =-\sin(\text{a}+\text{y})$
$\Big[\because\ \text{x}=-\sin\text{a}\frac{\cos(\text{a}+\text{y})}{\sin(\text{a}+\text{y})}\Big]$
$ \Rightarrow-\frac{\text{dy}}{\text{dx}}\Big[\frac{\sin\text{a}\cos^2(\text{a}+\text{y})+\sin\text{a}\sin^2(\text{a}+\text{y})}{\sin(\text{a}+\text{y})}\Big] \\ =-\sin(\text{a}+\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\sin(\text{a}+\text{y})\Big[\frac{\sin(\text{a}+\text{y})}{\sin\text{a}\{\cos^2(\text{a}+\text{y})+\sin^2(\text{a}+\text{y})\}}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sin^2(\text{a}+\text{y})}{\sin\text{a}}$
View full question & answer→Question 595 Marks
Differentiate the following functions with respect to x:
$\sin^{-1}\Big\{\frac{\sqrt{1+\text{x}}+\sqrt{1-\text{x}}}{2}\Big\},0<\text{x}<1$
AnswerLet $\text{y}=\sin^{-1}\Big\{\frac{\sqrt{1+\text{x}}+\sqrt{1-\text{x}}}{2}\Big\}$
Put $\text{x}=\cos2\theta,\text{ So}$
$=\sin^{-1}\Big\{\frac{\sqrt{1+\cos2\theta}+\sqrt{1-\cos2\theta}}{2}\Big\}$
$=\sin^{-1}\Big\{\frac{\sqrt{2\cos^2\theta}+\sqrt{2\sin^2\theta}}{2}\Big\}$
$=\sin^{-1}\Big\{\frac{\sqrt{2}\cos\theta+\sqrt{2}\sin\theta}{2}\Big\}$
$=\sin^{-1}\Big\{\cos\theta\Big(\frac{1}{\sqrt{2}}\Big)+\Big(\frac{1}{\sqrt{2}}\Big)\sin\theta\Big\}$
$=\sin^{-1}\Big\{\cos\theta\sin\Big(\frac{\pi}{4}\Big)+\cos\frac{\pi}{4}\sin\theta\Big\}$
$\text{y}=\sin^{-1}\Big\{\sin\big(\theta+\frac{\pi}{4}\Big)\Big\}\ .....(\text{i})$
Here, $0<\text{x}<1$
$\Rightarrow 0<\cos2\theta<1$
$\Rightarrow 0<2\theta<\frac{\pi}{2}$
$\Rightarrow 0 < \theta < \frac{\pi}{4}$
$\Rightarrow \frac{\pi}{4}<\Big(\theta+\frac{\pi}{4}\Big)<\frac{\pi}{2}$
So from eqaution (i),
$\text{y}=\theta+\frac{\pi}{4}\ \Big[\text{Since}, \sin^{-1}(\sin\theta)=\theta,\text{ if }\theta\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\Big]$
$\text{y}=\frac{1}{2}\cos^{-1}\text{x}+\frac{\pi}{4}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2}\Big(\frac{-1}{\sqrt{1-\text{x}^2}}\Big)+0$
$\frac{\text{dy}}{\text{dx}}=\frac{-1}{2\sqrt{1-\text{x}^2}}$
View full question & answer→Question 605 Marks
Find $\frac{\text{dy}}{\text{dx}},$ when
$\text{x}=\frac{2\text{t}}{1+\text{t}^2}\text{ and y}=\frac{1-\text{t}^2}{1+\text{t}^2}$
AnswerHere, $\text{x}=\frac{2\text{t}}{1+\text{t}^{2}}$
Differentiating it with respect to t using quotient rule,
$\frac{\text{dy}}{\text{dx}}=\bigg[\frac{(1+\text{t}^{2})\frac{\text{d}}{\text{dt}}(2\text{t})-2\text{t}\frac{\text{d}}{\text{dt}}(1+\text{t}^{2})}{(1+\text{t}^{2})^{2}}\bigg]$
$=\Big[\frac{(1+\text{t}^{2})(2)-2\text{t}(2\text{t})}{(1+\text{t}^{2})^{2}}\Big]$
$=\Big[\frac{2+2\text{t}^{2}-4\text{t}^{2}}{(1+\text{t}^{2})^{2}}\Big]$
$=\Big[\frac{2-2\text{t}^2}{(1+\text{t}^2)}\Big]$
$\frac{\text{dx}}{\text{dt}}=\frac{2(1-\text{t}^{2})}{(1+\text{t}^{2})^{2}}...(\text{i})$
And, $\text{y}=\frac{2(1-\text{t}^{2})}{(1+\text{t}^{2})^{2}}$
Differentiating it with respect to t using quotient rule,
$\frac{\text{dy}}{\text{dt}}=\bigg[\frac{(1+\text{t}^{2})\frac{\text{d}}{\text{dt}}(1-\text{t}^{2})-(1-\text{t}^{2})\frac{\text{d}}{\text{dt}}(1+\text{t}^{2})}{(1+\text{t}^{2})^{2}}\bigg]$
$=\Big[\frac{(1+\text{t}^{2})(-2\text{t})-(1-\text{t})^{2}(2\text{t})}{(1+\text{t}^{2})^{2}}\Big]$
$=\Big[\frac{-2\text{t}-2\text{t}^{3}-2\text{t}+2\text{t}^{3}}{(1+\text{t}^{2})^{2}}\Big]$
$\frac{\text{dy}}{\text{dx}}=\Big[\frac{-4\text{t}}{(1+\text{t}^{2})^{2}}\Big]...(\text{ii})$
Dividing equation (ii) by (i),
$\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{-4\text{t}}{(1-\text{t}^{2})^{2}}\times\frac{(1+\text{t}^{2})^{2}}{2(1+\text{t}^{2})}$
$=\frac{-2\text{t}}{(1-\text{t}^{2})}$
$\frac{\text{dy}}{\text{dx}}=-\frac{\text{x}}{\text{y}} $
$\Big[\text{Since}\frac{\text{x}}{\text{y}}=\frac{2\text{t}}{1+\text{t}^{2}}\times\frac{1+\text{t}^{2}}{1-\text{t}^{2}}=\frac{2\text{t}}{1-\text{t}^{2}}\Big]$
View full question & answer→Question 615 Marks
If $\text{y}=\log\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}+\frac{2}{\sqrt{3}}\tan^{-1}\Big(\frac{\sqrt{3}\text{x}}{1-\text{x}^2}\Big),$ find $\frac{\text{dy}}{\text{dx}}$
AnswerHere,
$\text{y}=\log\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}+\frac{2}{\sqrt{3}}\tan^{-1}\Big(\frac{\sqrt{3}\text{x}}{1-\text{x}^2}\Big)$
Differentiating it with respect to x using chain rule and quotient rule,
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\log\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}\Big)+\frac{2}{\sqrt{3}}\frac{\text{d}}{\text{dx}}\tan^{-1}\Big(\frac{\sqrt{3}\text{x}}{1-\text{x}^2}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}\Big)}\frac{\text{d}}{\text{dx}}{\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}\Big)}+\frac{2}{\sqrt{3}}\Bigg\{\frac{1}{1+\Big(\frac{\sqrt{3}\text{x}}{1-\text{x}^2}\Big)}\Bigg\}\frac{\text{d}}{\text{dx}}\Big(\frac{\sqrt{3}\text{x}}{1-\text{x}^2}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}\Big)\bigg(\frac{(\text{x}^2-\text{x}+1)\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{x}+1)-(\text{x}^2+\text{x}+1)\frac{\text{d}}{\text{dx}}(\text{x}^2-\text{x}+1)}{(\text{x}^2-\text{x}+1)^2}\bigg) \\ +\frac{2}{\sqrt{3}}\Big\{\frac{(1-\text{x})^2}{1+\text{x}^4-2\text{x}^2+3\text{x}^2}\Big\}\bigg\{\frac{(1-\text{x}^2)^2\frac{\text{d}}{\text{dx}}(\sqrt{3\text{x}})-\sqrt{3}\text{x}\frac{\text{d}}{\text{dx}}(1-\text{x})^2}{(1-\text{x}^2)^2}\bigg\}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\Big(\frac{1}{\text{x}^2-\text{x}+1}\Big)\Big(\frac{(\text{x}^2-\text{x}+1)(2\text{x}+1)-(\text{x}^2+\text{x}+1)(2\text{x}-1)}{(\text{x}^2-\text{x}+1)}\Big) \\ +\frac{2}{\sqrt{3}}\Big(\frac{(1-\text{x}^2)^2}{1+\text{x}^2+\text{x}^4}\Big)\Big(\frac{(1-\text{x}^2)(\sqrt{3})-\sqrt{3}\text{x}(-2\text{x})}{(1-\text{x}^2)^2}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\Big(\frac{2\text{x}^3-2\text{x}^2+2\text{x}+\text{x}^2-\text{x}+1-2\text{x}^3-2\text{x}^2-2\text{x}+\text{x}^2+\text{x}+1}{\text{x}^4+2\text{x}^2+1-\text{x}^2}\Big) \\ +\frac{2}{\sqrt{3}}\Big(\frac{\sqrt{3}-\sqrt{3}\text{x}^2\sqrt{3}\text{x}^2}{1+\text{x}^2+\text{x}^4}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\Big(\frac{-2\text{x}^2+2}{\text{x}^4+\text{x}^2+1}\Big)+\frac{2\sqrt{3}(\text{x}^2+1)}{\sqrt{3}(1+\text{x}^2+\text{x}^4)}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2(1-\text{x}^2)}{(\text{x}^4+\text{x}^2+1)}+\frac{2(\text{x}^2+1)}{1+\text{x}^2+\text{x}^4}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2(1-\text{x}^2+\text{x}^2+1)}{1+\text{x}^2+\text{x}^4}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{4}{1+\text{x}^2+\text{x}^4}$
View full question & answer→Question 625 Marks
Differentiate the following functions from first principles:
$\sin^{-1}(2\text{x}+3)$
AnswerLet $\text{f(x)}=\sin^{-1}(2\text{x}+3)$
$\Rightarrow\ \text{f}(\text{x}+\text{h})=\sin^{-1}(2(\text{x}+\text{h})+3)$
$\Rightarrow\ \text{f}(\text{x}+\text{h})=\sin^{-1}(2\text{x}+2\text{h}+3)$
$\therefore \frac{\text{d}}{\text{dx}}\{\text{f(x)}\}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f(x)}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\sin^{-1}(2\text{x}+2\text{h}+3)-\sin^{-1}(2\text{x}+3)}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\sin^{-1}\Big[(2\text{x}+2\text{h}+3)\sqrt{1+(2\text{x}+3)^2}-(2\text{x}+3)\sqrt{1-(2\text{x}+2\text{h}+3)^2}\Big]}{\text{h}}$
$\Big[\text{Since}, \sin^{-1}\text{x}-\sin^{-1}\text{y}=\sin^{-1}\big[\text{x}\sqrt{1-\text{y}^2}-\text{y}\sqrt{1-\text{x}^2}\big]\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\sin^{-1}\text{z}}{\text{z}}\times\frac{\text{z}}{\text{h}}$
Where, $\text{z}=(2\text{x}+2\text{h}+3)\sqrt{1-(2\text{x}+3)^2}-(2\text{x}+3)\sqrt{1-(2\text{x}+2\text{h}+3)^2}$
$\text{and }\lim\limits_{\text{h}\rightarrow0}\frac{\sin^{-1}\text{h}}{\text{h}}=1$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{z}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(2\text{x}+2\text{h}+3)\sqrt{1-(2\text{x}+3)^2}-(2\text{x}+3)\sqrt{1-(2\text{x}+2\text{h}+3)^2}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(2\text{x}+2\text{h}+3)^2-(2\text{x}+3)^2-(2\text{x}+3)^2\big(1-(2\text{x}+2\text{h}+3)^2\big)}{\big\{(2\text{x}+2\text{h}+3)\sqrt{1-(2\text{x}+3)^2}-(2\text{x}+3)\sqrt{1-(2\text{x}+2\text{h}+3)^2}\big\}}$
[Since, rationalizing numerator]
$=\lim\limits_{\text{h}\rightarrow0}\frac{\big[(2\text{x}+3)^2+4\text{h}^2+4\text{h}(2\text{x}+3)\big]\big(1-(2\text{x}+3)^2\big)-(2\text{x}+3)^2\big[1-(2\text{x}+3)^2-4\text{h}^2-4\text{h}(2\text{x}+3)\big]}{\text{h}\big\{(2\text{x}+2\text{h}+3)\sqrt{1-(2\text{x}+3)^2}+(2\text{x}+3)\sqrt{1-(2\text{x}+2\text{h}+3)^2}\big\}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\Big[(2\text{x}+3)^2+4\text{h}^2+4\text{h}(2\text{x}+3)-(2\text{x}+3)^4-4\text{h}^2(2\text{x}+3)^2-4\text{h}(2\text{x}+3)^3 -(2\text{x}+3)^2+(2\text{x}+3)^3+4\text{h}^2(2\text{x}+3)^2+4\text{h}(2\text{x}+3)^3\Big]}{\text{h}\big\{(2\text{x}+2\text{h}+3)\sqrt{1-(2\text{x}+3)^2}+(2\text{x}+3)\sqrt{1-(2\text{x}+2\text{h}+3)^2}\big\}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{4\text{h}\big[\text{h}+(2\text{x}+3)\big]}{\text{h}\Big\{(2\text{x}+2\text{h}+3)\sqrt{1-(2\text{x}+3)^2}+(2\text{x}+3)\sqrt{1-(2\text{x}+2\text{h}+3)^2}\Big\}}$
$=\frac{4\text{h}\big[\text{h}+(2\text{x}+3)\big]}{(2\text{x}+3)\sqrt{1-(2\text{x}+3)^2}+(2\text{x}+3)\sqrt{1-(2\text{x}+3)^2}}$
$=\frac{4(2\text{x}+3)}{2(2\text{x}+3)\sqrt{1-(2\text{x}+3)^2}}$
$=\frac{2}{\sqrt{1-(2\text{x}+3)^2}}$
So,
$\frac{\text{d}}{\text{dx}}\big(\sin^{-1}(2\text{x}+3)\big)=\frac{2}{\sqrt{1-(2\text{x}+3)^2}}$
View full question & answer→Question 635 Marks
Differentiate $\tan^{-1}\Big(\frac{\text{x}-1}{\text{x}+1}\Big)$ with respect to $\sin^{-1}\big(3\text{x}-4\text{x}^3\big),$ if $-\frac{1}{2}<\text{x}<\frac{1}{2}$
AnswerLet, $\text{u}=\tan^{-1}\Big(\frac{\text{x}-1}{\text{x}+1}\Big)$Put $\text{x}=\tan\theta$
$\Rightarrow \text{u}=\tan^{-1}\Big(\frac{\tan\theta-1}{\tan\theta+1}\Big)$
$\Rightarrow \text{u}=\tan^{-1}\bigg(\frac{\tan\theta-\tan\frac{\pi}{4}}{1+\tan\theta\tan\frac{\pi}{4}}\bigg)$
$\Rightarrow\text{u}\tan^{-1}\Big[\tan\big(\theta-\frac{\pi}{4}\big)\Big]\ .....(\text{i})$
Here, $-\frac{1}{2}<\text{x}<\frac{1}{2}$
$\Rightarrow-\frac{1}2{}<\tan\theta<\frac{1}{2}$
$\Rightarrow-\tan^{-1}\big(\frac{1}{2}\big)<\theta<\tan^{-1}\big(\frac{1}{2}\big)$
So, from equation (i)
$\text{u}=\theta-\frac{\pi}{4}$
$\Big[\text{Since,}\tan^{-1}(\tan\theta)=\theta,\text{if }\theta\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\Big]$
$\Rightarrow\text{u}=\tan^{-1}\text{x}-\frac{\pi}{4} \ [\text{Since,x}= \tan\theta]$
Differentiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=\frac{1}{1+\text{x}^2}-0$
$\Rightarrow \frac{\text{du}}{\text{dx}}=\frac{1}{1+\text{x}^2}\ ..... \text{(ii)}$
And,
Let, $\text{v}=\sin^{-1}(3\text{x}-4\text{x}^3)$
Put $\text{x}=\sin\theta$
$\Rightarrow\text{v}=\sin^{-1}(3\sin\theta-4\sin^3\theta)$
$\Rightarrow\text{v}=\sin^{-1}(\sin3\theta)\ .....\text{(iii)}$
Now, $-\frac{1}{2}<\text{x}<\frac{1}{2}$
$\Rightarrow-\frac{1}{2}<\sin\theta<\frac{1}{2}$
$\Rightarrow-\frac{1}{6}<\theta<\frac{\pi}{6}$
So, from equation (iii),
$\text{v}=3\theta\Big[\text{Since,}\sin^{-1}(\sin\theta)=\theta,\text{if }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\Rightarrow\text{v}=3\sin^{-1}\text{x}[\text{Since, x}=\sin\theta]$
Dirrerentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{3}{\sqrt{1-\text{x}^2}}\ .....\text{(iv)}$
Dividing equation (iii) by (iv),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{1}{1+\text{x}^2}\times\frac{\sqrt{1-\text{x}^2}}{3}$
$\therefore\frac{\text{du}}{\text{dv}}=\frac{\sqrt{1-\text{x}^2}}{3(1+\text{x}^2)}$
View full question & answer→Question 645 Marks
Differentiate $\sin^{-1}\Big(4\text{x}\sqrt{1-4\text{x}^2}\Big)$ with respect to $\sqrt{1-4\text{x}^2},$ if:
$\text{x}\in\Big(-\frac{1}{2\sqrt{2}},\frac{1}{\sqrt{2\sqrt{2}}}\Big)$
AnswerLet $\text{u}=\sin^{-1}\Big(4\text{x}\sqrt{1-4\text{x}^2}\Big)$
Put $2\text{x}=\cos\theta$
$\Rightarrow\text{u}=\sin^{-1}\Big(2\times\cos\theta\sqrt{1-\cos^2\theta}\Big)$
$\Rightarrow\text{u}=\sin^{-1}(2\cos\theta\sin\theta)$
$\Rightarrow\text{u}=\sin^{-1}(\sin2\theta)\ .....(\text{i})$
Let, $\text{v}=\sqrt{1-4\text{x}^2}\ .....(\text{ii})$
Here,
$\text{x}\in\Big(-\frac{1}{2\sqrt{2}},\frac{1}{2\sqrt{2}}\Big)$
$\Rightarrow2\text{x}\in\Big(-\frac{1}{\sqrt{2}}.\frac{1}{\sqrt{2}}\Big)$
$\Rightarrow\theta\in\Big(\frac{\pi}{4},\frac{3\pi}{4}\Big)$
So, from equation (i),
$\text{u}=\pi=2\theta$
$\Big[\text{Since},\sin^{-1}(\sin\theta)=\pi-\theta,\text{ if }\theta\in\Big(\frac{\pi}{2},\pi\Big)\Big]$
$\Rightarrow\text{u}=\pi-2\cos^{-1}(2\text{x})\big[\text{Since},2\text{x}=\cos\theta\big]$
Differentiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=0-2\bigg(\frac{-1}{\sqrt{1-(2\text{x})^2}}\bigg)\frac{\text{d}}{\text{dx}}(2\text{x})$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\frac{2}{\sqrt{1-4\text{x}^2}}(2)$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\frac{4}{\sqrt{1-4\text{x}^2}}\ .....(\text{iii})$
From equation (ii)
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{-4\text{x}}{\sqrt{1-4\text{x}^2}}$
But, $\text{x}\in\Big(-\frac{1}{2},\frac{1}{2\sqrt{2}}\Big)$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{-4(-\text{x})}{\sqrt{1-4(-\text{x})^2}}$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{4\text{x}}{\sqrt{1-4\text{x}^2}}\ .....(\text{iv})$
Diferentiating equation (ii) with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{1}{2\sqrt{1-4\text{x}^2}}\frac{\text{d}}{\text{dx}}(1-4\text{x}^2)$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{1}{2\sqrt{1-4\text{x}^2}}(-8\text{x})$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{-4\text{x}}{\sqrt{1-4\text{x}^2}}\ .....(\text{v})$
Dividing equation (iii) by (v)
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{4}{\sqrt{1-4\text{x}^2}}\times\frac{\sqrt{1-4\text{x}^2}}{-4\text{x}}$
$\therefore\frac{\text{du}}{\text{dv}}=-\frac{1}{\text{x}}$
View full question & answer→Question 655 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=\text{x}^{\log\text{x}}+(\log\text{x}^\text{x})$
AnswerLet $\text{y}=\text{x}^{\log\text{x}}+(\log\text{x}^\text{x})$
Also, let $\text{u}=(\log\text{x})^\text{x}\text{ and v}=\text{x}^{\log\text{x}}$
$\therefore\ \text{y}=\text{v}+\text{u}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{dv}}{\text{dx}}+\frac{\text{du}}{\text{dx}}\ .....(\text{i})$
Now, $\text{u}=(\log\text{x})^\text{x}$
$\Rightarrow\log\text{u}=\log\big[(\log\text{x})^\text{x}\big]$
$\Rightarrow\log\text{u}=\text{x}\log(\log\text{x})$
Differentiating both sides with respect to x,
$\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\log(\log\text{x})\frac{\text{d}}{\text{dx}}(\text{x})+\text{x}\frac{\text{d}}{\text{dx}}\big[\log(\log\text{x})\big]$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{u}\Big[\log(\log\text{x})+\text{x}\frac{1}{\log\text{x}}\frac{\text{d}}{\text{dx}}(\log\text{x})\Big]$
$\Rightarrow\frac{\text{du}}{\text{dx}}=(\log\text{x})^\text{x}\Big[\log(\log\text{x})+\frac{\text{x}}{\log\text{x}}\times\frac{1}{\text{x}}\Big]$
$\Rightarrow\frac{\text{du}}{\text{dx}}=(\log\text{x})^\text{x}\Big[\log(\log\text{x})+\frac{1}{\log\text{x}}\Big]\ .....(\text{ii})$
Also, $\text{v}=\text{x}^{\log\text{x}}$
$\Rightarrow\log\text{v}=\log\text{x}^{\log\text{x}}$
$\Rightarrow\log\text{v}=\log\text{x}\log\text{x}=(\log\text{x})^2$
Differentiating both sides with respect to x,
$\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[(\log)^2\big]$
$\Rightarrow\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=2(\log\text{x})\frac{\text{d}}{\text{dx}}(\log\text{x})$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=2\text{v}(\log\text{x})\frac{1}{\text{x}}$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=2\text{x}^{\log\text{x}}\frac{\log\text{x}}{\text{x}}$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=2\text{x}^{\log\text{x}}\frac{\log\text{x}}{\text{x}}\ .....(\text{iii})$
From (i), (ii) and (iii), we obtain
$\frac{\text{dy}}{\text{dx}}=2\text{x}^{\log\text{x}}\frac{\log\text{x}}{\text{x}}+(\log\text{x})^\text{x}\Big[\log(\log\text{x})+\frac{1}{\log\text{x}}\Big]$
View full question & answer→Question 665 Marks
If $\text{x}=\frac{\sin^3\text{t}}{\sqrt{\cos^2\text{t}}},\text{y}=\frac{\cos^3\text{t}}{\sqrt{\cos2\text{t}}},$ find $\frac{\text{dy}}{\text{dx}}$
AnswerWe have, $\text{x}=\frac{\sin^{3}\text{t}}{\sqrt{\cos2\text{t}}}$ and $\text{y}=\frac{\cos^{3}\text{t}}{\sqrt{\cos2\text{t}}}$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\Big[\frac{\sin^{3}\text{t}}{\sqrt{\cos2\text{t}}}\Big] $
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{\sqrt{\cos2\text{t}}\frac{\text{d}}{\text{dt}}(\sin^{3})-\sin^{3}\text{t}\frac{\text{d}}{\text{dt}}\sqrt{\cos2\text{t}}}{\cos2\text{t}}$
[Using quotient rule]
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{\sqrt{\cos2\text{t}}(3\sin^{3}\text{t})\frac{\text{d}}{\text{dt}}(\sin\text{t})-\sin^{3}\times\frac{1}{2\sqrt{\cos2\text{t}}}\frac{\text{d}}{\text{dt}}(\cos2\text{t})}{\cos2\text{t}}$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{3\sqrt{\cos2\text{t}}(\sin^{2}\text{t}\cos\text{t})-\frac{\sin^{3}\text{t}}{2\sqrt{\cos2\text{t}}}(-2\sin2\text{t})}{\cos2\text{t}}$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{3\cos2\text{t}\sin^{\text{2}}\text{t}\cos\text{t}+\sin^{3}\text{t}\sin2\text{t}}{\cos2\text{t}\sqrt{\cos2\text{t}}}$
Now, $\frac{\text{dy}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\Big[\frac{\cos^{3}}{\sqrt{\cos2\text{t}}}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{\sqrt{\cos2\text{t}} \frac{\text{d}}{\text{dt}}(\cos^{3}\text{t})-\cos^{3}\text{t}\frac{\text{d}}{\text{dt}}\sqrt{\cos2\text{t}}}{\cos2\text{t}}$
[Using quotient rule]
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{\sqrt{\cos2\text{t}}(3\cos^{2}\text{t})\frac{\text{d}}{\text{dt}}(\cos\text{t})-\cos^{3}\text{t}\times\frac{1}{2\sqrt{\cos2\text{t}}}\frac{\text{d}}{\text{dt}}(\cos2\text{t})}{\cos2\text{t}}$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{3\sqrt{\cos2\text{t}}\cos^{2}\text{t}-(\sin\text{t})-\frac{\cos^{3}\text{t}}{2\sqrt{\cos2\text{t}}}(-2\sin2\text{t})}{\cos2\text{t}}$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{-3\cos2\text{t}\cos^{2}\text{t}+\cos^{3}\text{t}\sin2\text{t}}{\cos2\text{t}\sqrt{\cos2\text{t}}}$
$\therefore\frac{\text{dy}}{\text{dt}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{-3\cos2\text{t}\cos^{2}\text{t}\sin\text{t}+\cos^{3}\text{t}\sin2\text{t}}{\cos2\text{t}\sqrt{\cos2\text{t}}}\times\frac{\cos2\text{t}\sqrt{\cos2\text{t}}}{3\cos2\text{t}\sin^{2}\text{t}\cos\text{t}+\sin^{3}\text{t}\sin2\text{t}}$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{\sin\text{t}\cos{\text{t}}[-3\cos2\text{t}\cos\text{t}+2\cos^{3}\text{t}]}{\sin\text{t}\cos\text{t}[3\cos2\text{t}\sin{\text{t}}+2\sin^{3}\text{t}]}$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{[-3(2\cos^{2}\text{t}-1)\cos\text{t}+2\cos^{3}\text{t}]}{[3(1-2\sin^{2}\text{t})\sin\text{t}+2\sin^{3}\text{t}]}$
$\begin{bmatrix} \cos2\text{t}=2\cos^2\text{t}-1 \\ \cos2\text{t}=1-2\sin^2\text{t} \end{bmatrix}$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{-4\cos^{3}\text{t}+3\cos\text{t}}{3\sin\text{t}-4\sin^{3}\text{t}}$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{-\cos3\text{t}}{\sin3\text{t}}$
$\begin{bmatrix} \cos3\text{t}=4\cos^3\text{t}-3\cos\text{t} \\ \sin3\text{t}=3\sin^2\text{t}-4\sin^3\text{t} \end{bmatrix}$
$\therefore\frac{\text{dy}}{\text{dx}}=-\cos3\text{t}$
View full question & answer→Question 675 Marks
Differentiate the following functions with respect to x:
$\sin^2\{\log(2\text{x}+3)\}$
AnswerConsider $\text{y}=\sin^2\{\log(2\text{x}+3)\}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[\sin^2\{\log(2\text{x}+3)\}\big]$
$=2\sin\big(\log(2\text{x}+3)\big)\frac{\text{d}}{\text{dx}}\sin\big(\log(2\text{x}+3)\big)$
[Using chain rule]
$=2\sin\big(\log(2\text{x}+3)\big)\cos\big(\log(2\text{x}+3)\big)\frac{\text{d}}{\text{dx}}(2\text{x}+3)$
$=\sin\big(2\log(2\text{x}+3)\big)\times\frac{1}{(2\text{x}+3)}\frac{\text{d}}{\text{dx}}(2\text{x}+3)$
$\big[\text{Since}, 2\sin\text{A}\cos\text{A}=\sin^2\text{A}\big]$
$=\sin\big(2\log(2\text{x}+3)\big)\times\frac{2}{(2\text{x}+3)}$
Hence, the solution is $\frac{\text{d}}{\text{dx}}\big(\sin^2\log(2\text{x}+3)\big)=\sin\big(2\log(2\text{x}+3)\big)\times\frac{2}{(2\text{x}+3)}$
View full question & answer→Question 685 Marks
Differentiate the following functions with respect to x:
$\sin(\text{x}^\text{x})$
AnswerLet $\text{y}=\sin(\text{x}^\text{x})\ .....(\text{i})$
Taking log on both sides,
$\log(\sin^{-1}\text{y})=\log\text{x}^\text{x}$
$\Rightarrow\ \log(\sin^{-1}\text{y})=\text{x}\log\text{x}$
Differentiating with respect to x,
$\Rightarrow\frac{1}{\sin^{-1}\text{y}}\frac{\text{dy}}{\text{dy}}(\sin^{-1}\text{y})=\text{x}\frac{\text{d}}{\text{dx}}\log\text{x}+\log\text{x}\frac{\text{d}}{\text{dx}}\text{x}$
$\Rightarrow\frac{1}{\sin^{-1}\text{y}}\times\Big(\frac{1}{\sqrt{1-\text{y}^2}}\Big)\frac{\text{dy}}{\text{dx}}=\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\sin^{-1}\text{y}\sqrt{1-\text{y}^2}(1+\log\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\sin^{-1}(\sin\text{x}^\text{x})\sqrt{1-(\sin\text{x}^\text{x})^2}(1+\log\text{x})$
$\therefore\frac{\text{dy}}{\text{dx}}=\text{x}^\text{x}\cos\text{x}^\text{x}(1+\log\text{x})$
[Using equation (i)]
View full question & answer→Question 695 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=\text{x}^{\text{x}}+\text{x}^\frac{1}{\text{x}}$
AnswerHere,
$\text{y}=\text{x}^{\text{x}}+\text{x}^\frac{1}{\text{x}}$
$=\text{e}^{\log\text{x}^\text{x}}+\text{e}^{\log\text{x}^\frac{1}{\text{x}}}$
$\text{y}=\text{e}^{\text{x}\log\text{x}}+\text{e}^{\big(\frac{1}{\text{x}}\log\text{x}\big)}$
$\big[\text{Since, e}^{\log\text{a}}=\text{a},\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating it with respect to x using chain rule and product rule,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}\log\text{x}}\big)+\frac{\text{d}}{\text{dx}}\Big(\text{e}^{\frac{1}{\text{x}}\log\text{x}}\Big)$
$=\text{e}^{\text{x}\log\text{x}}+\frac{\text{d}}{\text{dx}}(\text{x}\log\text{x})+\text{e}^{\frac{1}{\text{x}}\log\text{x}}\frac{\text{d}}{\text{dx}}\Big(\frac{1}{\text{x}}\log\text{x}\Big)$
$=\text{e}^{\text{x}\log\text{x}}\Big[\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big] \\ +\text{e}^{\log\text{x}^\frac{1}{\text{x}}}\Big[\frac{1}{\text{x}}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}\big(\frac{1}{\text{x}}\big)\Big]$
$=\text{x}^{\text{x}}\Big[\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}(1)\Big] \\ +\text{x}^\frac{1}{\text{x}}\Big[\Big(\frac{1}{\text{x}}\Big)\Big(\frac{1}{\text{x}}\Big)+\log\text{x}\Big(-\frac{1}{\text{x}^2}\Big)\Big]$
$=\text{x}^\text{x}[1+\log\text{x}]+\text{x}^{\frac{1}{\text{x}}}\Big(\frac{1}{\text{x}^2}-\frac{1}{\text{x}^2}\log\text{x}\Big)$
$\frac{\text{dy}}{\text{dx}}=\text{x}^\text{x}[1+\log\text{x}]+\text{x}^{\frac{1}{\text{x}}}\frac{(1-\log\text{x})}{\text{x}^2}$
View full question & answer→Question 705 Marks
If $\text{y}=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)+\sec^{-1}\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big),\text{x}>0,$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{4}{1+\text{x}^2}$
AnswerHere, $\text{y}=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)+\sec^{-1}\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)$
$\Rightarrow \text{y}=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)+\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$
Put $\text{x}=\tan\theta$
$\therefore \text{y}=\tan^{-1}\Big(\frac{2\tan\theta}{1-\tan^2\theta}\Big)+\cos^{-1}\Big(\frac{1-\tan^{2}\theta}{1+\tan^{2}\theta}\Big)$
$\Rightarrow \text{y}=\tan^{-1}(\tan2\theta)+\cos^{-1}(\cos2\theta)$
$\Rightarrow \text{y}=2\theta+2\theta$
$\Rightarrow \text{y}=4\theta$
$\Rightarrow \text{y}=4\tan^{-1} \text{x} \big[\text{using, x}=\tan\theta\big]$
Differentiate it with respect to x,
$\therefore \frac{\text{dy}}{\text{dx}}=\frac{4}{1+\text{x}^2}$
View full question & answer→Question 715 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=(\sin\text{x})^{\cos\text{x}}+(\cos\text{x})^{\sin\text{x}}$
AnswerWe have, $\text{y}=(\sin\text{x})^{\cos\text{x}}+(\cos\text{x})^{\sin\text{x}}$
$\Rightarrow\text{y}=\text{e}^{\log(\sin\text{x})^{\cos\text{x}}}+\text{e}^{\log(\cos\text{x})^{\sin\text{x}}}$
$\Rightarrow\text{y}=\text{e}^{\cos\text{x}\log\sin\text{x}}+\text{e}^{\sin\text{x}\log\cos\text{x}}$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\cos\text{x}\log\sin\text{x}}\big)+\frac{\text{d}}{\text{dx}}\big(\text{e}^{\sin\text{x}\log\cos\text{x}}\big)$
$=\text{e}^{\cos\text{x}\log\sin\text{x}}\frac{\text{d}}{\text{dx}}\big(\cos\text{x}\log\sin\text{x})+\text{e}^{\sin\text{x}\log\cos\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x}\log\cos\text{x})$
$=\text{e}^{\log(\sin\text{x})^{\cos\text{x}}}\Big[\cos\text{x}\frac{\text{d}}{\text{dx}}\log\sin\text{x}+\log\sin\text{x}\frac{\text{d}}{\text{dx}}(\cos\text{x})\Big] \\ +\text{e}^{\log(\cos\text{x})^{\sin\text{x}}}\Big[\sin\text{x}\frac{\text{d}}{\text{dx}}\log\cos\text{x}+\log\cos\text{x}\frac{\text{d}}{\text{dx}}(\sin\text{x})\Big]$
$=(\sin\text{x})^{\cos\text{x}}\Big[\cos\text{x}\frac{1}{\sin\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}\times(-\sin\text{x})\Big] \\ +(\cos\text{x})^{\sin\text{x}}\big[\sin\text{x}\frac{1}{\cos\text{x}}\frac{\text{d}}{\text{dx}}(\cos\text{x})+\log\cos\text{x}\times(\cos\text{x})\Big]$
$=(\sin\text{x})^{\cos\text{x}}\big[\cot\text{x}\cos\text{x}-\sin\text{x}\log\sin\text{x}\big] \\ +(\cos\text{x})^{\sin\text{x}}\big[\tan\text{x}(-\sin\text{x})+\cos\text{x}\log\cos\text{x}\big]$
$=(\sin\text{x})^{\cos\text{x}}\big[\cot\text{x}\cos\text{x}-\sin\text{x}\log\sin\text{x}\big] \\ +(\cos\text{x})^{\sin\text{x}}\big[\cos\text{x}\log\cos\text{x}-\sin\text{x}\tan\text{x}\big]$
View full question & answer→Question 725 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{\text{x}-\text{a}}{\text{x}+\text{a}}\Big)$
AnswerLet $\text{y}=\tan^{-1}\Big(\frac{\text{x}-\text{a}}{\text{x}+\text{a}}\Big)$
$=\tan^{-1}\bigg(\frac{\frac{\text{x}-\text{a}}{\text{x}}}{\frac{\text{x}+\text{a}}{\text{x}}}\bigg)$
$=\tan^{-1}\bigg(\frac{\frac{\text{x}}{\text{x}}-\frac{\text{x}}{\text{x}}}{\frac{\text{x}}{\text{x}}+\frac{\text{a}}{\text{x}}}\bigg)$
$=\tan^{-1}\bigg(\frac{1-\frac{\text{x}}{\text{x}}}{1+1\times\frac{\text{a}}{\text{x}}}\bigg)$
$\text{y}=\tan^{-1}(1)-\tan^{-1}\big(\frac{\text{a}}{\text{x}}\big)$
Differentiating it with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=0-\frac{1}{1+\big(\frac{\text{a}}{\text{x}}\big)^2}\frac{\text{d}}{\text{dx}}\big(\frac{\text{a}}{\text{x}}\big)$
$=-\frac{\text{x}^2}{\text{x}^2+\text{a}^2}\Big(\frac{-\text{a}}{\text{x}^2}\Big)$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{a}}{\text{a}^2+\text{x}^2}$
View full question & answer→Question 735 Marks
Differentiate the following functions with respect to x:
$(\text{x}^\text{x})\sqrt{\text{x}}$
AnswerLet $\text{y}=(\text{x}^\text{x})\sqrt{\text{x}}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=\log(\text{x}^\text{x}\sqrt{\text{x}})$
$\log\text{y}=\text{x}\log\text{x}+\frac{1}{2}\log\text{x}$
Differentiating it with respect to x,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x})+\frac{1}{2}\frac{\text{d}}{\text{dx}}(\log\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}(1)+\frac{1}{2}\Big(\frac{1}{\text{x}}\Big)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=1+\log\text{x}+\frac{1}{2\text{x}}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\text{y}\Big[1+\log\text{x}+\frac{1}{2\text{x}}\Big]$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\text{x}^\text{x}\sqrt{\text{x}}\Big[1+\log\text{x}+\frac{1}{2\text{x}}\Big]$
[Using equation (i)]
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}^{\text{x}+\frac{1}{2}}\Big[\Big(\frac{2\text{x}+1}{2\text{x}}\Big)+\log\text{x}\Big]$
View full question & answer→Question 745 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{\cos\text{x}+\sin\text{x}}{\cos\text{x}-\sin\text{x}}\Big), \frac{\pi}{4}<\text{x}<\frac{\pi}{4}$
AnswerLet $\text{y}=\tan^{-1}\Big[\frac{\cos\text{x}+\sin\text{x}}{\cos\text{x}-\sin\text{x}}\Big]$
$=\tan\bigg[\frac{\frac{\cos\text{x}+\sin\text{x}}{\cos\text{x}}}{\frac{\cos\text{x}-\sin\text{x}}{\cos\text{x}}}\bigg]$
$=\tan^{-1}\bigg[\frac{\frac{\cos\text{x}}{\cos\text{x}}+\frac{\sin\text{x}}{\cos\text{x}}}{\frac{\cos\text{x}}{\cos\text{x}}-\frac{\sin\text{x}}{\cos\text{x}}}\bigg]$
$=\tan^{-1}\Big[\frac{1+\tan\text{x}}{1-\tan\text{x}}\Big]$
$=\tan^{-1}\bigg[\frac{\frac{\tan\pi}{4}+\tan\text{x}}{1-\frac{\tan\pi}{4}\tan\text{x}}\bigg]$
$=\tan^{-1}\Big[\tan\Big(\frac{\pi}{4}+\text{x}\Big)\Big]$
$\text{y}=\frac{\pi}{4}+\text{x}$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=0+1$
$\frac{\text{dy}}{\text{dx}}=1$
View full question & answer→Question 755 Marks
If $\cos\text{y}=\text{x}\cos(\text{a}+\text{y}),$ where $\cos\text{a}\neq\pm1,$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\cos^2(\text{a}+\text{y})}{\sin\text{a}}$
Answerconsider the given function,
$\cos\text{y}=\text{x}\cos(\text{a}+\text{y}),$ where $\cos\text{a}\neq\pm1$
Differentiating both sides w.r.t. 'x' we get
$-\sin\text{y}\frac{\text{dy}}{\text{dx}}=\text{x}\Big(-\sin(\text{a}+\text{y})\frac{\text{dy}}{\text{dx}}\Big)+\cos(\text{a}+\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\big[\text{x}\sin(\text{a}+\text{y})-\sin\text{y}\big]=\cos(\text{a}+\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\cos(\text{a}+\text{y})}{\text{x}\sin(\text{a}+\text{y})-\sin\text{y}}$
Multiplying the numerator and the denominator
by $\cos(\text{a}+\text{y})$ on the R.H.S., we have,
$\frac{\text{dy}}{\text{dx}}=\frac{\cos^2(\text{a}+\text{y})}{\text{x}\cos(\text{a}+\text{y})\sin(\text{a}+\text{y})-\cos(\text{a}+\text{y})\sin\text{y}}$
$=\frac{\cos^2(\text{a}+\text{y})}{\text{x}\cos(\text{a}+\text{y})\sin(\text{a}+\text{y})-\cos(\text{a}+\text{y})\sin\text{y}}$
$\big[\because\cos\text{y}=\text{x}\cos(\text{a}+\text{y}),\text{given function}\big]$
$=\frac{\cos^2(\text{a}+\text{y})}{\sin\big[(\text{a}+\text{y})-\text{y}\big]}=\frac{\cos^2(\text{a}+\text{y})}{\sin\text{a}}$
View full question & answer→Question 765 Marks
Differentiate the following functions with respect to x:
$\log\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^3-\text{x}+1}\Big)$
AnswerLet, $\text{y}=\log\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^3-\text{x}+1}\Big)$
Differentiate with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big[\log\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^3-\text{x}+1}\Big)\Big]$
$=\frac{1}{\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}\Big)}\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}\Big)$
[Using chain rule and quotient rule]
$=\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^3-\text{x}+1}\Big)\bigg[\frac{\text{x}^2-\text{x}+1\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{x}+1)-(\text{x}^2+\text{x}+1)\frac{\text{d}}{\text{dx}}(\text{x}^2-\text{x}+1)}{(\text{x}^2-\text{x}+1)^2}\bigg]$
$=\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^3-\text{x}+1}\Big)\bigg[\frac{(\text{x}^2-\text{x}+1)(2\text{x}+1)-(\text{x}^2+\text{x}+1)(2\text{x}-1)}{(\text{x}^2-\text{x}+1)^2}\bigg]$
$=\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^3-\text{x}+1}\Big)\bigg[\frac{2\text{x}^3-2\text{x}^2+2\text{x}+\text{x}^3-\text{x}+1-2\text{x}^3-2\text{x}^2-2\text{x}+\text{x}^2+\text{x}+1}{(\text{x}^2-\text{x}+1)^2}\bigg]$
$=\frac{-4\text{x}^2+2\text{x}^3+2}{(\text{x}^2+\text{x}+1)(\text{x}^2+\text{x}+1)}$
$=\frac{-4\text{x}^2+2\text{x}^3+2}{(\text{x}^2+1)^2-(\text{x})^2}$
$=\frac{-2(\text{x}^2-1)}{\text{x}^4+1+2\text{x}^2-\text{x}^2}$
$=\frac{-2(\text{x}^2-1)}{\text{x}^4+\text{x}^2+1}$
So,
$\frac{\text{d}}{\text{dx}}\Big\{\log\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}\Big)\Big\}=\frac{-2(\text{x}^2-1)}{\text{x}^4+\text{x}^2+1}$
View full question & answer→Question 775 Marks
Differentiate the following functions with respect to x:
$\sin^{-1}\Big\{\frac{\sin\text{x}+\cos\text{x}}{\sqrt{2}}\Big\},-\frac{3\pi}{4}<\text{x}<\frac{\pi}{4}$
AnswerLet $\text{y}=\sin^{-1}\Big\{\frac{\sin\text{x}+\cos\text{x}}{\sqrt{2}}\Big\}$
$=\sin^{-1}\bigg\{\sin\text{x}\Big(\frac{1}{\sqrt{2}}\Big)+\cos\text{x}\Big(\frac{1}{\sqrt{2}}\Big)\bigg\}$
$=\sin^{-1}\Big\{\sin{\text{x}}\cos\frac{\pi}{4}+\cos\text{x}\times\sin\frac{\pi}{4}\Big\}$
$\text{y}=\sin^{-1}\Big\{\sin\Big(\text{x}+\frac{\pi}{4}\Big)\Big\}$
Here, $\frac{-3\pi}{4}<\text{x}<\frac{\pi}{4}$
$\Rightarrow\Big(\frac{-3\pi}{4}+\frac{\pi}{4}\Big)$
$\Big[\text{Since},\sin^{-1}(\sin\theta)=\theta, \text{ if }\theta\in\Big[\frac{-\pi}{2},\frac{\pi}{2}\Big]\Big]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=1+0$
$\frac{\text{dy}}{\text{dx}}=1$
View full question & answer→Question 785 Marks
Find the derivative of the function f(x) given by $f(x) = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8)$ and hence find $f(1).$
AnswerHere,
$f(x) = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8)$
Differentiating with respect to x using product rule and chain rule,
$\Rightarrow\text{f}'\text{(x)}=(1+\text{x})\big(1+\text{x}^2\big) \frac{\text{d}}{\text{dx}}\big(1+\text{x}^8\big)+(1+\text{x})\big(1+\text{x}^2\big)\big(1+\text{x}^8\big) \\ \frac{\text{d}}{\text{dx}}\big(1+\text{x}^4\big)+(1+\text{x})\big(1+\text{x}^4\big)\big(1+\text{x}^8\big) \\ \frac{\text{d}}{\text{dx}}\big(1+\text{x}^2\big)+\big(1+\text{x}^2\big)\big(1+\text{x}^4\big)\big(1+\text{x}^8\big) \frac{\text{d}}{\text{dx}}(1+\text{x})$
$\Rightarrow\text{f}'\text{(x)}=(1+\text{x})\big(1+\text{x}^2\big)\big(1+\text{x}^4\big)8\text{x}^7+(1+\text{x})\big(1+\text{x}^2\big)\big(1+\text{x}^2)\big(1+\text{x}^8\big)\big(4\text{x}^3\big) \\ +(1+\text{x})\big(1+\text{x}^4\big)\big(1+\text{x}^8\big)(2\text{x})+\big(1+\text{x}^2\big)\big(1+\text{x}^4\big)\big(1+\text{x}^8\big)(1)$
$f'(1)=(1+1)(1+1)(8) + (1+1)(1+1)(1+1)(4) + (1+1)(1+1)(1+1)(2) + (1+1)(1+1)(1+1)$
$f'(1) = (2)(2)(2)(8) + (2)(2)(2)(4) + (2)(2)(2)(2) + (2)(2)(2)$
$= 64 + 32 + 16 + 8$
$=120$
So,
$f'(1) = 120$
View full question & answer→Question 795 Marks
If $x^{16}y^9 = (x + y)^{17}, $ prove that $\text{x}\frac{\text{dy}}{\text{dx}}=2\text{y}$
AnswerHere,$x^{16}y^9 = (x + y)^{20}$
Taking log on both the siede,
$\log(\text{x}^{16}\times\text{y}^{19})=\log(\text{x}^2+\text{y})^{17}$
$\big[\text{since}, \log(\text{AB})=\log\text{A}+\log\text{B},\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
$16\log\text{x}+9\log\text{y}=17\log(\text{x}^2+\text{y})$
Differentiating it with respect to x using chain rule
$16\frac{\text{d}}{\text{dx}}(\log\text{x})+9\frac{\text{d}}{\text{dx}}(\log\text{y})=17\frac{\text{d}}{\text{dx}}\log(\text{x}^2+\text{y})$
$\frac{16}{\text{x}}+\frac{9}{\text{y}}\frac{\text{dy}}{\text{dx}}=17\frac{1}{(\text{x}^2+\text{y})}\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{y})$
$\frac{16}{\text{x}}+\frac{9}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{17}{\text{x}^2+\text{y}}\Big[2\text{x}+\frac{\text{dy}}{\text{dx}}\Big]$
$\frac{9}{\text{y}}\frac{\text{dy}}{\text{dx}}-\frac{17}{(\text{x}^2+\text{y})}\frac{\text{dy}}{\text{dx}}=\Big(\frac{34\text{x}}{\text{x}^2+\text{y}}\Big)-\frac{16}{\text{x}}$
$\frac{\text{dy}}{\text{dx}}\Big[\frac{9}{\text{y}}-\frac{17}{(\text{x}^2+\text{y})}\Big]=\frac{34\text{x}^2-16\text{x}^2-16\text{y}}{\text{x}(\text{x}^2+\text{y})}$
$\frac{\text{dy}}{\text{dx}}\Big[\frac{9\text{x}^2+9\text{y}-17\text{y}}{\text{y}(\text{x}^2+\text{y})}\Big]=\frac{18\text{x}^2-16\text{y}}{\text{x}(\text{x}^2+\text{y})}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}\Big(\frac{2(9\text{x}^2-8\text{y})}{9\text{x}^2-8\text{y}}\Big)$
$\frac{\text{dy}}{\text{dx}}=\frac{2\text{y}}{\text{x}}$
$\text{x}\frac{\text{dy}}{\text{dx}}=2\text{y}$
View full question & answer→Question 805 Marks
If $\text{x}=10(\text{t}-\sin\text{t}),\text{y}=12(1-\cos\text{t}),$ find $\frac{\text{dy}}{\text{dx}}.$
AnswerHere, $\text{x}=10(\text{t}-\sin\text{t}),\text{y}=12(1-\cos\text{t})$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=10(1-\cos\text{t})\ ...(\text{i})$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=12(\sin\text{t})\ ...(\text{ii})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{12(\sin\text{t})}{10(1-\cos\text{t})}$ From equation (i) and (ii)
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{12\sin\frac{\text{t}}{2}\cdot\cos\frac{\text{t}}{2}}{10\sin^2\frac{\text{t}}{2}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{6}{5}\cot\frac{\text{t}}{2}$
View full question & answer→Question 815 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=(\sin\text{x})^\text{x}+\sin^{-1}\sqrt{\text{x}}$
AnswerHere,
$\text{y}=(\sin\text{x})^{\text{x}}+\sin^{-1}\sqrt{\text{x}}$
$=\text{e}^{\log(\sin\text{x})^\text{x}}+\sin^{-1}\sqrt{\text{x}}$
$\text{y}=\text{e}^{\text{x}\log\sin\text{x}}+\sin^{-1}\sqrt{\text{x}}$
$\big[\text{Since},\log_\text{e}^\text{e}=1,\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentitating it with respect to x using chain rule and product rule,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}\log\sin\text{x}}\big)+\frac{\text{d}}{\text{dx}}\sin^{-1}\big(\sqrt{\text{x}}\big)$
$=\text{e}^{\text{x}\log\sin\text{x}}\frac{\text{d}}{\text{dx}}(\text{x}\log\sin\text{x})+\frac{1}{\sqrt{1-\big(\sqrt{\text{x}}\big)^2}}\frac{\text{d}}{\text{dx}}\big(\sqrt{\text{x}}\big)$
$=\text{e}^{\log(\sin\text{x})^\text{x}}\Big[\text{x}\frac{\text{d}}{\text{dx}}\log\sin\text{x}+\log\sin\text{x}\frac{\text{d}}{\text{dx}}(\text{x})+\frac{1}{\sqrt{1-\text{x}}}\times\frac{1}{2\sqrt{\text{x}}}\Big]$
$=(\sin\text{x})^\text{x}\Big[\text{x}\times\frac{1}{\sin\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}(1)\Big]+\frac{1}{2\sqrt{\text{x}-\text{x}^2}}$
$=(\sin\text{x})^\text{x}\Big[\frac{\text{x}}{\sin\text{x}}(\cos\text{x})+\log\sin\text{x}\Big]+\frac{1}{2\sqrt{\text{x}-\text{x}^2}}$
$\frac{\text{dy}}{\text{dx}}=(\sin\text{x})^\text{x}\Big[\text{x}\cot\text{x}+\log\sin\text{x}\Big]+\frac{1}{2\sqrt{\text{x}-\text{x}^2}}$
View full question & answer→Question 825 Marks
Differentiate the following functions with respect to x:
$\log\Big(\frac{\sin\text{x}}{1+\cos\text{x}}\Big)$
AnswerLet, $\log\Big(\frac{\sin\text{x}}{1+\cos\text{x}}\Big)$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\log\Big(\frac{\sin\text{x}}{1+\cos\text{x}}\Big)$
$=\frac{1}{\Big(\frac{\sin\text{x}}{1+\cos\text{x}}\Big)}\times\frac{\text{d}}{\text{dx}}\Big(\frac{\sin\text{x}}{1+\cos\text{x}}\Big)$
[Using chain rule]
$=\Big(\frac{1+\cos\text{x}}{\sin\text{x}}\Big)\Bigg[\frac{(1+\cos\text{x})\frac{\text{d}}{\text{dx}}(\sin\text{x})-\sin\text{x}\frac{\text{d}}{\text{dx}}(1+\cos\text{x})}{(1+\cos\text{x})^2}\Bigg]$
[Using quotient rule]
$=\frac{(1+\cos\text{x})}{\sin\text{x}}\bigg[\frac{(1+\cos\text{x})(\cos\text{x})-\sin\text{x}(-\sin\text{x})}{(1+\cos\text{x})^2}\bigg]$
$=\frac{(1+\cos\text{x})}{\sin\text{x}}\Big[\frac{\cos\text{x}+\cos^2\text{x}+\sin^2\text{x}}{(1+\cos\text{x})^2}\Big]$
$=\frac{(1+\cos\text{x})}{\sin\text{x}}\Big[\frac{(1+\cos\text{x})}{(1+\cos\text{x})^2}\Big]$
$=\frac{1}{\sin\text{x}}$
$=\text{cosec x}$
So,
$\frac{\text{d}}{\text{dx}}\Big(\log\Big(\frac{\sin\text{x}}{1+\cos\text{x}}\Big)\Big)=\text{cosec x}$
View full question & answer→Question 835 Marks
Differentiate the following functions from first principles:
$\log\text{cosec x}$
AnswerLet $\text{f(x)}=\log\text{cosec x}$
$\Rightarrow\ \text{f}(\text{x}+\text{h})=\log\text{cosec x}$
$\therefore \frac{\text{d}}{\text{dx}}\{\text{f(x)}\}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f(x)}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log\text{cosec}(\text{x}+\text{h})-\log\text{cosec x}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log\Big\{\frac{\text{cosec}(\text{x}+\text{h})}{\text{cosec x}}\Big\}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log\Big\{1+\Big(\frac{\sin\text{x}}{\sin(\text{x}+\text{h})}-1\Big)\Big\}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\begin{Bmatrix}\frac{\log\Big\{1+\Big(\frac{\sin\text{x}-\sin(\text{x}+\text{h})}{\sin(\text{x}+\text{h})}\Big)\Big\}}{\Big\{\frac{\sin\text{x}-\sin(\text{x}+\text{h})}{\sin(\text{x}+\text{h})}\Big\}} \end{Bmatrix}\frac{\Big\{\frac{\sin\text{x}-\sin(\text{x}+\text{h})}{\sin(\text{x}+\text{h})}\Big\}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{2\cos\Big(\frac{\text{x}+\text{x}+\text{h}}{2}\Big)\sin\Big(\frac{\text{x}-\text{x}-\text{h}}{2}\Big)}{\sin(\text{x}+\text{h})\text{h}}$
$\Big[\because\ \lim\limits_{\text{x}\rightarrow0}\frac{\log(1+\text{x})}{\text{x}}=1\text{ and }\sin\text{A}-\sin\text{B} \\ =2\cos\Big(\frac{\text{A}+\text{B}}{2}\Big)\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{2\cos\Big(\frac{2\text{x}+\text{h}}{2}\Big)}{\sin(\text{x}+\text{h})(-2)}\bigg\{\frac{\sin\big(-\frac{\text{h}}{2}\big)}{-\frac{\text{h}}{2}}\bigg\}$
$=-\cot\text{x}$
$\therefore\ \frac{\text{d}}{\text{dx}}(\log\text{cosec x})=-\cot\text{x}$
View full question & answer→Question 845 Marks
If $\text{y}=(\text{x}-1)\log(\text{x}-1)-(\text{x}+1)\log(\text{x}+1)$ prove that $\frac{\text{dy}}{\text{dx}}=\log\Big(\frac{\text{x}-1}{1+\text{x}}\Big)$
AnswerWe have, $\text{y}=(\text{x}-1)\log(\text{x}-1)-(\text{x}+1)\log(\text{x}+1)$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[(\text{x}-1)\log(\text{x}-1)-(\text{x}+1)\log(\text{x}+1)\big]$
$=\Big[(\text{x}-1)\frac{\text{d}}{\text{dx}}\log(\text{x}-1)+\log(\text{x}-1)\frac{\text{d}}{\text{dx}}(\text{x}-1)\Big] \\ -\Big[(\text{x}+1)\frac{\text{d}}{\text{dx}}\log(\text{x}+1)+\log(\text{x}+1)\frac{\text{d}}{\text{dx}}(\text{x}+1)\Big]$
$=\Big[(\text{x}-1)\times\frac{1}{(\text{x}-1)}\frac{\text{d}}{\text{dx}}(\text{x}-1)+\log(\text{x}-1)\times(1)\Big] \\ -\Big[(\text{x}+1)\times\frac{1}{(\text{x}+1)}\times\frac{\text{d}}{\text{dx}}(\text{x}+1)+\log(\text{x}+1)(1)\Big]$
$=\big[1+\log(\text{x}-1)\big]-\big[1+\log(\text{x}+1)\big]$
$=\log(\text{x}-1)-\log(\text{x}+1)$
$=\log\frac{(\text{x}-1)}{(\text{x}+1)}$
So,
$\frac{\text{dy}}{\text{dx}}=\log\frac{(\text{x}-1)}{(\text{x}+1)}$
View full question & answer→Question 855 Marks
Differentiate the following functions with respect to x:
$(\log\text{x})^{\log\text{x}}$
AnswerLet $\text{y}=(\log\text{x})^{\log\text{x}}\ .....(\text{i})$
Taking logarithm on both the sides, we obtain
$\log\text{y}=\log\text{x}.\log(\log\text{x})$
Differentiating both sides with resepect to x, we obtain
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[\log\text{x}.\log(\log\text{x})\big]$
$\Rightarrow\ \frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\log(\log\text{x}).\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}.\frac{\text{d}}{\text{dx}}[\log(\log\text{x})]$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\text{y}\Big[\log(\log\text{x}).\frac{1}{\text{x}}+\log\text{x}.\frac{1}{\log\text{x}}.\frac{\text{d}}{\text{dx}}(\log\text{x})\Big]$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{1}{\text{x}}\log(\log\text{x})+\frac{1}{\text{x}}\Big]$
$\therefore\ \frac{\text{dy}}{\text{dx}}=(\log\text{x})^{\log\text{x}}\Big[\frac{1}{\text{x}}+\frac{\log(\log\text{x})}{\text{x}}\Big]$
View full question & answer→Question 865 Marks
If $(\sin\text{x})^{\text{y}}=(\cos\text{y})^{\text{x}},$ Prove that $\frac{\text{dy}}{\text{dx}}=\frac{\log\cos\text{y}-\text{y}\cot\text{x}}{\log\sin\text{x}+\text{x}\tan\text{y}}$
AnswerWe have, $(\sin\text{x})^{\text{y}}=(\cos\text{y})^{\text{x}}$
Taking log on both sides,
$\log(\sin\text{x})^\text{y}=\log(\cos\text{y})^{\text{x}}$
$\Rightarrow\text{y}\log(\sin\text{x})=\text{x}\log(\cos\text{y})$
Differentiating with respect to x,
$\frac{\text{d}}{\text{dx}}\big[\text{y}\log\sin\text{x}\big]=\frac{\text{d}}{\text{dx}}\big[\text{x}\log\cos\text{y}\big]$
$\Rightarrow\text{y}\frac{\text{d}}{\text{dx}}(\log\sin\text{x})+\log\sin\text{x}\frac{\text{dy}}{\text{dx}} \\ =\text{x}\frac{\text{dy}}{\text{dx}}(\log\cos\text{y})+\log\cos\text{y}\frac{\text{d}}{\text{dx}}(\text{x})$
$\Rightarrow\text{y}\Big(\frac{1}{\sin\text{x}}\Big)\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}\frac{\text{dy}}{\text{dx}} \\ =\frac{\text{x}}{\cos\text{y}}\frac{\text{x}}{\text{dx}}(\cos\text{y})+\log\cos\text{y}(1)$
$\Rightarrow\frac{\text{y}}{\sin\text{x}}(\cos\text{x})+\log\sin\text{x}\frac{\text{dy}}{\text{dx}} \\ =\frac{\text{x}}{\cos\text{y}}(-\sin\text{y})\frac{\text{dy}}{\text{dx}}+\log\cos\text{y}$
$\Rightarrow \text{y}\cot\text{x}+\log\sin\text{x}\frac{\text{dy}}{\text{dx}} \\ =-\text{x}\tan\text{y}\frac{\text{dy}}{\text{dx}}+\log\cos\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}(\log\sin\text{x}+\text{x}\tan\text{y}) \\ =\log\cos\text{y}-\text{y}\cot\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\log\cos\text{y}-\text{y}\cot\text{x}}{\log\sin\text{x}+\text{x}\tan\text{y}}$
View full question & answer→Question 875 Marks
Differentiate the following functions with respect to x:
$(\sin^{-1}\text{x})^\text{x}$
AnswerLet $\text{y}=(\sin^{-1}\text{x})^\text{x}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=\log(\sin^{-1}\text{x})^\text{x}$
$\log\text{y}=\text{x}\log(\sin^{-1}\text{x})\ \big[\text{Since},\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating it with respect to x using product rule and chain rule,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}(\log\sin^{-1}\text{x})+\log\sin(-1)\times\frac{\text{d}}{\text{dx}}(\text{x})$
$=\text{x}\times\frac{1}{\sin^{-1}\text{x}}\frac{\text{d}}{\text{dx}}\big(\sin^{-1}\text{x}\big)+\log\sin^{-1}\text{x}(1)$
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\sin^{-1}\text{x}}\Big(\frac{1}{\sqrt{1-\text{x}^2}}\Big)+\log\sin^{-1}\text{x}$
$\frac{\text{dy}}{\text{dx}}=\text{y}\bigg[\log\sin^{-1}\text{x}+\frac{\text{x}}{\sin^{-1}\text{x}\big(\sqrt{1-\text{x}^2}\big)}\bigg]$
$\frac{\text{dy}}{\text{dx}}=(\sin^{-1}\text{x})^2\Big[\log\sin^{-1}\text{x}+\frac{\text{x}}{\sin^{-1}\text{x}\sqrt{1-\text{x}^2}}\Big]$
[Using equation (i)]
View full question & answer→Question 885 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{\sqrt{1+\text{a}^2\text{x}^2-1}}{\text{ax}}\Big),\text{x}\neq0$
AnswerLet $\text{y}=\tan^{-1}\Big(\frac{\sqrt{1+\text{a}^2\text{x}^2}-1}{\text{ax}}\Big)$
Put $\text{ax}=\tan\theta$
$\text{y}=\tan^{-1}\Big(\frac{\sqrt{1+\text{a}^2\text{x}^2}-1}{\text{ax}}\Big)$
$=\tan^{-1}\Big(\frac{\sec\theta-1}{\tan\theta}\Big)$
$=\tan^{-1}\Big(\frac{1-\cos\theta}{\sin\theta}\Big)$
$=\tan^{-1}\bigg(\frac{\frac{2\sin^2\theta}{2}}{\frac{2\sin\theta}{2}\frac{\cos\theta}{2}}\bigg)$
$\text{y}=\tan^{-1}\Big(\frac{\tan\theta}{2}\Big)$
$=\frac{\theta}{2}$
$\text{y}=\frac{1}{2}\tan^{-1}(\text{ax})$
Differentiating it with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2}\times\Big(\frac{1}{1+(\text{ax})^2}\Big)\frac{\text{d}}{\text{dx}}(\text{ax})$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2(1+\text{a}^2\text{x}^2)}(\text{a})$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{a}}{2(1+\text{a}^2\text{x}^2)}$
View full question & answer→Question 895 Marks
Differentiate the following functions with respect to x:
$\log\sqrt{\frac{\text{x}-1}{\text{x}+1}}$
AnswerLet $\text{y}=\log\sqrt{\frac{\text{x}-1}{\text{x}+1}}$
$\Rightarrow\text{y}=\log\Big(\frac{\text{x}-1}{\text{x}+1}\Big)^\frac{1}{2}$
$\Rightarrow\text{y}=\frac{1}{2}\log\Big(\frac{\text{x}-1}{\text{x}+1}\Big)$
$\Rightarrow\text{y}=\frac{1}{2}\big[\log(\text{x}-1)-\log(\text{x}+1)\big]$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2}\Big[\frac{\text{d}}{\text{dx}}\big\{\log(\text{x}-1)\big\}-\frac{\text{d}}{\text{dx}}\big\{\log(\text{x}+1)\big\}\Big]$
$=\frac{1}{2}\Big(\frac{1}{\text{x}-1}-\frac{1}{\text{x}+1}\Big)$
$=\frac{1}{2}\Big(\frac{2}{\text{x}^2-1}\Big)$
$=\frac{2}{\text{x}^2-1}$
So,
$\frac{\text{dy}}{\text{dx}}=\frac{2}{\text{x}^2-1}$
View full question & answer→Question 905 Marks
Differentiate the following functions from first principles:
$x^2e^x.$
AnswerLet $f(x) = x^2e^x$
$\Rightarrow f(x + h) = (x + h)^2 e^{(x+h)}$
$\therefore \frac{\text{d}}{\text{dx}}\{\text{f(x)}\}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f(x)}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(\text{x}+\text{h})^2\text{e}^{(\text{x}+\text{h})}-\text{x}^2\text{e}^{\text{x}}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\text{x}^2\text{e}^{(\text{x}+\text{h})}-\text{x}^2\text{e}^\text{x}}{\text{h}}+\frac{2\text{xhe}^{(\text{x}+\text{h})}}{\text{h}}+\frac{\text{h}^2\text{e}^{(\text{x}+\text{h})}}{\text{h}}\Big)$
$=\lim\limits_{\text{x}\rightarrow0}\bigg(\frac{\text{x}^2\text{e}^\text{x}\big(\text{e}^{(\text{x}+\text{h}-\text{x})}-1\big)}{\text{x}}+2\text{xe}^{(\text{x}+\text{h})}+\text{he}^{(\text{x}+\text{h})}\bigg)$
$=\lim\limits_{\text{h}\rightarrow0}\bigg[\text{x}^2\text{e}^{\text{x}}\frac{\big(\text{e}^\text{h}-1\big)}{\text{h}}+2\text{xe}^{(\text{x}+\text{h})}+\text{h}^{\text{e}}(\text{x}+\text{h})\bigg]$
$=\text{x}^2\text{e}^\text{x}+2\text{xe}^\text{x}+0\times\text{e}^\text{x}\ \Big[\text{Since,}\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{x}-1}{\text{x}}=1\Big]$
So,
$\frac{\text{d}}{\text{dx}}(\text{x}^2\text{e}^\text{x})=\text{e}^\text{x}(\text{x}^2+2\text{x})$
View full question & answer→Question 915 Marks
If $\log\sqrt{\text{x}^2+\text{y}^2}=\tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big),$ prove that $\frac{\text{dx}}{\text{dx}}=\frac{\text{x}+\text{y}}{\text{x}-\text{y}}$
AnswerHere,
$\log\sqrt{\text{x}^2+\text{y}^2}=\tan^{-1}\Big(\frac{\text{x}}{\text{y}}\Big)$
$\Rightarrow\log(\text{x}^2+\text{y}^2)^{\frac{1}{2}}=\tan^{-1}\Big(\frac{\text{x}}{\text{y}}\Big)$
$\Rightarrow\frac{1}{2}\log(\text{x}^2+\text{y}^2)=\tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big)$
Differentiating with respect to x,
$\Rightarrow\frac{1}{2}\frac{\text{d}}{\text{dx}}\log(\text{x}^2+\text{y}^2)=\frac{\text{d}}{\text{dx}}\tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big)$
$\Rightarrow\frac{1}{2}\times\Big(\frac{1}{\text{x}^2+\text{y}^2}\Big)\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{y}^2)=\frac{1}{1+\Big(\frac{\text{y}}{\text{x}}\Big)^2}\frac{\text{d}}{\text{dx}}\Big(\frac{\text{y}}{\text{x}}\Big)$
$\Rightarrow\frac{1}{2}\Big(\frac{1}{\text{x}^2+\text{y}^2}\Big)\Big[2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}\Big]=\frac{\text{x}^2}{(\text{x}^2+\text{y}^2)}\bigg[\frac{\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}\frac{\text{d}}{\text{dx}}(\text{x})}{\text{x}^2}\bigg]$
$ \Rightarrow\frac{1}{2}\Big(\frac{1}{\text{x}^2+\text{y}^2}\Big)\times2\Big(\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}\Big)=\frac{\text{x}^2}{(\text{x}^2+\text{y}^2)}\bigg[\frac{\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}(1)}{\text{x}^2}\bigg]$
$\Rightarrow\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}$
$\Rightarrow\text{y}\frac{\text{dy}}{\text{dx}}-\text{x}\frac{\text{dy}}{\text{dx}}=-\text{y}-\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}(\text{y}-\text{x})=-(\text{y}+\text{x})$
View full question & answer→Question 925 Marks
Differentiate the following functions with respect to x:
$(\cos\text{x})^\text{x}+(\sin\text{x})^\frac{1}{\text{x}}$
AnswerLet $\text{y}=(\cos\text{x})^\text{x}+(\sin\text{x})^\frac{1}{\text{x}}$
$\Rightarrow\text{y}=\text{e}^{\log(\cos\text{x})^\text{x}}+\text{e}^{\log(\sin\text{x})^\frac{1}{\text{x}}}$
$\Rightarrow\text{y}=\text{e}^{\text{x}\log(\cos\text{x})}+\text{e}^{\frac{1}{\text{x}}\log\sin\text{x}}$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}\log\cos\text{x}}\big)+\frac{\text{d}}{\text{dx}}\big(\text{e}^{\frac{1}{\text{x}}\log\sin\text{x}}\big)$
$=\text{e}^{\log\cos\text{x}}\times\frac{\text{d}}{\text{dx}}(\text{x}\log\cos\text{x})+\text{e}^{\frac{1}{\text{x}}\log\sin}\frac{\text{d}}{\text{dx}}\big(\frac{1}{\text{x}}\log\sin\text{x}\big)$
$=\text{e}^{\log(\cos\text{x})^\text{x}}\times\Big[\text{x}\frac{\text{d}}{\text{dx}}(\log\cos\text{x})+\log\cos\text{x}\times\frac{\text{d}}{\text{dx}}(\text{x})\Big] \\ +\text{e}^{\log(\sin\text{x})^\frac{1}{\text{x}}}\times\Big[\frac{1}{\text{x}}\frac{\text{d}}{\text{dx}}(\log\sin\text{x})+\log\sin\text{x}\frac{\text{d}}{\text{dx}}\big(\frac{1}{\text{x}}\big)\Big]$
$=(\cos\text{x})^\text{x}\Big[\text{x}\big(\frac{1}{\cos\text{x}}\big)\frac{\text{d}}{\text{dx}}(\cos\text{x})+\log\cos\text{x}+\log\cos\text{x}(1)\Big] \\ +(\sin)^\frac{1}{\text{x}}\Big[\frac{1}{\text{x}}\times\frac{1}{\sin\text{x}}\times\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}\Big(-\frac{1}{\text{x}^2}\Big)\Big]$
$=(\cos\text{x})^\text{x}\Big[\text{x}\Big(\frac{1}{\cos\text{x}}\Big)(-\sin\text{x})+\log\cos\text{x}\Big] \\ +(\sin\text{x})^\frac{1}{\text{x}}\Big[\frac{1}{\text{x}}\times\frac{1}{\sin\text{x}}(\cos\text{x})-\frac{1}{\text{x}^2}\log\sin\text{x}\Big]$
$=(\cos\text{x})^\text{x}\big[\log\cos\text{x}-\text{x}\tan\text{x}\big](\sin\text{x})^\frac{1}{\text{x}} \\ \Big[\frac{\cot\text{x}}{\text{x}}-\frac{1}{\text{x}^2}\log\sin\text{x}\Big]$
View full question & answer→Question 935 Marks
Differentiate the following functions with respect to x:
$\text{e}^\text{x}\log\sin2\text{x}$
AnswerLet $\text{y}=\text{e}^\text{x}\log\sin2\text{x}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[\text{e}^\text{x}\log\sin2\text{x}\big]$
$=\text{e}^\text{x}\frac{\text{d}}{\text{dx}}\log\sin2\text{x}+\log\sin2\text{x}\frac{\text{d}}{\text{dx}}\big(\text{e}^\text{x}\big)$
[Using product rule and chain rule]
$=\text{e}^\text{x}\frac{1}{\sin2\text{x}}\frac{\text{d}}{\text{dx}}(\sin2\text{x})+\log\sin2\text{x}\big(\text{e}^\text{x}\big)$
$=\frac{\text{e}^\text{x}}{\sin2\text{x}}\cos2\text{x}\frac{\text{d}}{\text{dx}}(2\text{x})+\text{e}^\text{x}\log\sin2\text{x}$
$=\frac{2\cos2\text{xe}^\text{x}}{\sin2\text{x}}+\text{e}^\text{x}\log\sin2\text{x}$
$=\text{e}^\text{x}(2\cot2\text{x}+\log\sin2\text{x})$
So,
$\frac{\text{d}}{\text{dx}}\big(\text{e}^\text{x}\log\sin2\text{x}\big)=\text{e}^\text{x}(2\cot2\text{x}+\log\sin2\text{x})$
View full question & answer→Question 945 Marks
If $\text{y}\sqrt{\text{x}^2+1}=\log\Big(\sqrt{\text{x}^2+1}-\text{x}\Big),$ prove that $\big(\text{x}^2+1\big)\frac{\text{dx}}{\text{dx}}+\text{xy}+1=0$
AnswerWe have, $\text{y}\sqrt{\text{x}^2+1}=\log\Big(\sqrt{\text{x}^2+1}-\text{x}\Big)$Differentiating with respect to x, we get,
$\Rightarrow\frac{\text{d}}{\text{dx}}\Big(\text{y}\sqrt{\text{x}^2+1}\Big)=\frac{\text{d}}{\text{dx}}\log\Big(\sqrt{\text{x}^2+1}-\text{x}\Big)$
[Using Product rule and chain rule]
$\Rightarrow\text{y}\frac{\text{d}}{\text{dx}}\big(\sqrt{\text{x}^2+1}\big)+\sqrt{\text{x}^2+1}\frac{\text{dy}}{\text{dx}} \\ =\frac{1}{\big(\sqrt{\text{x}^2+1}-\text{x}\big)}\times\frac{\text{d}}{\text{dx}}\big(\sqrt{\text{x}^2+1}-\text{x}\big)$
$\Rightarrow\frac{\text{y}}{2\sqrt{\text{x}^2+1}}\times\frac{\text{d}}{\text{dx}}(\text{x}^2+1)+\sqrt{\text{x}^2+1}\frac{\text{d}}{\text{dx}} \\ =\frac{1}{\big(\sqrt{\text{x}^2+1}-\text{x}\big)}\times\Big[\frac{1}{2\sqrt{\text{x}^2+1}}\frac{\text{d}}{\text{dx}}(\text{x}^2+1)-1\Big]$
$\Rightarrow\ \frac{2\text{xy}}{2\sqrt{\text{x}^2+1}}+\sqrt{\text{x}^2+1}\frac{\text{dy}}{\text{dx}} \\ =\frac{1}{\big(\sqrt{\text{x}^2+1}-\text{x}\big)}\Big[\frac{2\text{x}}{2\sqrt{\text{x}^2+1}}-1\Big]$
$\Rightarrow\sqrt{\text{x}^2+1}\frac{\text{dy}}{\text{dx}} \\ =\Big[\frac{1}{\sqrt{\text{x}^2+1}-\text{x}}\Big]\Big[\frac{\text{x}-\sqrt{\text{x}^2+1}}{\sqrt{\text{x}^2+1}}\Big]-\frac{\text{xy}}{\sqrt{\text{x}^2+1}}$
$\Rightarrow\sqrt{\text{x}^2+1}\frac{\text{dy}}{\text{dx}}=\frac{-(1+\text{xy})}{\sqrt{\text{x}^2+1}}$
$\Rightarrow\big(\text{x}^2+1\big)\frac{\text{dy}}{\text{dx}}=-(1+\text{xy})$
$\Rightarrow(\text{x}^2+1)\frac{\text{dy}}{\text{dx}}+1+\text{xy}=0$
View full question & answer→Question 955 Marks
If $y^x = e^{y-x},$ Prove that $\frac{\text{dy}}{\text{dx}}=\frac{(1+\log\text{y})^2}{\log\text{y}}$
AnswerHere,
$y^x = e^{y-x}$
Taking log on both the sides,
$\log\text{y}=\log\text{e}^{(\text{y}-\text{x})}$
$\big[\text{Since},\log\text{a}^{\text{b}}=\text{b}\log\text{a and}\log_\text{e}\text{e}=1\big]$
$\text{x}\log\text{y}=(\text{y}-\text{x})\log\text{e}$
$\text{x}\log\text{y}=\text{y}-\text{x}\ .....(\text{i})$
Differentiating it with respect to x using product rule,
$\frac{\text{d}}{\text{dx}}(\text{x}\log\text{y})=\frac{\text{d}}{\text{dx}}(\text{y}-\text{x})$
$\Big[\text{x}\frac{\text{d}}{\text{dx}}(\log\text{y})+\log\text{y}\frac{\text{d}}{\text{dx}}(\text{x})\Big]=\frac{\text{dy}}{\text{dx}}-1$
$\text{x}\Big(\frac{\text{x}}{\text{y}}\Big)\frac{\text{dy}}{\text{dx}}+\log\text{y}(1)=\frac{\text{dy}}{\text{dx}}-1$
$\frac{\text{dx}}{\text{dx}}\Big(\frac{\text{x}}{\text{y}}-1\Big)=-1-\log\text{y}$
$\frac{\text{dy}}{\text{dx}}\Big(\frac{\text{y}}{(1+\log\text{y})\text{y}}\Big)=-(1+\log\text{y})$
$\Big[\text{Since, from equation (i), x}=\frac{\text{y}}{(1+\log\text{y})}\Big]$
$\frac{\text{dy}}{\text{dx}}\Big[\frac{1-1-\log\text{y}}{(1+\log\text{y})}\Big]=-(1+\log\text{y})$
$\frac{\text{dy}}{\text{dx}}=-\frac{(1+\log\text{y})^2}{-\log\text{y}}$
$\frac{\text{dy}}{\text{dx}}=\frac{(1+\log\text{y})^2}{\log\text{y}}$
View full question & answer→Question 965 Marks
Differentiate the following functions with respect to x:
$\log(3\text{x}+2)-\text{x}^2\log(2\text{x}-1)$
AnswerLet $\text{y}=\log(3\text{x}+2)-\text{x}^2\log(2\text{x}-1)$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[\log(3\text{x}+2)-\text{x}^2\log(2\text{x}-1)\big]$
$\frac{\text{d}}{\text{dx}}\log(3\text{x}+2)-\frac{\text{d}}{\text{dx}}\big(\text{x}^2\log(2\text{x}-1)\big)$
$=\frac{1}{3\text{x}+2}\frac{\text{d}}{\text{dx}}(3\text{x}+2)-\Big[\text{x}^2\frac{\text{d}}{\text{dx}}\log(2\text{x}-1)+\log(2\text{x}-1)\frac{\text{d}}{\text{dx}}\big(\text{x}^2\big)\Big]$
[Using product rule and chain rule]
$=\frac{3}{3\text{x}+2}\Big[\text{x}^2\times\frac{1}{2\text{x}-1}\frac{\text{d}}{\text{dx}}(2\text{x}-1)+\log(2\text{x}-1)\times2\text{x}\Big]$
$=\frac{3}{3\text{x}+2}-\frac{2\text{x}^2}{2\text{x}-1}-2\text{x}\log(2\text{x}-1)$
So,
$\frac{\text{d}}{\text{dx}}\big(\log(3\text{x}+2)-\text{x}^2\log(2\text{x}-1)\big) \\ =\frac{3}{3\text{x}+2}-\frac{2\text{x}^2}{2\text{x}-1}-2\text{x}\log(2\text{x}-1)$
View full question & answer→Question 975 Marks
If $x^x + y^x = 1$, prove that $\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}(\text{y}+\text{x}\log\text{y})}{\text{x}(\text{y}\log\text{x}+\text{x})}$
AnswerHere,
$x^x + y^x = 1$
Taking on bith sides,
$\log(\text{x}^\text{y}\times\text{y}^\text{x})=\log(1)$
$\text{y}=\log\text{x}+\text{x}\log\text{y}=\log1$
$\big[\text{Since}, \log(\text{AB})=\log\text{A}+\log\text{B},\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating it with respect to x using product rule,
$\frac{\text{d}}{\text{dx}}(\text{y}\log\text{x})+\frac{\text{d}}{\text{dx}}(\text{x}\log\text{y})=\frac{\text{d}}{\text{dx}}(\log1)$
$\Big[\text{y}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{dy}}{\text{dx}}\Big]+\Big[\text{x}\frac{\text{d}}{\text{dx}}(\log\text{y})+\log\text{y}\frac{\text{d}}{\text{dx}}\text{(x)}\Big]=0$
$\Big[\text{y}\Big(\frac{1}{\text{x}}\Big)\log\text{x}\frac{\text{dy}}{\text{dx}}\Big]+\Big[\text{x}\Big(\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}\Big)+\log\text{y}(1)\Big]=0$
$\frac{\text{y}}{\text{x}}+\log\text{x}\log\frac{\text{dy}}{\text{dx}}+\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}+\log\text{y}=0$
$\frac{\text{dy}}{\text{dx}}\Big(\log\text{x}+\frac{\text{x}}{\text{y}}\Big)=-\Big[\log\text{y}+\frac{\text{y}}{\text{x}}\Big]$
$\frac{\text{dy}}{\text{dx}}\Big[\frac{\text{y}\log\text{x}+\text{x}}{\text{y}}\Big]=-\Big[\frac{\text{x}\log\text{y}+\text{y}}{\text{x}}\Big]$
$\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}}{\text{x}}\Big[\frac{\text{x}\log\text{y}+\text{y}}{\text{y}\log\text{x}+\text{x}}\Big]$
View full question & answer→Question 985 Marks
If $\text{x}=\text{a}\sin2\text{t}(1+\cos 2\text{t})$ and $\text{y}=\text{b}\cos\text{t}(1-\cos2\text{t}),$ show that at $\text{t}=\frac{\pi}{4},\frac{\text{dy}}{\text{dx}}=\frac{\text{b}}{\text{a}}\text{ t}=\frac{\pi}{4},\frac{\text{dy}}{\text{dx}}=\frac{\text{b}}{\text{a}}$
AnswerConsider the given functions,
$\text{x}=\text{a}\sin(2\text{t})(1+\cos2\text{t})\text{ and y}=\text{b}\cos2\text{t}(1-\cos2\text{t})$
Rewriting the above function, we have,
$\text{x}=\text{a}\sin2\text{t}+\frac{\text{a}}{2}\sin4\text{t}$
Differentiating the above function w.r.t. 't', we have,
$\frac{\text{dx}}{\text{dx}}=2\text{a}\cos2\text{t}+2\text{a}\cos4\text{t}\ .....(\text{i})$
$\text{y}=\text{b}\cos2\text{t}(1-\cos2\text{t})$
$\text{y}=\text{b}\cos2\text{t}-\text{b}\cos^22\text{t}$
$\frac{\text{dy}}{\text{dt}}=-2\text{b}\sin2\text{t}+2\text{b}\cos2\text{t}\sin2\text{t} \\ =-2\text{b}\sin2\text{t}+\text{b}\sin4\text{t}\ .....(\text{ii})$
From (1) and (2),
$\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{-2\text{b}\sin2\text{t}+\text{b}\sin4\text{t}}{2\text{a}\cos2\text{t}+2\text{a}\cos4\text{t}}$
$\therefore\frac{\text{dy}}{\text{dx}}\Big|_{\frac{\pi}{4}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}\Bigg|_{\text{t}=\frac{\pi}{4}}=\frac{-2\text{b}}{-2\text{a}}=\frac{\text{b}}{\text{a}}$
View full question & answer→Question 995 Marks
If $\text{y}=\text{x}\sin(\text{a}+\text{y}),$ prove that $\frac{\text{dx}}{\text{dx}}=\frac{\sin^2(\text{a}+\text{y})}{\sin(\text{a}+\text{y})-\text{y}\cos(\text{a}+\text{y})}$
AnswerWe have, $\text{y}=\text{x}\sin(\text{a}+\text{y})$
Differentiate with respect to y,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[\text{x}\sin(\text{a}+\text{y})\big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}\{\sin(\text{a}+\text{y})\}+\sin(\text{a}+\text{y})\frac{\text{d}}{\text{dx}}(\text{x})$
[Using product rule and chain rule]
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}\cos(\text{a}+\text{y})\frac{\text{d}}{\text{dx}}+\sin(\text{a}+\text{y})(1)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\{1-\text{x}\cos(\text{a}+\text{y})\}=\sin(\text{a}+\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sin(\text{a}+\text{y})}{1-\text{x}\cos(\text{a}+\text{y})}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sin(\text{a}+\text{y})}{1-\frac{\text{y}}{\sin(\text{a}+\text{y})}\cos(\text{a}+\text{y})}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sin^2(\text{a}+\text{y})}{\sin(\text{a}+\text{y}){-\text{y}\cos(\text{a}+\text{y})}}$
Hence, proved.
View full question & answer→Question 1005 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$y = e^x + 10^x + x^x$
AnswerHere, $\text{y}=\text{e}^\text{x}+10^\text{x}+\text{x}^\text{x}$
$\text{e}^{\text{x}}+10^{\text{x}}+\text{e}^{\log\text{x}^\text{x}}$
$\big[\text{Since, e}^{{\log}_\text{a}^\text{b}}=\text{a},\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
$\text{y}=\text{e}^{\text{x}}+10^{\text{x}}+\text{e}^{\log\text{x}^\text{x}}$
Differentiating it with respect to x using product rule, chain rule,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{e}^\text{x})+\frac{\text{d}}{\text{dx}}(10^{\text{x}})+\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}\log\text{x}}\big)$
$=\text{e}^\text{x}+10^\text{x}\log10+\text{e}^{\text{x}\log\text{x}}\frac{\text{d}}{\text{dx}}(\text{x}\log\text{x})$
$=\text{e}^\text{x}+10^\text{x}\log10+\text{e}^{\text{x}\log\text{x}}\Big[\text{x}\times\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big]$
$=\text{e}^\text{x}+10^\text{x}\log10+\text{e}^{\text{x}\log\text{x}}\Big[\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}(1)\Big]$
$=\text{e}^\text{x}+10^\text{x}\log10+\text{x}^{\text{x}}[1+\log\text{x}]$
$=\text{e}^\text{x}+10^\text{x}\log10+\text{x}^{\text{x}}[\log\text{e}+\log\text{x}] \big[\text{Since}, \log_\text{e}\text{e}=1\big]$
$\frac{\text{dy}}{\text{dx}}=\text{e}^\text{x}+10^\text{x}\log10+\text{x}^\text{x}(\log\text{ex})\ \big[\text{Since}\log\text{A}+\log\text{B}=\log\text{AB}]$
View full question & answer→Question 1015 Marks
If $\text{x}=\Big(\text{t}+\frac{1}{\text{t}}\Big)^\text{a},\text{y}=\text{a}^{\text{t}+\frac{1}{\text{t}}},$ find $\frac{\text{dy}}{\text{dx}}$
AnswerHere, $\text{x}=\Big(\text{t}+\frac{1}{\text{t}}\Big)^{\text{a}}$
Differentiating it with respect to t using chain rule,
$\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\left[\Big(\text{t}+\frac{1}{\text{t}}\Big)^{\text{a}}\right]$
$=\text{a}\Big(\text{t}+\frac{1}{\text{t}}\Big)^{\text{a-1}}\frac{\text{d}}{\text{dt}}\Big(\text{t}+\frac{1}{\text{t}}\Big)$
$\frac{\text{dx}}{\text{dt}}=\text{a}\Big(\text{t}+\frac{1}{\text{t}}\Big)^{\text{a-1}}\Big(\text{1}-\frac{1}{\text{t}^{2}}\Big)\ .....(\text{i})$
And, $\text{y}=\text{a}^{(\text{t}+\frac{1}{\text{t}})}$
Differentiating it with respect to t using chain rule,
$\frac{\text{dy}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\bigg[\text{a}^{\big(\text{t}+\frac{1}{\text{t}}\big)}\bigg]$
$=\frac{\text{d}}{\text{dt}}\text{a}^{\big(\text{t}+\frac{1}{\text{t}}\big)}\times\log\frac{\text{d}}{\text{dt}}\left(\text{t}+\frac{1}{\text{t}}\right)$
$\frac{\text{dy}}{\text{dt}}=\text{a}^{\big(\text{t}+\frac{1}{\text{t}}\big)}\times\log\text{a}\Big(\text{t}-\frac{1}{\text{t}^{2}}\Big)\ .....(\text{ii})$
Dividing equation (ii) by (i),
$\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{\text{a}^{\big(\text{t}+\frac{1}{\text{t}}\big)}\times\log\text{a}\left(1-\frac{1}{\text{t}^{2}}\right)}{\text{a}\big(\text{t}+\frac{1}{\text{t}}\big)^{\text{a}-1}\big(1-\frac{1}{\text{t}}^{2}\big)}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{a}^{\big(\text{t}+\frac{1}{\text{t}}\big)}\times\log\text{a}}{\text{a}\big(\text{t}+\frac{1}{\text{t}}\big)^{\text{a}-1}}$
View full question & answer→Question 1025 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=\frac{\text{e}^{\text{ax}}\sec\text{x}\times\log\text{x}}{\sqrt{1-2\text{x}}}$
AnswerHere,
$\text{y}=\frac{\text{e}^{\text{ax}}\sec\text{x}\times\log\text{x}}{\sqrt{1-2\text{x}}}\ .....(\text{i})$
$\Rightarrow\text{y}=\frac{\text{e}^{\text{ax}}\times\sec^\text{x}\times\log\text{x}}{(1-2\text{x})^\frac{1}{2}}$
Taking log on both the sides,
$\log\text{y}=\log\text{e}^{\text{ax}}+\log{\sec\text{x}}+\log\log\text{x}-\frac{1}{2}\log(1-2\text{x}) \\ \begin{bmatrix} \text{Since}, \log\Big(\frac{\text{A}}{\text{B}}\Big)=\log\text{A}-\log\text{B},\\ \log(\text{AB})=\log\text{A}+\log\text{B} \end{bmatrix}$
$\log\text{y}=\text{ax}+\log{\sec\text{x}}+\log\log\text{x}-\frac{1}{2}\log(1-2\text{x}) \\ \big[\text{Since}, \log\text{a}^\text{b}=\text{b}\log\text{a and }\log_\text{e}\text{e}=1\big]$
Differentiating it with respect to x using chain rule,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{ax})+\frac{\text{d}}{\text{dx}}(\log\sec\text{x})+\frac{\text{d}}{\text{dx}}(\log\log\text{x})-\frac{1}{2}\log(1-2\text{x})$
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{a}+\frac{1}{\sec\text{x}}\frac{\text{d}}{\text{dx}}(\sec\text{x})+\frac{1}{\log\text{x}}\frac{\text{d}}{\text{dx}}-\frac{1}{2}\Big(\frac{1}{1-2\text{x}}\Big)\frac{\text{d}}{\text{dx}}(1-2\text{x})$
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{a}+\frac{\sec\text{x}\tan\text{x}}{\sec\text{x}}+\frac{1}{(\log\text{x})}\Big(\frac{1}{\text{x}}\Big)-\frac{1}{2}\Big(\frac{1}{1-2\text{x}}\Big)(-2)$
$\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\text{a}+\tan\text{x}+\frac{1}{\text{x}\log\text{x}}+\frac{1}{1-2\text{x}}\Big]$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^{\text{ax}}\sec\text{x}\log\text{x}}{\sqrt{1-2\text{x}}}\Big[\text{a}+\tan\text{x}+\frac{1}{\text{x}\log\text{x}}+\frac{1}{1-2\text{x}}\Big]$
[Using equation (i)]
View full question & answer→Question 1035 Marks
If $x^{13} y^7 = (x + y)^{20}$, prove that $\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}$
AnswerHere,
$x^{13}y^7 = (x + y)^{20}$
Taking log on both the sides,
$\log(\text{x}^{13}\text{y}^7)=\log(\text{x}+\text{y})^{20}$
$13\log\text{x}+7\log\text{y}=20\log(\text{x}+\text{y})$
$\big[\text{Since},\log(\text{AB})=\log\text{A}+\log\text{B},\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating it with respect to x using chain rule,
$13\frac{\text{d}}{\text{dx}}(\log\text{x})+7\frac{\text{d}}{\text{dx}}(\log\text{y})=20\frac{\text{d}}{\text{dx}}\log(\text{x}+\text{y})$
$\frac{13}{\text{x}}+\frac{7}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{20}{\text{x}+\text{y}}\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})$
$\frac{13}{\text{x}}+\frac{7}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{20}{(\text{x}+\text{y})}\Big[1+\frac{\text{dy}}{\text{dx}}\Big]$
$\frac{7}{\text{y}}\frac{\text{dy}}{\text{dx}}-\frac{20}{(\text{x}+\text{y})}=\frac{20}{(\text{x}+\text{y})}-\frac{13}{\text{x}}$
$\frac{\text{dy}}{\text{dx}}\Big[\frac{\text{7}}{\text{y}}-\frac{20}{(\text{x}+\text{y})}\Big]=\frac{20}{(\text{x}+\text{y})}-\frac{13}{\text{x}}$
$\frac{\text{dy}}{\text{dx}}\Big[\frac{2(\text{x}+\text{y})-20\text{y}}{\text{y}(\text{x}+\text{y})}\Big]=\Big[\frac{20\text{x}-13(\text{x}+\text{y})}{\text{x}(\text{x}+\text{y})}\Big]$
$\frac{\text{dy}}{\text{dx}}=\Big[\frac{20\text{x}-13\text{x}-13\text{y}}{\text{x}(\text{x}+\text{y})}\Big]\Big(\frac{\text{y}(\text{x}+\text{y})}{7\text{x}+7\text{y}-20\text{y}}\Big)$
$=\frac{\text{y}}{\text{x}}\Big(\frac{7\text{x}-13\text{y}}{7\text{x}-13\text{y}}\Big)$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}$
View full question & answer→Question 1045 Marks
If $\tan^{-1}\Big(\frac{\text{x}^2-\text{y}^2}{\text{x}^2+\text{y}^2}\Big)=\text{a}$ prove that $\frac{\text{dx}}{\text{dx}}=\frac{\text{y}}{\text{x}}\frac{(1-\tan\text{a})}{(1+\tan\text{a})}$
AnswerWe have, $\tan^{-1}\Big(\frac{\text{x}^2-\text{y}^2}{\text{x}^2+\text{y}^2}\Big)=\text{a}$
$\Rightarrow\frac{\text{x}^2-\text{y}^2}{\text{x}^2+\text{y}^2}=\tan\text{a}$
$\Rightarrow\text{x}^2-\text{y}^2=\tan\text{a}(\text{x}^2+\text{y}^2)$
Differentiating with respect to x, we get
$\Rightarrow\frac{\text{d}}{\text{dx}}(\text{x}^2-\text{y}^2)=\tan\text{a}\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{y}^2)$
$\Rightarrow\Big(2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}\Big)=\tan\text{a}\Big(2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}\Big)$
$\Rightarrow2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}=2\text{x}\tan\text{a}+2\text{y}\tan\text{a}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow2\text{y}\tan\text{a}\frac{\text{dy}}{\text{dx}}+2\text{y}\frac{\text{dy}}{\text{dx}}=2\text{x}-2\text{x}\tan\text{a}$
$\Rightarrow2\text{y}(1+\tan\text{a})\frac{\text{dy}}{\text{dx}}=2\text{x}(1-\tan\text{a})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\text{y}}\Big(\frac{1-\tan\text{a}}{1+\tan\text{a}}\Big)$
View full question & answer→Question 1055 Marks
Differentiate the following functions with respect to x:
$\text{e}^{\sin\text{x}}+(\tan\text{x})^\text{x}$
AnswerLet $\text{y}=\text{e}^{\sin\text{x}}+(\tan\text{x})^\text{x}$
$\Rightarrow\text{y}=\text{e}^{\sin\text{x}}+\text{e}^{\log(\tan\text{x})^\text{x}}$
$\Rightarrow\text{y}=\text{e}^{\sin\text{x}}+\text{e}^{\text{x}\log(\tan\text{x})}$
Differentiating with resepect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\sin\text{x}}\big)+\frac{\text{d}}{\text{dx}}\big\{\text{e}^{\text{x}\log(\tan\text{x})}\big\}$
$=\text{e}^{\sin\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x})+\text{e}^{\text{x}\log(\tan\text{x})}\frac{\text{d}}{\text{dx}}(\text{x}\log\tan\text{x})$
$=\text{e}^{\sin\text{x}}(\cos\text{x})+\text{e}^{\log(\tan\text{x})^\text{x}}\Big[\text{x}\frac{\text{d}}{\text{dx}}(\log\tan\text{x})+\log\tan\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big]$
$=\text{e}^{\sin\text{x}}(\cos\text{x})+(\tan\text{x})^\text{x}\Big[\frac{\text{x}}{\tan\text{x}}(\sec^2\text{x})+\log\tan\text{x}\Big]$
$=\text{e}^{\sin\text{x}}(\cos\text{x})+(\tan\text{x})^\text{x}\big[\text{x}\sec\text{x cosec x}+\log\tan\text{x}\big]$
View full question & answer→Question 1065 Marks
If $\text{y}=\cot^{-1}\Big\{\frac{\sqrt{1+\sin\text{x}}+\sqrt{1-\sin\text{x}}}{\sqrt{1+\sin\text{x}}-\sqrt{1-\sin\theta}}\Big\},$ show that $\frac{\text{dy}}{\text{dx}}$ is independent of x.
AnswerLet $\text{y}=\cot^{-1}\Big[\frac{\sqrt{1+\sin\text{x}}+\sqrt{1-\sin\text{x}}}{\sqrt{1+\sin\text{x}}-\sqrt{1-\sin\theta}}\Big] \ .....(\text{i})$
Then, $\frac{\sqrt{1+\sin\theta}+\sqrt{1-\sin\text{x}}}{\sqrt{1+\sin\text{x}}-\sqrt{1-\sin\text{x}}}$
$=\frac{\big(\sqrt{1+\sin\text{x}}+\sqrt{1-\sin\text{x}}\big)^2}{\big(\sqrt{1+\sin\text{x}}-\sqrt{1-\sin\text{x}}\big)\big(\sqrt{1+\sin\text{x}}+\sqrt{1-\sin\text{x}}\big)} $
$=\frac{(1+\sin\text{x})+(1-\sin\text{x})+2\sqrt{(1-\sin\text{x})(1+\sin\text{x})}}{(1+\sin\text{x})-(1-\sin\text{x})}$
$=\frac{2+2\sqrt{1-\sin^2\text{c}}}{2\sin\text{x}}$
$=\frac{1+\cos\text{x}}{\sin\text{x}}$
$=\frac{2\cos^2\frac{\text{x}}{2}}{2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}$
$=\cot\frac{\text{x}}{2}$
Therefore, equation (i) becomes
$\text{y}=\cot^{-1}\Big(\cot\frac{\text{x}}{2}\Big)$
$\Rightarrow\ \text{y}=\frac{\text{x}}{2}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{1}{2}\frac{\text{d}}{\text{dx}}(\text{x})$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\frac{1}{2}$
View full question & answer→Question 1075 Marks
If $(\cos\text{x})^{\text{y}}=(\tan\text{y})^{\text{x}},$ Prove that $\frac{\text{dy}}{\text{dx}}=\frac{\log\tan\text{y}-\text{y}\tan\text{x}}{\log\cos\text{x}-\text{x}\sec\text{y cosec y}}$
AnswerHere,
$(\cos\text{x})^{\text{y}}=(\tan\text{y})^{\text{x}}$
Taking log on both sides,
$\log(\cos\text{x})^{\text{y}}=\log(\tan\text{y})^{\text{x}}$
$\text{y}\log(\cos\text{x})=\text{x}\log(\tan\text{y})$
$\big[\text{Since}, \log\text{e}^{\text{b}}=\text{b}\log\text{a}\big]$
Differentiating it with respect to x using chain rule and product rule,
$\frac{\text{d}}{\text{dx}}(\text{y}\log\cos\text{x})=\frac{\text{d}}{\text{dx}}(\text{x}\log\tan\text{y})$
$\Big(\text{y}\frac{\text{d}}{\text{dx}}\log\cos\text{x}+\log\cos\text{x}\frac{\text{dy}}{\text{dx}}\Big) \\ =\Big(\text{x}\frac{\text{d}}{\text{dx}}\log\tan\text{y}+\log\tan\text{y}\frac{\text{d}}{\text{dx}}(\text{x})\Big)$
$\Big(\text{y}\big(\frac{1}{\cos\text{x}}\big)\frac{\text{d}}{\text{dx}}(\cos\text{x})+\log\cos\text{x}\frac{\text{dy}}{\text{dx}}\Big) \\ =\Big(\text{x}\frac{1}{\tan\text{y}}\frac{\text{d}}{\text{dx}}(\tan\text{y})+\log\tan\text{y}(1)\Big)$
$\Big(\frac{\text{y}}{\cos\text{x}}(-\sin\text{x})+\log\cos\text{x}\frac{\text{dy}}{\text{dx}}\Big)\\ =\Big(\frac{\text{x}}{\tan\text{y}}(\sec^2\text{y})\Big)\frac{\text{dy}}{\text{dx}}+\log\tan\text{y}-\text{y}\tan\text{x}+\log\cos\text{x}\frac{\text{dy}}{\text{dx}} \\ =\Big(\sec\text{y cosec y}\times\text{y}\frac{\text{dy}}{\text{dx}}+\log\tan\text{y}\Big)$
$\frac{\text{dy}}{\text{dx}}\big[\log\cos\text{x}-\text{x}\sec\text{y cosec y}\big] \\ =\log\tan\text{y}+\text{y}\tan\text{x}$
$\frac{\text{dy}}{\text{dx}}=\Big[\frac{\log\tan\text{x}+\text{y}\tan\text{x}}{\log\cos\text{x}-\text{x}\sec\text{y cosec y}}\Big]$
View full question & answer→Question 1085 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=(\tan\text{x})^{\log\text{x}}+\cos^2\big(\frac{\pi}{4}\big)$
AnswerHere,
$\text{y}=(\tan\text{x})^{\log\text{x}}+\cos^2\big(\frac{\pi}{4}\big)$
$\text{y}=\text{e}^{\log(\tan\text{x})^{\log\text{x}}}+\cos^2\big(\frac{\pi}{2}\big)$
$\text{y}=\text{e}^{\log\text{x}\log\tan\text{x}}+\cos\text{x}^2\big(\frac{\pi}{4}\big)$
$\big[\text{Since, e}^{\log\text{a}}=\text{a and}\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating ti using chain rule and product rule,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\log\text{x}\log\tan\text{x}})+\frac{\text{d}}{\text{dx}}\cos^2\big(\frac{\pi}{4}\big)$
$=\text{e}^{\log\text{x}\log\tan\text{x}}\frac{\text{d}}{\text{dx}}(\log\text{x}\log\tan\text{x})+0$
$=\text{e}^{\log(\tan\text{x})^{\log\text{x}}}\Big[\log\times\frac{\text{d}}{\text{dx}}(\log\tan\text{x})+\log\tan\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})\Big]$
$=(\tan\text{x})^{\log\text{x}}\Big[\log\times\Big(\frac{1}{\tan\text{x}}\Big)\frac{\text{d}}{\text{dx}}(\tan\text{x})+\log\tan\text{x}\Big(\frac{1}{\text{x}}\Big)\Big]$
$=(\tan\text{x})^{\log\text{x}}\Big[\log\times\Big(\frac{1}{\tan\text{x}}\Big)\big(\sec^2\text{x}\big)+\frac{\log\tan\text{x}}{\text{x}}\Big]$
$\frac{\text{dy}}{\text{dx}}=(\tan\text{x})^{\log\text{x}}\Big[\log\times\Big(\frac{\sec^2\text{x}}{\tan\text{x}}\Big)+\frac{\log\tan\text{x}}{\text{x}}\Big]$
View full question & answer→Question 1095 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{\text{a}+\text{b}\tan\text{x}}{\text{b}-\text{a}\tan\text{x}}\Big)$
AnswerLet, $\text{y}=\tan^{-1}\Big[\frac{\text{a}+\text{b}\tan\text{x}}{\text{b}-\text{a}\tan\text{x}}\Big]$
$\Rightarrow\text{y}=\tan^{-1}\bigg[\frac{\frac{\text{a}+\text{b}\tan\text{x}}{\text{b}}}{\frac{\text{b}-\text{a}\tan\text{x}}{\text{b}}}\bigg]$
$\Rightarrow\text{y}=\tan^{-1}\bigg[\frac{\frac{\text{a}}{\text{b}}+\tan\text{x}}{1-\frac{\text{a}}{\text{b}}\tan\text{x}}\bigg]$
$\Rightarrow\text{y}=\tan^{-1}\Bigg[\frac{\tan\Big(\tan^{-1}\frac{\text{a}}{\text{b}}\Big)+\tan\text{x}}{1-\tan\Big(\tan^{-1}\frac{\text{a}}{\text{b}}\Big)\times\tan\text{x}}\Bigg]$
$\Rightarrow\text{y}=\tan^{-1}\Big[\tan\Big(\tan^{-1}\frac{\text{a}}{\text{b}}+\text{x}\Big)\Big]$
$\Rightarrow\text{y}=\tan^{-1}\big(\frac{\text{a}}{\text{b}}\big)+\text{x}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=0+1$
$\therefore\ \frac{\text{dy}}{\text{dx}}=1$
View full question & answer→Question 1105 Marks
Differentiate the following functions with respect to x:
$(\sin\text{x})^{\log\text{x}}$
AnswerLet $\text{y}=(\sin\text{x})^{\log\text{x}}\ .....(\text{i})$
Taking log on both sides,
$\log\text{y}=\log(\sin\text{x})^{\log\text{x}}$
$\Rightarrow\log\text{y}=\log\text{x}\log\sin\text{x}$
Differentiating with respect to x,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\log\text{x}\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})$
$\Rightarrow \frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\log\text{x}\frac{1}{\sin\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}\Big(\frac{1}{\text{x}}\Big)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\log\text{x}}{\sin\text{x}}(\cos\text{x})+\frac{\log\sin\text{x}}{\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\log\text{x}\cot\text{x}+\frac{\log\sin\text{x}}{\text{x}}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=(\sin\text{x})^{\log\text{x}}\Big[\log\text{x}\cot\text{x}+\frac{\log\sin\text{x}}{\text{x}}\Big]$
[Using equation (i)]
View full question & answer→Question 1115 Marks
If $\text{xy}\log(\text{x}+\text{y})=1,$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\text{y}(\text{x}^2\text{y}+\text{x}+\text{y})}{\text{x}(\text{xy}^2+\text{x}+\text{y})}$
AnswerHere,
$\text{xy}\log(\text{x}+\text{y})=1\ .....(\text{i})$
Differentiaitng with respect to x using chain rula, product rule,
$\frac{\text{dy}}{\text{dx}}(\text{xy}\log(\text{x}+\text{y}))=\frac{\text{d}}{\text{dx}}(1)$
$\text{xy}\frac{\text{d}}{\text{dx}}\log(\text{x}+\text{y})+\text{x}\log(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}+\text{y}\log(\text{x}+\text{y})\frac{\text{d}}{\text{dx}}(\text{x})=0$
$\frac{\text{xy}}{(\text{x}+\text{y})}\Big(1+\frac{\text{dy}}{\text{dx}}\Big)+\text{x}\log(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}+\text{y}\log(\text{x}+\text{y})(1)=0$
$\Big(\frac{\text{xy}}{\text{x}+\text{y}}\Big)\Big(1+\frac{\text{dy}}{\text{dx}}\Big)+\text{x}\log(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}+\text{x}\log(\text{x}+\text{y})=0$
$\Big(\frac{\text{xy}}{\text{x}+\text{y}}\Big)\frac{\text{dy}}{\text{dx}}+\frac{\text{xy}}{\text{x}+\text{y}}+\text{x}\Big(\frac{1}{\text{xy}}\Big)\frac{\text{dy}}{\text{dx}}+\text{y}\Big(\frac{1}{\text{xy}}\Big)=0$
[Using equation (i)]
$\frac{\text{dy}}{\text{dx}}\Big[\frac{\text{xy}}{\text{x}+\text{y}}+\frac{1}{\text{y}}\Big]=-\Big[\frac{1}{\text{x}}+\frac{\text{xy}}{\text{x}+\text{y}}\Big]$
$\frac{\text{dy}}{\text{dx}}\Big[\frac{\text{xy}^2+\text{x}+\text{y}}{(\text{x}+\text{y})\text{y}}\Big]=-\Big[\frac{\text{x}+\text{y}+\text{x}^2\text{y}}{\text{x}(\text{x}+\text{y})}\Big]$
$\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}}{\text{x}}\Big(\frac{\text{x}+\text{y}+\text{x}^2\text{y}}{\text{x}+\text{y}+\text{xy}^2}\Big)$
View full question & answer→Question 1125 Marks
Differentiate the following functions with respect to x:
$(\log\text{x})^\text{x}$
AnswerLet $\text{y}=(\log\text{x})^\text{x}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=\log(\log\text{x})^\text{x}$
$\Rightarrow\log\text{y}=\text{x}\log(\log\text{x})\ \big[\text{Since}, \log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating with respect to x, using product rule, chain rule,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}\log(\log\text{x})+\log\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x})$
$=\text{x}\frac{1}{\log\text{x}}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\log\text{x}(1)$
$=\frac{\text{x}}{\log\text{x}}\Big(\frac{1}{\text{x}}\Big)+\log\log\text{x}$
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{1}{\log\text{x}}+\log\log\text{x}$
$\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{1}{\log\text{x}}+\log\log\text{x}\Big]$
$\frac{\text{dy}}{\text{dx}}=(\log\text{x})^\text{x}\Big[\frac{1}{\log\text{x}}+\log\log\text{x}\Big]$
[Using equation (i)]
View full question & answer→Question 1135 Marks
If $(\sin\text{x})^{\text{y}}=\text{x}+\text{y},$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{1-(\text{x}+\text{y})\text{y}\cot\text{x}}{(\text{x}+\text{y})\log\sin\text{x}-1}$
AnswerHere,
$(\sin\text{x})^{\text{y}}=\text{x}+\text{y}$
Taking log on both the sides,
$\log(\sin\text{x})^\text{y}=\log(\text{x}+\text{y})$
$\text{y}\log(\sin\text{x})=\log(\text{x}+\text{y})\ \big[\text{Since},\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating it with respect to x using chain rule, product rule,
$\frac{\text{d}}{\text{dx}}(\text{y}\log(\sin\text{x}))=\frac{\text{d}}{\text{dx}}\log(\text{x}+\text{y})$
$\text{y}\frac{\text{d}}{\text{dx}}\log\sin\text{x}+\log\sin\text{x}\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}+\text{y}}\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})$
$\frac{\text{y}}{\sin\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}\frac{\text{dy}}{\text{dx}}=\frac{1}{(\text{x}+\text{y})}\Big[1+\frac{\text{dy}}{\text{dx}}\Big]$
$\frac{\text{y}(\cos\text{x})}{(\sin\text{x})}+\log\sin\text{x}\frac{\text{dy}}{\text{dx}}=\frac{1}{(\text{x}+\text{y})}+\frac{1}{(\text{x}+\text{y})}\frac{\text{dy}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}\Big(\log\sin\text{x}-\frac{1}{\text{x}+\text{y}}\Big)=\frac{1}{(\text{x}+\text{y})}-\text{y}\cot\text{x}$
$\frac{\text{dy}}{\text{dx}}\Big(\frac{(\text{x}+\text{y})\log\sin\text{x}-1}{(\text{x}+\text{y})}\Big)=\Big(\frac{1-\text{y}(\text{x}+\text{y})\cot\text{x}}{\text{x}+\text{y}}\Big)$
$\frac{\text{dy}}{\text{dx}}=\Big(\frac{1-\text{y}(\text{x}+\text{y})\cot\text{x}}{(\text{x}+\text{y})\log\sin\text{x}-1}\Big)$
View full question & answer→Question 1145 Marks
If $\text{y}=(\sin\text{x}-\cos\text{x})^{\sin\text{x}-\cos\text{x}},\frac{\pi}{4}<\text{x}<\frac{3\pi}{4},$ find $\frac{\text{dy}}{\text{dx}}$
AnswerWe have, $\text{y}=(\sin\text{x}-\cos\text{x})^{\sin\text{x}-\cos\text{x}}\ .....(\text{i})$
Taking log on both sides,
$\log\text{y}=\log(\sin\text{x}-\cos\text{x})^{\sin\text{x}-\cos\text{x}}$
$\Rightarrow\log\text{y}=(\sin\text{x}-\cos\text{x})^{\sin\text{x}-\cos\text{x}}$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\log(\sin\text{x}-\cos\text{x})\frac{\text{d}}{\text{dx}}(\sin\text{x}-\cos\text{x}) \\ +(\sin\text{x}-\cos\text{x})\frac{\text{d}}{\text{dx}}\log(\sin\text{x}-\cos\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=(\sin\text{x}-\cos\text{x})(\sin\text{x}-\cos\text{x}) \\ +\frac{(\sin\text{x}-\cos\text{x})}{(\sin\text{x}-\cos\text{x})}\frac{\text{d}}{\text{dx}}(\sin\text{x}-\cos\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=(\cos\text{x}+\sin\text{x})\log(\sin\text{x}-\cos\text{x})+(\cos\text{x}+\sin\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=(\cos\text{x}+\sin\text{x})[1+\log(\sin\text{x}-\cos\text{x})]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}\big[(\cos\text{x}+\sin\text{x})\big\{1+\log(\sin\text{x}-\cos\text{x})\big\}\big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=(\sin\text{x}-\cos\text{x})^{(\sin\text{x}-\cos\text{x})} \\ \big[(\cos\text{x}+\sin\text{x})\big\{1+\log(\sin\text{x}-\cos\text{x})\big\}\big]$
View full question & answer→Question 1155 Marks
Differentiate the following functions with respect to x:
$\frac{\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}}{\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}}$
AnswerLet, $\text{y}=\frac{\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}}{\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}}$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big[\frac{\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}}{\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}}\Big]$
$=\Bigg[\frac{\big(\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}\big)\frac{\text{d}}{\text{dx}}\big(\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}\big)-\big(\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}\big)\frac{\text{d}}{\text{dx}}\big(\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}\big)}{\big(\text{e}^{2\text{x}}-\text{e}^{4\text{x}}\big)^2}\Bigg]$
[Using quotient rule and chain rule]
$=\frac{(\text{e}^{2\text{x}}-\text{e}^{-2\text{x}})\Big[\text{e}^{2\text{x}}\frac{\text{d}}{\text{dx}}(2\text{x})+\text{e}^{-2\text{x}}\frac{\text{d}}{\text{dx}}(-2\text{x})\Big]-\big(\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}\big)\Big[\text{e}^{2\text{x}}\frac{\text{d}}{\text{dx}}(2\text{x})-\text{e}^{-2\text{x}}\frac{\text{d}}{\text{dx}}(-2\text{x})\Big]}{\big(\text{e}^{2\text{x}}-\text{e}^{2\text{x}}\big)^2}$
$=\frac{\big(\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}\big)\big(2\text{e}^{2\text{x}}-2\text{e}^{-2\text{e}}\big)-\big(\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}\big)\big(2\text{e}^{2\text{e}}+2\text{e}^{-2\text{x}}\big)}{\big(\text{e}^{2\text{e}}-\text{e}^{-2\text{x}}\big)^2}$
$=\frac{2\big(\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}\big)^2-2\big(\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}\big)^2}{\big(\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}\big)^2}$
$=\frac{2\big[\text{e}^{4\text{x}}+\text{e}^{-4\text{x}}-2\text{e}^{2\text{x}}\text{e}^{-2\text{x}}-\text{e}^{4\text{x}}-\text{e}^{-4\text{x}}-2\text{e}^{2\text{x}}\text{e}^{-2\text{x}}\big]}{\big(\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}\big)^2}$
$=\frac{-8}{\big(\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}\big)^2}$
So,
$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}}{\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}}\Big)=\frac{-8}{\big(\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}\big)^2}$
View full question & answer→Question 1165 Marks
If $\text{x}=\text{e}^{\cos2\text{t}}$ and $\text{y}=\text{e}^{\sin2\text{t}},$ prove that $\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}\log\text{x}}{\text{x}\log\text{y}}$
AnswerWe have, $\text{x}=(\text{e}^{\cos2\text{t}})$ and $\text{y}=\text{e}^{\sin2\text{t}}$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\big(\text{e}^{\cos2\text{t}}\big)$ and $\frac{\text{dy}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\big(\text{e}^{\sin2\text{t}}\big)$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\text{e}^{\cos2\text{t}}\frac{\text{d}}{\text{dt}}(\cos2\text{t})$ and $\frac{\text{dx}}{\text{dt}}=\text{e}^{\cos2\text{t}}\frac{\text{d}}{\text{dt}}(\cos2\text{t})$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\text{e}^{\cos2\text{t}}(-\sin2\text{t})\frac{\text{d}}{\text{dt}}(2\text{t})$ and $\frac{\text{dy}}{\text{dt}}=\text{e}^{\sin2\text{t}}(\cos2\text{t})\frac{\text{d}}{\text{dt}}(2\text{t})$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=-2\sin 2\text{t}\text{e}^{\cos2\text{t}}$ and $\frac{\text{dy}}{\text{dt}}=2\cos 2\text{t}\text{e}^{\sin2\text{t}}$
$\because\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{2\cos2\text{te}^{\sin2\text{t}}}{-2\sin2\text{te}^{\cos2\text{t}}}$
$\begin{bmatrix} \because\text{x}=\text{e}^{\cos2\text{t}}\Rightarrow\log\text{x}=\cos2\text{t} \\ \text{y}=\text{e}^{\sin2\text{t}}\Rightarrow\log\text{y}=\sin2\text{t} \end{bmatrix}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}\log\text{x}}{\text{x}\log\text{y}}$
View full question & answer→Question 1175 Marks
If $e^y= y^x,$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{(\log\text{y})^2}{\log\text{y}-1}$
AnswerWe have, $e^y = y^x$
Taking log on both sides,
$\log\text{e}^{\text{y}}=\log\text{y}^\text{x}$
$\Rightarrow\text{y}\log\text{e}=\text{x}\log\text{y}$
$\Rightarrow\text{y}=\text{x}\log\text{y}\ .....(\text{i})$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x}\log\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{dy}}{\text{dx}}(\log\text{y})+\log\text{y}\frac{\text{d}}{\text{dx}}(\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}+\log\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\Big(1-\frac{\text{x}}{\text{y}}\Big)=\log\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\big(\frac{\text{y}-\text{x}}{\text{y}}\big)=\log\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}\log\text{y}}{\text{y}-\text{x}}$
$\Rightarrow\frac{\text{y}\log\text{y}}{\Big(\text{y}-\frac{\text{y}}{\log\text{y}}\Big)}$
[Using equation (i)]
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}\log\text{y}(\log\text{y})}{\text{y}\log\text{y}-\text{y}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}(\log\text{y})^2}{\text{y}(\log\text{y}-1)}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(\log\text{y})^2}{(\log\text{y}-1)}$
View full question & answer→Question 1185 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$y = x^n + n^x + x^x + n^n$
AnswerWe have, $y = x^n + n^x + x^x + n^x$
$\Rightarrow\text{y}=\text{x}^\text{n}+\text{n}^\text{x}+\text{e}^{\log\text{x}^\text{x}}+\text{n}^\text{n}$
$\Rightarrow\text{y}=\text{x}^\text{n}+\text{n}^\text{x}+\text{e}^{\text{x}\log\text{x}}+\text{n}^\text{n}$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x}^\text{n})+\frac{\text{d}}{\text{dx}}(\text{n}^\text{x})+\frac{\text{d}}{\text{dx}}(\text{e}^{\text{x}\log\text{x}})+\frac{\text{d}}{\text{dx}}(\text{n}^\text{n})$
$=\text{nx}^{\text{n}-1}+\text{n}^\text{x}\log\text{n}=\text{e}^{\log\text{x}^\text{x}}\Big[\text{x}\frac{\text{d}}{\text{dx}}\log\text{x}+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big]$
$=\text{nx}^{\text{n}-1}+\text{n}^\text{x}\log\text{n}=\text{x}^{\text{x}}\Big[\text{x}\big(\frac{1}{\text{x}}\big)+\log\text{x}\Big]$
$=\text{nx}^{\text{n}-1}+\text{n}^\text{x}\log\text{n}=\text{x}^{\text{x}}\big[1+\log\text{x}\big]$
$=\text{nx}^{\text{n}-1}+\text{n}^\text{x}\log\text{n}=\text{x}^{\text{x}}\big[\log\text{e}+\log\text{x}\big] \\ \big[\because\log_\text{e}\text{e}=1\text{ and }\log\text{A}+\log\text{B}=\log(\text{AB})\big]$
$=\text{nx}^{\text{n}-1}+\text{n}^\text{x}\log\text{n}=\text{x}^{\text{x}}\log\big(\text{ex}\big)$
View full question & answer→Question 1195 Marks
If $\text{x}=\text{a}\Big(\text{t}+\frac{1}{\text{t}}\Big)\text{ and y}=\text{a}\Big(\text{t}-\frac{1}{\text{t}}\Big),$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\text{y}}$
AnswerHere, $ \text{x}=\text{a}\Big(\text{t}+\frac{1}{\text{t}}\Big)$
Differentiating it with respect to t,
$\frac{\text{dx}}{\text{dt}}=\text{a}\frac{\text{d}}{\text{dt}}\Big(\text{t}+\frac{1}{\text{t}}\Big)$
$=\text{a}\Big(\text{t}-\frac{1}{\text{t}^{2}}\Big)$
$\frac{\text{dx}}{\text{dt}}=\text{a}\Big(\frac{\text{t}^{2}-1}{\text{t}^{2}}\Big)\ .....(\text{i})$
And, $\text{y}=\text{a}\Big(\text{t}-\frac{1}{\text{t}}\Big)$
Differetiating it with respect to t,
$\frac{\text{dy}}{\text{dt}}=\text{a}\frac{\text{d}}{\text{dt}}\Big(\text{t}-\frac{1}{\text{t}}\Big)$
$=\text{a}\Big(1+\frac{1}{\text{t}^{2}}\Big)$
$\frac{\text{dy}}{\text{dt}}=\text{a}\Big(\frac{\text{t}^{2}+1}{\text{t}^{2}}\Big)\ .....(\text{ii})$
Dividing equation (ii) by (i),
$\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\text{a}\frac{(\text{t}^{2}+1)}{\text{t}^{2}}\times\frac{\text{t}^{2}}{\text{a}(\text{t}^{2}-1)}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{t}^{2}+1}{\text{t}^{2}-1}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\text{y}}$
$\Big[\text{Since},\frac{\text{x}}{\text{y}}=\frac{\text{a}(\text{t}^{2}+1)}{\text{t}}\times\frac{\text{t}}{\text{a}(\text{t}^{2}-1)}=\Big(\frac{\text{t}^{2}+1}{\text{t}^{2}-1}\Big)\Big]$
View full question & answer→Question 1205 Marks
Differentiate $\tan^{-1}\Big(\frac{\text{x}}{\sqrt{1-\text{x}^2}}\Big)$ with respect to $\sin^{-1}\Big(2\text{x}\sqrt{1-\text{x}^2}\Big),$ if $-\frac{1}{\sqrt{2}}<\text{x}<\frac{1}{\sqrt{2}}$
AnswerLet, $\text{u}=\tan^{-1}\Big(\frac{\text{x}}{\sqrt{1-\text{x}^2}}\Big)$
Put $\text{x}=\sin\theta$
$\Rightarrow\theta=\sin^{-1}\text{x}$
$\Rightarrow\text{u}=\tan^{-1}\Big(\frac{\sin\theta}{\sqrt{1-\sin^{2}\theta}}\Big)$
$\Rightarrow\text{u}=\tan^{-1}\Big(\frac{\sin\theta}{\cos\theta}\Big)$
$\Rightarrow\text{u}=\tan^{-1}(\tan\theta)\ .....(\text{i})$
And
Let, $\text{v}=\sin^{-1}(2\text{x}\sqrt{1-\text{x}^2})$
$\text{v}=\sin^{-1}(2\sin\theta\sqrt{1-\sin^{2}\theta})$
$\text{v}-\sin^{-1}(2\sin\theta\cos\theta)$
$\text{v}=\sin^{-1}(\sin2\theta)\ .....(\text{ii})$
Here,
$-\frac{1}{\sqrt{2}}<\text{x}<\frac{1}{\sqrt{2}}$
$\Rightarrow-\frac{1}{\sqrt{2}}<\sin\theta<\frac{1}{\sqrt{2}}$
$\Rightarrow-\frac{\pi}{4}<\theta<\frac{\pi}{4}$
So, from equation (i)
$\text{u}=\theta\Big[\text{since,}\tan^{-1}(\tan\theta)=\theta,\text{if }\theta\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\Big]$
$\Rightarrow\text{u}=\sin^{-1}\text{x}$
Differentiatiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}\ .....(\text{iii})$
From equation (ii),
$\text{v}=2\theta\Big[\text{Since,}\sin^{-1}(\sin\theta)=\theta,\text{if }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\Rightarrow\text{v}=2\sin^{-1}\text{x}$
Differentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{2}{\sqrt{1-\text{x}^2}}\ .....(\text{iv})$
Dividing equation (iii) by (iv),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\Big(\frac{1}{\sqrt{1-\text{x}^2}}\Big)\Big(\frac{\sqrt{1-\text{x}^2}}{2}\Big)$
$\therefore\frac{\text{du}}{\text{dv}}=\frac{1}{2}$
View full question & answer→Question 1215 Marks
Differentiate $\tan^{-1}\Big(\frac{1-\text{x}}{1+\text{x}}\Big)$ with respect to $\sqrt{1-\text{x}^2},$ if -1 < x < 1.
AnswerLet, $\text{u}=\tan^{-1}\Big(\frac{1-\text{x}}{1+\text{x}}\Big)$
Put $\text{x}=\tan\theta$
$\Rightarrow\theta=\tan^{-1}\text{x}$
$\Rightarrow\text{u}=\tan^{-1}\Big(\frac{1-\tan\theta}{1+\tan\theta}\Big)$
$\Rightarrow\text{u}=\tan^{-1}\Big[\tan\Big(\frac{\pi}{4}-\theta\Big)\Big]\ .....(\text{i})$
Here,
$-1<\text{x}<1$
$\Rightarrow-1<\tan\theta<1$
$\Rightarrow-\frac{\pi}{4}<\theta<\frac{\pi}{4}$
$\Rightarrow\frac{\pi}{4}>-\theta>\frac{\pi}{4}$
$\Rightarrow-\frac{\pi}{4}<-\theta<\frac{\pi}{4}$
$\Rightarrow0<\frac{\pi}{4}-\theta<\frac{\pi}{2}$
So, from equation (i)
$\text{u}=\frac{\pi}{4}-\theta$
$\Big[\text{Since,}\tan^{-1}(\tan\theta)=\theta,\text{if }\theta\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\Big]$
$\Rightarrow\text{u}=\frac{\pi}{4}-\tan^{-1}\text{x}$
$\frac{\text{du}}{\text{dx}}=0-\Big(\frac{1}{1+\text{x}}\Big)$
$\Rightarrow\frac{\text{du}}{\text{dx}}=-\frac{1}{1+\text{x}^2}\ .....(\text{ii})$
And
Let, $\text{v}=\sqrt{1-\text{x}^2}$
$\frac{\text{dv}}{\text{dx}}=\frac{1}{2\sqrt{1-\text{x}^2}}\times\frac{\text{d}}{\text{dx}}(1-\text{x}^2)$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{1}{2\sqrt{1-\text{x}^2}}(-2\text{x})$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{-\text{x}}{\sqrt{1-\text{x}^2}}\ .....(\text{iii})$
Dividing equation (ii) by (iii),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=-\frac{1}{1+\text{x}^2}\times\frac{\sqrt{1-\text{x} ^2}}{-\text{x}}$
$\therefore\frac{\text{du}}{\text{dv}}=\frac{\sqrt{1-\text{x}^2}}{\text{x}(1+\text{x}^2)}$
View full question & answer→Question 1225 Marks
Differentiate the following functions with respect to x:
$\text{e}^{\sqrt{\cot\text{x}}}$
AnswerLet, $\text{y}=\text{e}^\sqrt{{\cot\text{x}}}$
$\Rightarrow\ \text{y}=\text{e}^{(\cot\text{x})^\frac{1}{2}}$
Differentiate with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\text{e}^{(\cot\text{x})^\frac{1}{2}}\Big)$
$=\text{e}^{(\cot\text{x})^\frac{1}{2}}\frac{\text{d}}{\text{dx}}(\cot\text{x})^\frac{1}{2}$
[Using chain rule]
$=\text{e}^\sqrt{\cot\text{x}}\times\frac{1}{2}(\cot\text{x})^{\frac{1}{2}-1}\frac{\text{d}}{\text{dx}}(\cot\text{x})$
$=-\frac{\text{e}^\sqrt{\cot\text{x}}\times\text{cosec}^2\text{x}}{2\sqrt{\cot\text{x}}}$
So,
$\frac{\text{d}}{\text{dx}}\Big(\text{e}^\sqrt{\cot\text{x}}\Big)=-\frac{\text{e}^\sqrt{\cot\text{x}}\times\text{cosec}^2\text{x}}{2\sqrt{\cot\text{x}}}$
View full question & answer→Question 1235 Marks
Differentiate $\sin^{-1}\Big(4\text{x}\sqrt{1-4\text{x}^2}\Big)$ with respect to $\sqrt{1-4\text{x}^2},$ if:
$\text{x}\in\Big(-\frac{1}{2},-\frac{1}{2\sqrt{2}}\Big)$
AnswerLet $\text{u}=\sin^{-1}\Big(4\text{x}\sqrt{1-4\text{x}^2}\Big)$
Put $2\text{x}=\cos\theta \text{ So},$
$\Rightarrow\text{u}=\sin^{-1}\Big(2\times\cos\theta\sqrt{1-\cos^2\theta}\Big)$
$\Rightarrow\text{u}=\sin^{-1}(2\cos\theta\sin\theta)$
$\Rightarrow\text{u}=\sin^{-1}(\sin2\theta)\ .....(\text{i})$
Let, $\text{v}=\sqrt{1-4\text{x}^2}\ .....(\text{ii})$
Here,
$\text{x}\in\Big(\frac{1}{2},-\frac{1}{2\sqrt{2}}\Big)$
$\Rightarrow2\text{x}\in\Big(-1,-\frac{1}{\sqrt{2}},\Big)$
$\Rightarrow\theta\in\Big(\frac{3\pi}{4},\pi\Big)$
So, from equation (i),
$\text{u}=\pi-2\theta$
$\Big[\text{Since},\sin^{-1}(\sin\theta)=\pi-\theta,\text{ if }\theta\in\Big[-\frac{\pi}{2},\frac{3\pi}{2}\Big]\Big]$
$\Rightarrow\text{u}=\pi-2\cos^{-1}(2\text{x})\big[\text{Since},2\text{x}=\cos\theta\big]$
Differentiating it with respect to x using chain rule,
$\frac{\text{du}}{\text{dx}}=0-2\bigg(\frac{-1}{\sqrt{1-(2\text{x})^2}}\bigg)\frac{\text{d}}{\text{dx}}(2\text{x})$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\frac{-2}{\sqrt{1-4\text{x}^2}}(2)$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\frac{-4}{\sqrt{1-4\text{x}^2}}\ .....(\text{vi})$
From equation (iv)
$\frac{\text{dv}}{\text{dx}}=\frac{4}{\sqrt{1-4\text{x}^2}}$
But, $\text{x}\in\Big(-\frac{1}{2},-\frac{1}{2\sqrt{2}}\Big)$
$\frac{\text{dv}}{\text{dx}}=\frac{-4(-\text{x})}{\sqrt{1-4(-\text{x})^2}}$
$\frac{\text{dv}}{\text{dx}}=\frac{4\text{x}}{\sqrt{1-4\text{x}^2}}\ .....(\text{vii})$
Dividing equation (vi) by (vii),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{-4}{\sqrt{1-4\text{x}^2}}\times\frac{\sqrt{1-4\text{x}^2}}{4\text{x}}$
$\frac{\text{du}}{\text{dv}}=-\frac{1}{\text{x}}$
View full question & answer→Question 1245 Marks
If $\text{x}=\cot\text{t and y}=\sin\text{t},$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{3}}\text{ at t}=\frac{2\pi}{3}$
AnswerWe have, $\text{x}=\cos\text{t}$ and $\text{y}=\sin\text{t}$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\cos\text{t}) $ and $\frac{\text{dy}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\sin\text{t}) $
$\Rightarrow\frac{\text{dx}}{\text{dt}}=-\sin\text{t}$ and $\frac{\text{dy}}{\text{dt}}=\cos\text{t}$
$\therefore\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{\cos\text{t}}{-\sin\text{t}}=-\cot{\text{t}}$
Now, $\big(\frac{\text{dy}}{\text{dx}}\big)_{\text{t}=\frac{2\pi}{3}}=-\cot\big(\frac{2\pi}{3}\big)=\frac{1}{\sqrt{3}}$
View full question & answer→Question 1255 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{\text{x}}{1+6\text{x}^3}\Big)$
AnswerLet $\text{y}=\tan^{-1}\Big(\frac{\text{x}}{1+6\text{x}^3}\Big)$
$\Rightarrow\ \text{y}=\tan^{-1}\Big(\frac{3\text{x}-2\text{x}}{1+(3\text{x})(2\text{x})}\Big)$
$\Rightarrow\text{y}=\tan^{-1}3\text{x}-\tan^{-1}2\text{x}$
$\Big[\text{Since}, \tan^{-1}\text{x}-\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}-\text{y}}{1+\text{xy}}\Big)\Big]$
Differentiate it with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{1+(3\text{x})^2}\frac{\text{d}}{\text{dx}}(3\text{x})-\frac{1}{1+(2\text{x})^3}\frac{\text{d}}{\text{dx}}(2\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{1+9\text{x}^2}(3)-\frac{1}{1+4\text{x}^2}(2)$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{3}{1+9\text{x}^2}-\frac{2}{1+4\text{x}^2}$
View full question & answer→Question 1265 Marks
If $\sec\Big(\frac{\text{x}+\text{y}}{\text{x}-\text{y}}\Big)=\text{a}$ prove that $\frac{\text{dx}}{\text{dx}}=\frac{\text{y}}{\text{x}}$
AnswerWe have, $\sec\Big(\frac{\text{x}+\text{y}}{\text{x}-\text{y}}\Big)=\text{a}$
$\Rightarrow\frac{\text{x}+\text{y}}{\text{x}-\text{y}}=\sec^{-1}({\text{a}})$
Differentiate with respect to x, we get,
$\Rightarrow\bigg[\frac{(\text{x}-\text{y})\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})-(\text{x}+\text{y})\frac{\text{d}}{\text{dx}}(\text{x}-\text{y})}{(\text{x}-\text{y}}\bigg]=0$
$\Rightarrow(\text{x}-\text{y})\Big(1+\frac{\text{d}}{\text{dx}}\Big)-(\text{x}+\text{y})\Big(1-\frac{\text{d}}{\text{dx}}\Big)=0$
$\Rightarrow(\text{x}-\text{y})+(\text{x}-\text{y})\frac{\text{dy}}{\text{dx}}-(\text{x}+\text{y})+(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}[\text{x}-\text{y}+\text{x}+\text{y}]=\text{x}+\text{y}-\text{x}+\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}(2\text{x})=2\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}$
View full question & answer→Question 1275 Marks
Differentiate $\sin^{-1}\sqrt{1-\text{x}^2}$ with respect to $\cos^{-1}\text{x},$ if
$\text{x}\in(-1,0)$
AnswerLet $\text{u}=\sin^{-1}\sqrt{1-\text{x}^2}$
Put $\text{x}=\cos\theta, \text{So},$
$\Rightarrow\text{u}=\sin^{-1}\sqrt{1-\cos^2\theta}$
$\Rightarrow\text{u}=\sin^{-1}(\sin\theta)\ .....(\text{i})$
And, $\text{v}=\cos^{-1}\text{x}\ .....(\text{ii})$
Here,
$\text{x}\in(-1,0)$
$\Rightarrow\cos\theta\in(-1, 0)$
$\Rightarrow\theta\in\Big(\frac{\pi}{2},\pi\Big)$
So, from equation (i),
$\text{u}=\pi-\theta$
$\Big[\text{Since},\sin^{-1}(\sin\theta)=\pi-\theta,\theta\in\Big(\frac{\pi}{2},\frac{3\pi}{2}\Big)\Big]$
$\text{u}=\pi-\cos^{-1}\text{x }\big[\text{Since}\text{x}=\cos\theta\big]$
Differentiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=0-\Big(\frac{-1}{\sqrt{1-\text{x}^2}}\Big)$
$\frac{\text{du}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}\ .....(\text{iii})$
And, from equation (ii),
$\text{v}=\cos^{-1}\text{x}$
Differentaiting it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{-1}{\sqrt{1-\text{x}^2}}\ .....(\text{iv})$
Dividing equation (iii) by (iv)
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{1}{\sqrt{1-\text{x}^2}}\times\frac{\sqrt{1-\text{x}^2}}{-1}$
$\frac{\text{du}}{\text{dv}}=-1$
View full question & answer→Question 1285 Marks
If $\text{y}=\sqrt{\text{x}^2+\text{a}^2},$ prvoe that $\text{y}\frac{\text{dy}}{\text{dx}}-\text{x}=0$
AnswerHere, $\text{y}=\sqrt{\text{x}^2+\text{a}^2}$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\sqrt{\text{x}^2+\text{a}^2}\big)$
$=\frac{1}{2\sqrt{\text{x}^2+\text{a}^2}}\frac{\text{d}}{\text{dx}}\big(\text{x}^2+\text{a}^2\big)$
[Using chain rule]
$=\frac{1}{2\sqrt{\text{x}^2+\text{a}^2}}\times(2\text{x})$
$=\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\text{y}}\ \Big[\text{Since},\sqrt{\text{x}^2+\text{a}^2}=\text{y}\Big]$
$\Rightarrow \text{y}\frac{\text{dy}}{\text{dx}}=\text{x}$
$\Rightarrow\text{y}\frac{\text{dy}}{\text{dx}}-\text{x}=0$
View full question & answer→Question 1295 Marks
If $\text{y}=\text{x}\sin(\text{a}+\text{y}),$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\sin^2(\text{a}+\text{y})}{\sin(\text{a}+\text{y})-\text{y}\cos(\text{a}+\text{y})}$
AnswerHere,
$\text{y}=\text{x}\sin(\text{a}+\text{y})$
Differentiating with respect to x using the chain rule and product rule,
$\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}\sin(\text{a}+\text{y})+\sin(\text{a}+\text{y})\frac{\text{dx}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}=\text{x}\cos(\text{a}+\text{y})\frac{\text{dy}}{\text{dx}}+\sin(\text{a}+\text{y})$
$(1-\text{x}\cos(\text{a}+\text{y}))\frac{\text{dy}}{\text{dx}}=\sin(\text{a}+\text{y})$
$\frac{\text{dy}}{\text{dx}}=\frac{\sin(\text{a}+\text{y})}{(1-\text{x}\cos(\text{a}+\text{y}))}$
$\frac{\text{dy}}{\text{dx}}=\frac{\sin(\text{a}+\text{y})}{\Big(1-\frac{\text{y}}{\sin(\text{a}+\text{y})}\cos(\text{a}+\text{y})\Big)}\ \Big[\text{Since}\frac{\text{y}}{\sin(\text{a}+\text{y})}=\text{x}\Big]$
$\frac{\text{dy}}{\text{dx}}=\frac{\sin^2(\text{a}+\text{y})}{\sin(\text{a}+\text{y})-\text{y}\cos(\text{a}+\text{y})}$
View full question & answer→Question 1305 Marks
If $\text{y}=\text{e}^{\text{x}}+\text{e}^{-\text{x}},$ prvoe that $\frac{\text{dy}}{\text{dx}}=\sqrt{\text{y}^2-4}$
AnswerDifferentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}}+\text{e}^{-\text{x}}\big)$
$=\frac{\text{d}}{\text{dx}}\text{e}^{\text{x}}+\frac{\text{d}}{\text{dx}}{\text{e}}^{-\text{x}}$
$=\text{e}^{\text{x}}+\text{e}^{-\text{x}}\frac{\text{d}}{\text{dx}}\big(-\text{x}\big)$
[Using chain rule]
$=\text{e}^{\text{x}}+\text{e}^{-\text{x}}(-1)$
$=\big(\text{e}^{\text{x}}-\text{e}^{-\text{x}}\big)$
$=\sqrt{\big(\text{e}^{\text{x}}-\text{e}^{-\text{x}}\big)^2-4\text{e}^{\text{x}}\times\text{e}^{-\text{x}}}$
$\Big[\text{Since},(\text{a}-\text{b}=\sqrt{(\text{a}+\text{b})^2-4\text{ab}}\Big]$
$=\sqrt{\text{y}^2-4}$
$\big[\text{Since e}^\text{x}+\text{e}^{-\text{x}}=\text{y}\big]$
Hence, the solution is, $\frac{\text{dy}}{\text{dx}}=\sqrt{\text{y}^2-4}$
View full question & answer→Question 1315 Marks
If $\text{y}=(\sin\text{x})^{(\sin\text{x})^{(\sin\text{x})^{....\infty}}},$ prove that $\frac{\text{y}^2\cot\text{x}}{(1-\text{y}\log\sin\text{x})}$
AnswerHere,
$\text{y}=(\sin\text{x})^{(\sin\text{x})^{(\sin\text{x})^{....\infty}}}$
$\Rightarrow\text{y}=(\sin\text{x})^\text{y}$
Taking log on both sides,
$\log\text{y}=\log(\sin\text{x})^{\text{y}}$
$\log\text{y}=\text{y}(\log\sin\text{x})$
Differentiating it with respect to x, using product rule,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{y}\frac{\text{d}}{\text{dx}}(\log\sin\text{x})+\log\sin\text{x}\frac{\text{dy}}{\text{dx}}$
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{y}\frac{1}{\sin\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}\frac{\text{dy}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}\Big(\frac{1}{\text{y}}-\log\sin\text{x}\Big)=\frac{\text{y}}{\sin\text{x}}(\cot\text{x})$
$\frac{\text{dy}}{\text{dx}}\Big(\frac{1-\text{y}\log\sin\text{x}}{\text{y}}\Big)=\text{y}\cot\text{x}$
$\frac{\text{dy}}{\text{dx}}=\Big(\frac{\text{y}^2\cot\text{x}}{1-\text{y}\log\sin\text{x}}\Big)$
View full question & answer→Question 1325 Marks
Differentiate the following functions with respect to x:
$(\tan\text{x})^\frac{1}{\text{x}}$
AnswerLet $\text{y}=(\tan\text{x})^\frac{1}{\text{x}}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=\log(\tan\text{x})^\frac{1}{\text{x}}$
$\log\text{y}=\frac{1}{\text{x}}\log(\tan\text{x})\ \big[\text{Since}, \log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating it with respect to x using product rule and chain rule,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}\frac{\text{d}}{\text{dx}}\log(\tan\text{x})+\log(\tan\text{x})\frac{\text{d}}{\text{dx}}\Big(\frac{1}{\text{x}}\Big)$
$=\frac{1}{\text{x}}\times\frac{1}{\tan\text{x}}\frac{\text{d}}{\text{dx}}(\tan\text{x})+\log(\tan\text{x})\Big(-\frac{1}{\text{x}^2}\Big)$
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}\tan\text{x}}(\sec^2\text{x})-\frac{\log(\tan\text{x})}{\text{x}^2}$
$\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{\sec^2\text{x}}{\text{x}\tan\text{x}}-\frac{\log(\tan\text{x})}{\text{x}^2}\Big]$
$\frac{\text{dy}}{\text{dx}}=(\tan\text{x})^\frac{1}{\text{x}}\Big[\frac{\sec^2\text{x}}{\text{x}\tan\text{x}}-\frac{\log(\tan\text{x})}{\text{x}^2}\Big]$
[Using equation (i)]
View full question & answer→Question 1335 Marks
Differentiate the following functions with respect to x:
$\sin\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)$
AnswerLet $\text{y}=\sin\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dy}}\Big(\sin\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)\Big)$
$=\cos\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)\frac{\text{d}}{\text{dx}}\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)$
[Using chain rule]
$=\cos\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)\Bigg[\frac{(1-\text{x}^2)\frac{\text{d}}{\text{dx}}(1+\text{x}^2)-(1+\text{x}^2)\frac{\text{d}}{\text{dx}}(1-\text{x}^2)}{(1-\text{x})^2}\Bigg]$
[Using chain rule]
$=\cos\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)\bigg[\frac{(1-\text{x}^2)(2\text{x})-(1+\text{x}^2)(-2\text{x})}{(1-\text{x}^2)^2}\bigg]$
$=\cos\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)\Big[\frac{2\text{x}-2\text{x}^3+2\text{x}+2\text{x}^3}{(1-\text{x}^2)^2}\Big]$
$=\frac{4\text{x}}{\big(1-\text{x}^2\big)^2}\cos\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)$
So,
$\frac{\text{d}}{\text{dx}}\Big(\sin\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)\Big)=\frac{4\text{x}}{\big(1-\text{x}^2\big)^2}\cos\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)$
View full question & answer→Question 1345 Marks
Differentiate the following functions with respect to x:
$\sin^{-1}\Big(\frac{\text{x}}{\sqrt{\text{x}^2+\text{x}^2}}\Big)$
AnswerLet $\text{y}=\sin^{-1}\Big(\frac{\text{x}}{\sqrt{\text{x}^2+\text{x}^2}}\Big)$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big\{\sin^{-1}\Big(\frac{\text{x}}{\sqrt{\text{x}^2+\text{x}^2}}\Big)\Big\}$
$=\frac{1}{\sqrt{1-\Big(\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}}\Big)}}\times\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}}\Big)$
[Using chain rule and quotient rule]
$=\frac{1}{\sqrt{1-\Big(\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}}\Big)^2}}\times\Bigg[\frac{(\text{x}^2+\text{a}^2)^\frac{1}{2}\frac{\text{d}}{\text{dx}}(\text{x})-\text{x}\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{a}^2)^\frac{1}{2}}{\Big[(\text{x}^2+\text{a}^2)^\frac{1}{2}\Big]^2}\Bigg]$
$=\frac{\sqrt{\text{x}^2+\text{a}^2}}{\sqrt{\text{x}^2+\text{a}^2-\text{x}^2}}\times\Bigg[\frac{\sqrt{\text{x}^2+\text{a}^2}-\frac{\text{x}}{2\sqrt{\text{x}^2+\text{a}^2}}\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{a}^2)}{(\text{x}^2+\text{a}^2)}\Bigg]$
$=\frac{\sqrt{\text{x}^2+\text{a}^2}}{\text{a}(\text{x}^2+\text{a}^2)}\times\Big[\sqrt{\text{x}^2+\text{a}^2}-\frac{\text{x}}{2\sqrt{\text{x}^2+\text{a}^2}}\times2\text{x}\Big]$
$=\frac{\sqrt{\text{x}^2+\text{a}^2}}{\text{a}(\text{x}^2+\text{a}^2)}\times\Big[\frac{\text{x}^2+\text{a}^2-\text{x}^2}{\sqrt{\text{x}^2+\text{a}^2}}\Big]$
$=\frac{\text{a}^2}{\text{a}(\text{x}^2+\text{a}^2)}$
$=\frac{\text{a}}{(\text{x}^2+\text{a}^2)}$
So,
$\frac{\text{d}}{\text{dx}}\Big\{\sin^{-1}\Big(\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}}\Big)\Big\}=\frac{\text{a}^2}{\text{a}(\text{x}^2+\text{a}^2)}$
View full question & answer→Question 1355 Marks
If $(\text{x}-\text{y})\text{e}^{\frac{\text{x}}{\text{x}-\text{y}}}=\text{a},$ prove that $\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}=2\text{y}$
AnswerConsider the given function, $(\text{x}-\text{y})\text{e}^{\frac{\text{x}}{\text{x}-\text{y}}}=\text{a}.$
We need to prove that $\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}=2\text{y}.$
Differentiating the given equation w.r.t 'x' we get
$(\text{x}-\text{y})\Bigg[\text{e}^{\frac{\text{x}}{\text{x}-\text{y}}}\Bigg(\frac{(\text{x}-\text{y})-\text{x}\big(1-\frac{\text{dy}}{\text{dx}}\big)}{(\text{x}-\text{y})^2}\Bigg)\Bigg]+\text{e}^\frac{\text{x}}{\text{x}-\text{y}}\Big(1-\frac{\text{dy}}{\text{dx}}\Big)=0$
$\Rightarrow\frac{(\text{x}-\text{y})-\text{x}\Big(1-\frac{\text{dy}}{\text{dx}}\Big)}{(\text{x}-\text{y})}+\Big(1-\frac{\text{dy}}{\text{dx}}\Big)=0$
$\Rightarrow\Big(1+\frac{\text{dy}}{\text{dx}}\Big)\Big(1-\frac{\text{x}}{\text{x}-\text{y}}\Big)+1=0$
$\Rightarrow\Big(1+\frac{\text{dy}}{\text{dx}}\Big)\Big(\frac{-\text{y}}{\text{x}-\text{y}}\Big)+1=0$
$\Rightarrow-\text{y}+\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}-\text{y}=0$
$\Rightarrow\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}=2\text{y}$
View full question & answer→Question 1365 Marks
If $\text{x}=\frac{1+\log\text{t}}{\text{t}^2},\text{y}=\frac{3+2\log\text{t}}{\text{t}},$ find $\frac{\text{dy}}{\text{dx}}$
Answer$\text{x}=3\sin\text{t}-\sin3\text{t},\text{y}=3\cos\text{t}-\cos3\text{t}$
$\frac{\text{dx}}{\text{dt}}=3\cos\text{t}-3\cos3\text{t}$
$\frac{\text{dy}}{\text{dt}}=-3\sin\text{t}+3\sin3\text{t}$
$\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{3\sin\text{t}+3\sin3\text{t}}{3\cos\text{t}-3\cos3\text{t}}$
When $\text{t}=\frac{\pi}{3}$
$\frac{\text{dy}}{\text{dx}}=\frac{-3\sin\big(\frac{\pi}{3}\big)+3\sin(\pi)}{3\cos\big(\frac{\pi}{3}\big)-3\cos(\pi)}=\frac{3\times\frac{\sqrt{3}}{2}+0}{3\times\frac{1}{2}-3(-1)}=-\frac{1}{\sqrt{3}}$
View full question & answer→Question 1375 Marks
Differentiate the following functions from first principles:$e^{ax+b}.$
AnswerLet $f(x) = e^{ax+b}\Rightarrow f(x + h) = e^{a(x+h)+b}$
$\therefore\frac{\text{d}}{\text{dx}}(\text{f(x)})=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{\text{a}(\text{x}+\text{h})+\text{b}}-\text{e}^{(\text{ax}+\text{b})}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{\text{ax}+\text{b}}\text{e}^{\text{ah}}-\text{e}^{\text{ax}+\text{b}}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\text{ e}^{\text{ax}+\text{b}}\left\{\frac{(\text{e}^{\text{ah}}-1)}{\text{ah}}\right\}\times\text{a}$ $=\text{ae}^{\text{ax}+\text{b}} \lim\limits_{\text{h}\rightarrow0}\left\{\frac{(\text{e}^{\text{ah}}-1)}{\text{ah}}\right\}$ $=\text{ae}^{\text{ax}+\text{b}}$ So, $\frac{\text{d}}{\text{dx}}(\text{e}^{\text{ax}}+\text{b})=\text{ae}^{\text{ax}+\text{b}}$
View full question & answer→Question 1385 Marks
If $\tan(\text{x}+\text{y})+\tan(\text{x}+\text{y})=1,$ find $\frac{\text{dy}}{\text{dx}}$
AnswerWe have, $\tan(\text{x}+\text{y})+\tan(\text{x}-\text{y})=1$
Differentiating with respect to x, we get,
$\Rightarrow\ \frac{\text{d}}{\text{dx}}\tan(\text{x}+\text{y})+\frac{\text{d}}{\text{dx}}\tan(\text{x}+\text{y})=\frac{\text{d}}{\text{dx}}(1)$
$\Rightarrow\ \sec^2(\text{x}+\text{y})\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})+\sec^2(\text{x}-\text{y})\frac{\text{d}}{\text{dx}}(\text{x}-\text{y})=0$
$\Rightarrow\ \sec^2(\text{x}+\text{y})\Big[1+\frac{\text{dy}}{\text{dx}}\Big]+\sec^2(\text{x}-\text{y})\Big[1-\frac{\text{dy}}{\text{dx}}\Big]=0$
$\Rightarrow\ \sec^2(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}-\sec^2(\text{x}-\text{y})\frac{\text{dy}}{\text{dx}} \\ =-\big[\sec^2(\text{x}-\text{y})+\sec^2(\text{x}-\text{y})\big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\big[\sec^2(\text{x}+\text{y})-\sec^2(\text{x}-\text{y})\big] \\ =-\big[\sec^2(\text{x}+\text{y})+\sec^2(\text{x}-\text{y})\big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sec^2(\text{x}+\text{y})+\sec^2(\text{x}-\text{y})}{\sec^2(\text{x}-\text{y})-\sec^2(\text{x}+\text{y})}$
View full question & answer→Question 1395 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=\text{x}^{\cos\text{x}}+(\sin\text{x})^{\tan\text{x}}$
AnswerWe have, $\text{y}=\text{x}^{\cos\text{x}}+(\sin\text{x})^{\tan\text{x}}$
$\text{y}=\text{e}^{\log\text{x}^{\cos\text{x}}}+\text{e}^{\log(\sin\text{x})^{\tan\text{x}}}$
$\text{y}=\text{e}^{\cos\text{x}\log\text{x}}+\text{e}^{\tan\text{x}\log\sin\text{x}}$
Differentiating with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\cos\text{x}\log\text{x}}\big)+\frac{\text{d}}{\text{dx}}\big(\text{e}^{\tan\text{x}\log\sin\text{x}}\big)$
$=\text{e}^{\cos\text{x}\log\text{x}}\frac{\text{d}}{\text{dx}}(\cos\text{x}\log\text{x}) \\ +\text{e}^{\tan\text{x}\log\sin\text{x}}\frac{\text{d}}{\text{dx}}(\tan\text{x}\log\sin\text{x})$
$=\text{e}^{\log\text{x}^{\cos\text{x}}}\Big[\cos\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\cos\text{x})\Big] \\ +\text{e}^{\log(\sin\text{x})^{\tan\text{x}}}\Big[\tan\text{x}\frac{\text{d}}{\text{dx}}\log\sin\text{x}+\log\sin\text{x}\frac{\text{d}}{\text{dx}}(\tan\text{x})\Big]$
$=\text{x}^{\cos\text{x}}\Big[\cos\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}(-\sin\text{x})\Big] \\ +(\sin\text{x})^{\tan\text{x}}\Big[\tan\text{x}\Big(\frac{1}{\sin\text{x}}\Big)\frac{\text{d}}{\text{dx}}(\sin)\text{x}+\log\sin\text{x}(\sec^2\text{x})\Big]$
$=\text{x}^{\cos\text{x}}\Big[\frac{\cos\text{x}}{\text{x}}-\sin\text{x}\log\text{x}\Big] \\ +(\sin\text{x})^{\tan\text{x}}\Big[\tan\text{x}\Big(\frac{1}{\sin\text{x}}\Big)(\cos\text{x})+\sec^2\text{x}\log\sin\text{x}\Big]$
$=\text{x}^{\cos\text{x}}\Big[\frac{\cos\text{x}}{\text{x}}-\sin\text{x}\log\text{x}\Big] \\ +(\sin\text{x})^{\tan\text{x}}\big[1+\sec^2\text{x}\log\sin\text{x}\big]$
View full question & answer→Question 1405 Marks
Differentiate $\sin^{-1}\Big(2\text{x}\sqrt{1-\text{x}^2}\Big)$ with respect to $\tan^{-1}\Big(\frac{\text{x}}{\sqrt{1-\text{x}^2}}\Big),$ if $-\frac{1}{\sqrt{2}}<\text{x}<\frac{1}{\sqrt{2}}$
AnswerLet, $\text{u}=\sin^{-1}(2\text{x}\sqrt{1-\text{x}^2})$
Put $\text{x}=\sin\theta$
$\Rightarrow\text{u}=\sin^{-1}\Big(2\sin\theta\sqrt{1-\sin^2\theta}\Big)$
$\Rightarrow\text{u}=\sin^{-1}(2\sin\theta\cos\theta)$
$\Rightarrow\text{u}=\sin^{-1}(\sin2\theta)\ .....(\text{i})$
Let $\text{v}=\tan^{-1}\Big(\frac{\text{x}}{\sqrt{1-\text{x}}}\Big)$
$\Rightarrow\text{v}=\tan^{-1}\Big(\frac{\sin\theta}{\sqrt{1-\sin^2\theta}}\Big)$
$\Rightarrow\text{v}=\tan^{-1}\Big(\frac{\sin\theta}{\cos\theta}\Big)$
$\Rightarrow\text{v}=\tan^{-1}(\tan\theta)\ .....(\text{ii})$
Here, $-\frac{1}{\sqrt{2}}<\text{x}<\frac{1}{\sqrt{2}}$
$\Rightarrow-\frac{1}{\sqrt{2}}<\sin\theta<\frac{1}{\sqrt{2}}$
$\Rightarrow-\frac{\pi}{4}<\theta<\frac{\pi}{4}$
So, from equation (i),
$\text{u}= 2\theta\bigg[\text{Since,}\sin^{-1}(\sin\theta)=\theta,\text{if }\theta\in\bigg[-\frac{\pi}{2},\frac{\pi}{2}\bigg]\bigg]$
$\Rightarrow\text{u}=2\sin^{-1}\text{x}[\text{Since, x}=\sin\theta]$
Differentiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=\frac{2}{\sqrt{1-\text{x}^2}}\ .....(\text{iii})$
From equation (ii),
$\text{v}=\theta\bigg[\text{Since,}\tan^{-1}(\tan\theta)=\theta,\text{if }\theta\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\bigg]$
$\Rightarrow\text{v}=\sin^{-1}\text{x}[\text{since, x}=\sin\theta]$
Differentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}\ .....\text{(iv)}$
Dividing equation (iii) by (iv)
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{2}{\sqrt{1-\text{x}^2}}\times\frac{\sqrt{1-\text{x}^2}}{1}$
$\therefore\frac{\text{du}}{\text{dv}}=2$
View full question & answer→Question 1415 Marks
If $\text{y}=\tan^{-1}\Big(\frac{\sqrt{1+\text{x}}-\sqrt{1-\text{x}}}{\sqrt{1+\text{x}}+\sqrt{1+\text{x}}}\Big),$ find $\frac{\text{dy}}{\text{dx}}.$
AnswerHere, $\text{y}=\tan^{-1}\Big(\frac{\sqrt{1+\text{x}}-\sqrt{1-\text{x}}}{\sqrt{1+\text{x}}+\sqrt{1+\text{x}}}\Big)$
Put $\text{x}=\cos2\theta$
$\therefore \text{y}=\tan^{-1}\Big(\frac{\sqrt{1+\cos2\theta}-\sqrt{1-\cos2\theta}}{\sqrt{1+\cos2\theta}+\sqrt{1-\cos2\theta}}\Big)$
$=\tan^{-1}\Big(\frac{\sqrt{2\cos^2\theta}-\sqrt{2\sin^2\theta}}{\sqrt{2\cos^2\theta}+\sqrt{2\sin^2\theta}}\Big)$
$=\tan^{-1}\Big(\frac{\sqrt{2}(\cos\theta-\sin\theta)}{\sqrt{2}(\cos\theta+\sin\theta)}\Big)$
$=\tan^{-1}\bigg(\frac{\frac{\cos\theta-\sin\theta}{\cos\theta}}{\frac{\cos\theta+\sin\theta}{\cos\theta}}\bigg)$
[Dividing numerator and denomainator by $\cos\theta$]
$=\tan^{-1}\bigg(\frac{\frac{\cos\theta}{\cos\theta}-\frac{\sin\theta}{\cos\theta}}{\frac{\cos\theta}{\cos\theta}+\frac{\sin\theta}{\cos\theta}}\bigg)$
$=\tan^{-1}\Big(\frac{1-\tan\theta}{1+\tan\theta}\Big)$
$=\tan^{-1}\bigg(\frac{\tan\frac{\pi}{4}-\tan\theta}{1+\tan\frac{\pi}{4}\times\tan\theta}\bigg)$
$=\tan^{-1}\Big[\tan\big(\frac{\pi}{4}-\theta\big)\Big]$
$=\frac{\pi}{4}-\theta$
$=\frac{\pi}{4}-\frac{1}{2}\cos^{-1}\text{x}\ (\text{Using x}=\cos2\theta)$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=0-\frac{1}{2}\Big(\frac{-1}{\sqrt{1-\text{x}^2}}\Big)$
$\therefore \frac{\text{dy}}{\text{dx}}=\frac{1}{2\sqrt{1-\text{x}^2}}$
View full question & answer→Question 1425 Marks
Differentiate the following functions with respect to x:
$\sqrt{\frac{1+\text{x}}{1-\text{x}}}$
AnswerLet $\text{y}=\sqrt{\frac{1+\text{x}}{1-\text{x}}}$
$\Rightarrow\ \text{y}=\Big(\frac{1+\text{x}}{1-\text{x}}\Big)^\frac{1}{2}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\frac{1+\text{x}}{1-\text{x}}\Big)^\frac{1}{2}$
$=\frac{1}{2}\Big(\frac{1+\text{x}}{1-\text{x}}\Big)^{\frac{1}{2}-1}\frac{\text{d}}{\text{dx}}\Big(\frac{1+\text{x}}{1-\text{x}}\Big)$
[Using chain rule]
$=\frac{1}{2}\Big(\frac{1+\text{x}}{1-\text{x}}\Big)^{\frac{-1}{2}}\bigg[\frac{(1-\text{x})\frac{\text{d}}{\text{dx}}(1+\text{x})-(1+\text{x})\frac{\text{d}}{\text{dx}}(1-\text{x})}{(1-\text{x})^2}\bigg]$
[Using chain rule]
$=\frac{1}{2}\Big(\frac{1+\text{x}}{1-\text{x}}\Big)^{\frac{1}{2}}\bigg[\frac{(1-\text{x})(1)-(1+\text{x})(-1)}{(1-\text{x})^2}\bigg]$
$=\frac{1}{2}\Big(\frac{1+\text{x}}{1-\text{x}}\Big)^{\frac{1}{2}}\bigg[\frac{1-\text{x}+1+\text{x}}{(1-\text{x})^2}\bigg]$
$=\frac{1}{2}\frac{1+\text{x}^\frac{1}{2}}{1-\text{x}^\frac{1}{2}}^{\frac{1}{2}}\times\frac{1}{(1-\text{x})^2}$
$=\frac{1}{\sqrt{1+\text{x}}(1-\text{x})^\frac{3}{2}}$
So,
$\frac{\text{d}}{\text{dx}}\Big(\sqrt{\frac{1+\text{x}}{1-\text{x}}}\Big)=\frac{1}{\sqrt{1+\text{x}}(1-\text{x})^\frac{3}{2}}$
View full question & answer→Question 1435 Marks
Differentiate the following functions with respect to x:
$\sqrt{\tan^{-1}\big(\frac{\text{x}}{2}\big)}$
AnswerLet, $\text{y}=\sqrt{\tan^{-1}\big(\frac{\text{x}}{2}\big)}$
$\Rightarrow\ \text{y}=\Big(\tan^{-1}\big(\frac{\text{x}}{2}\big)\Big)^\frac{1}{2}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\tan^{-1}\big(\frac{\text{x}}{2}\big)\Big)^\frac{1}{2}$
$=\frac{1}{2}\Big(\tan^{-1}\frac{\text{x}}{2}\Big)^{\frac{1}{2}-1}\frac{\text{d}}{\text{dx}}\Big(\tan^{-1}\frac{\text{x}}{2}\Big)$
$=\frac{1}{2}\Big(\tan^{-1}\frac{\text{x}}{2}\Big)^{\frac{-1}{2}}\times\frac{1}{1+\big(\frac{\text{x}}{2}\big)^2}\times\frac{\text{d}}{\text{dx}}\big(\frac{\text{x}}{2}\big)$
$=\frac{4}{4\sqrt{\tan^{-1}\big(\frac{\text{x}}{2}\big)}\big(4+\text{x}^2\big)}$
$=\frac{1}{\big(4+\text{x}^2\big)\sqrt{\tan^{-1}\big(\frac{\text{x}}{2}\big)}}$
So,
$\frac{\text{d}}{\text{dx}}\bigg(\sqrt{\tan^{-1}\big(\frac{\text{x}}{2}\big)}\bigg)=\frac{1}{\big(4+\text{x}^2\big)\sqrt{\tan^{-1}\big(\frac{\text{x}}{2}\big)}}$
View full question & answer→Question 1445 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{\sqrt{\text{x}}+\sqrt{\text{a}}}{1-\sqrt{\text{xa}}}\Big)$
AnswerLet $\text{y}=\tan^{-1}\Big(\frac{\sqrt{\text{x}}+\sqrt{\text{a}}}{1-\sqrt{\text{xa}}}\Big)$$\text{y}=\tan^{-1}\sqrt{\text{x}}+\tan^{-1}\sqrt{\text{a}}$
$\Big[\text{Since},\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\frac{\text{x}+\text{y}}{1-\text{xy}}\Big]$
Differentiating it with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\tan^{-1}\sqrt{\text{x}})+\frac{\text{d}}{\text{dx}}(\tan^{-1}\sqrt{\text{a}})$
$=\frac{1}{1+\big(\sqrt{\text{x}}\big)^2}\times\frac{\text{d}}{\text{dx}}\big(\sqrt{\text{x}}\big)+0$
$=\Big(\frac{1}{1+\text{x}}\Big)\Big(\frac{1}{2\sqrt{\text{x}}}\Big)$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2\sqrt{\text{x}}(1+\text{x})}$
View full question & answer→Question 1455 Marks
If $\sin^2\text{y}+\cos\text{xy}=\text{k},$ find $\frac{\text{dy}}{\text{dx}}$ at $\text{x}=1,\text{y}=\frac{\pi}{4}$
AnswerHere, $\text{e}^{\text{x}}+\text{e}^\text{y}=\text{e}^{\text{x}+\text{y}}$
Differentiating with respect to x using chain rule,
$\Rightarrow \frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}}\big)+\frac{\text{d}}{\text{dx}}\text{e}^{\text{y}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}+\text{y}}\big)$
$\Rightarrow \text{e}^\text{x}+\text{e}^{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}\frac{\text{d}}{\text{dy}}(\text{a}+\text{y})$
$\Rightarrow \text{e}^{\text{x}}+\text{e}^\text{y}\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}\Big[1+\frac{\text{dy}}{\text{dx}}\Big]$
$\Rightarrow\text{e}^{\text{x}}\frac{\text{dy}}{\text{dx}}-\text{e}^{\text{x}+\text{y}}\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}-\text{e}^{\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\big(\text{e}^{\text{y}}-\text{e}^{\text{x}+\text{y}}\big)=\text{e}^{\text{x}+\text{y}}-\text{e}^{\text{x}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\Big(\frac{\text{e}^\text{x}\times\text{e}^\text{y}-\text{e}^\text{x}}{\text{e}^\text{y}-\text{e}^\text{x}\times\text{e}^\text{y}}\Big)$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\frac{\text{e}^\text{x}\big(\text{e}^\text{y}-1\big)}{\text{e}^\text{y}\big({1-\text{e}}^\text{x}\big)}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=-\frac{\text{e}^\text{x}\big(\text{e}^\text{y}-1\big)}{\text{e}^\text{y}\big({\text{e}^\text{x}-1}\big)}$
View full question & answer→Question 1465 Marks
Differentiate the following with respect to x:
$\cos^{-1}\Big(\frac{1-\text{x}}{1+\text{x}}\Big)$
AnswerLet $\text{y}=\cot^{-1}\Big(\frac{1-\text{x}}{1+\text{x}}\Big)$
Put $\text{x}=\tan\theta,\text{ So}$
$\text{y}=\cot^{-1}\Big(\frac{1-\tan\theta}{1+\tan\theta}\Big)$
$=\cot^{-1}\bigg(\frac{\tan\frac{\pi}{4}-\tan\theta}{1+\tan\frac{\pi}{4}\tan\theta}\bigg)$
$=\cot^{-1}\Big[\tan\Big(\frac{\pi}{4}-\theta\Big)\Big]$
$=\cot^{-1}\Big[\cot\Big(\frac{\pi}{2}-\frac{\pi}{4}+\theta\Big)\Big]$
$=\frac{\pi}{4}+\theta$
$\text{y}=\frac{\pi}{4}+\tan^{-1}\text{x}\ \big[\text{Since x}=\tan\theta\big]$
Differentiating it with respect do x,
$\frac{\text{dy}}{\text{dx}}=0+\frac{1}{1+\text{x}^2}$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{1+\text{x}^2}$
View full question & answer→Question 1475 Marks
If the derivative of $\tan^{-1} (a + bx)$ takes the value 1 at $x = 0$, prove that $1 + a^2 = b.$
AnswerHere, $\frac{\text{d}}{\text{dx}}\big[\tan^{-1}(\text{a}+\text{bx})\big]=1\text{ at x}=0$
So, using chain rule,
$\Big[\Big\{\frac{1}{1+(\text{a}+\text{bx})^2}\Big\}\frac{\text{d}}{\text{dx}}(\text{a}+\text{bx})\Big]_{\text{x}=0}=0$
$\Big[\frac{1}{1+(\text{a}+\text{bx})^2}\times(\text{b})\Big]_{\text{x}=0}=1$
$\Rightarrow \frac{\text{b}}{1+(\text{a}+0)^2}=1$
$\Rightarrow \text{b}=1+\text{a}^2$
View full question & answer→Question 1485 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=\sin\text{x}\sin2\text{x}\sin3\text{x}\sin4\text{x}$
AnswerHere,
$\text{y}=\sin\text{x}\cdot\sin2\text{x}\cdot\sin3\text{x}\cdot\sin4\text{x}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=\log(\sin\text{x}\cdot\sin2\text{x}\cdot\sin3\text{x}\cdot\sin4\text{x})$
$\log\text{y}=\log\sin\text{x}+\log\sin2\text{x}+\log\sin3\text{x}+\log\sin4\text{x}$
Differentiating it with respect to x using chain rule,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\log\sin\text{x}+\frac{\text{d}}{\text{dx}}\log\sin2\text{x}+\frac{\text{d}}{\text{dx}}\log\sin3\text{x}+\frac{\text{d}}{\text{dx}}\log\sin4\text{x}$
$=\frac{1}{\sin\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x})+\frac{1}{\sin2\text{x}}\frac{\text{d}}{\text{dx}}(\sin2\text{x})+\frac{1}{\sin3\text{x}}\frac{\text{d}}{\text{dx}}(\sin3\text{x})+\frac{1}{\sin4\text{x}}\frac{\text{d}}{\text{dx}}(\sin4 \text{x})$
$=\frac{1}{\sin\text{x}}(\cos\text{x})+\frac{1}{\sin2\text{x}}(\cos2\text{x})\frac{\text{d}}{\text{dx}}(2\text{x}) \\ +\frac{1}{\sin3\text{x}}(\cos3\text{x})\frac{\text{d}}{\text{dx}}(3\text{x})+\frac{1}{\sin4\text{x}}(\cos4\text{x})\frac{\text{d}}{\text{dx}}(4\text{x})$
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\big[\cot\text{x}+\cot2\text{x}(2)+\cot3\text{x}(3)+\cot4\text{x}(4)\big]$
$\frac{\text{dy}}{\text{dx}}=\text{y}\big[\cot\text{x}+2\cot2\text{x}+3\cot\text{x}3\text{x}+4\cot4\text{x}\big]$
$\frac{\text{dy}}{\text{dx}}=(\sin\text{x}\sin2\text{x}\sin3\text{x}\sin4\text{x}) \\ \big[\cot\text{x}+2\cot2\text{x}+3\cot\text{x}3\text{x}+4\cot4\text{x}\big]$
[Using equation (i)]
View full question & answer→Question 1495 Marks
If $\text{y}=(\cos\text{x})^{\cos\text{x}^{\cos\text{x}^{.....\infty}}},$ prove that $\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}^2\tan\text{x}}{(1-\text{y}\log\cos\text{x})}$
AnswerHere,
$\text{y}=(\cos\text{x})^{\cos\text{x}^{\cos\text{x}^{.....\infty}}}$
$\text{y}=(\cos\text{x})^\text{y}$
Taking log on both the sides,
$\log\text{y}=\log(\cos\text{x})^\text{y}$
$\log\text{y}=\text{y}\log(\cos\text{x}),\big\{\text{Since},\log\text{a}^\text{b}=\text{b}\log\text{a}\big\}$
Differentiating it with resepect to x using product rule and chain rule,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{y}\frac{\text{d}}{\text{dx}}\log(\cos\text{x})+\log\cos\text{x}\frac{\text{dy}}{\text{dx}}$
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{y}\Big(\frac{1}{\cos\text{x}}\Big)\frac{\text{d}}{\text{dx}}(\cos\text{x})+\log\cos\text{x}\frac{\text{dy}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}\Big(\frac{1}{\text{y}}-\log\cos\text{x}\Big)=\frac{\text{y}}{\cos\text{x}}(-\sin\text{x})$
$\frac{\text{dy}}{\text{dx}}\Big(\frac{1-\log\cos\text{x}}{\text{y}}\Big)=-\text{y}\tan\text{x}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}^2\tan\text{x}}{(1-\log\cos\text{x})}$
View full question & answer→Question 1505 Marks
Differentiate the following functions with respect to x:
$\sin^{-1}\Big(\frac{1}{\sqrt{1+\text{x}^2}}\Big)$
AnswerLet $\text{f(x)}=\sin^{-1}\Big(\frac{2^{\text{x}+1}}{1+4^\text{x}}\Big)$ To find the domain, we need to find all x such that $-1\leq\frac{2^{\text{x}+1}}{1+4^\text{x}}\leq1$ Since the quantity in the middle is always psitive, we need to find all x such that $\frac{2^{\text{x}+1}}{1+4^\text{x}}\leq1$ i.e. all x such that $2^{\text{x}+1}\leq1+4^\text{x}$ We may req. write as $2\leq\frac{1}{2^\text{x}}+2^\text{x},$ which is true for all x Hence, the function is defined at all real numbers. Putting $2^\text{x}=\tan\theta$$\text{f(x)}=\sin^{-1}\Big(\frac{2^{\text{x}+1}}{1+4^\text{x}}\Big)=\sin^{-1}\Big(\frac{2^\text{x}.2}{1+(2^\text{x})^2}\Big)$
$=\sin^{-1}\Big[\frac{2\tan\theta}{1+\tan^2\theta}\Big]=\sin^{-1}(\sin2\theta)=2\theta=2\tan^{-1}(2^\text{x})$ Thus, $\text{f(x)}=2\frac{1}{1+(2^\text{x})^2}\frac{\text{d}}{\text{dx}}(2^\text{x})$ $=\frac{2}{1+4^\text{x}}(2^\text{x})\log2=\frac{2^{\text{x}+1}\log2}{1+4^\text{x}}$
View full question & answer→Question 1515 Marks
If $\text{y}=\text{e}^{\text{x}^{\text{e}^\text{x}}}+\text{x}^{\text{e}^{\text{e}^\text{x}}}+\text{e}^{\text{x}^{\text{x}^{\text{e}}}},$ prove that $\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}^{\text{e}^\text{x}}}\times\text{x}^{\text{e}^{\text{x}}}\Big\{\frac{\text{e}^\text{x}}{\text{x}}+\text{e}^\text{x}\log\text{x}\Big\}+\text{e}^{\text{x}^{\text{e}^{\text{x}}}}\times\text{e}^{\text{e}^\text{x}}\Big\{\frac{1}{\text{x}}+\text{e}^\text{x}\times\log\text{x}\Big\}+\text{e}^{\text{x}^{\text{x}^\text{e}}}\text{x}^{\text{x}^{\text{e}}}\times\text{x}^{\text{e}-1}\Big\{\text{x}+\text{e}\log\text{x}\Big\}$
AnswerWe have, $\text{y}=\text{e}^{\text{x}^{\text{e}^\text{x}}}+\text{x}^{\text{e}^{\text{e}^\text{x}}}+\text{e}^{\text{x}^{\text{x}^{\text{e}}}}$
$\Rightarrow\text{y}=\text{u}+\text{v}+\text{w}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{du}}{\text{dx}}+\frac{\text{dv}}{\text{dx}}+\frac{\text{dw}}{\text{dx}}\ .....(\text{i})$
Where $\text{u}=\text{e}^{\text{x}^{\text{e}^{\text{x}}}},\text{v}=\text{x}^{\text{e}^{\text{e}^{\text{x}}}}\text{ and w}=\text{e}^{\text{x}^{\text{x}^{\text{e}}}}$
Now, $\text{u}=\text{e}^{\text{x}^{\text{e}^{\text{x}}}}\ .....(\text{ii})$
Taking log on both sides,
$\log\text{u}=\log\text{e}^{\text{x}^{\text{e}^{\text{x}}}}$
$\Rightarrow\log\text{u}=\text{x}^{\text{e}^\text{x}}\log\text{e}$
$\Rightarrow\log\text{u}=\text{x}^{\text{e}^\text{x}}\ .....(\text{iii})$
Taking $\log$ on both sides,
$\log\log\text{u}=\log\text{x}^{\text{e}^\text{x}}$
$\Rightarrow\log\log\text{u}=\text{e}^\text{x}\log\text{x}$
Differentiating with respect to x,
$\Rightarrow\frac{1}{\log\text{u}}\frac{\text{d}}{\text{dx}}(\log\text{u})=\text{e}^\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{e}^\text{x})$
$\Rightarrow\frac{1}{\log\text{u}}\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\frac{\text{e}^\text{x}}{\text{x}}+\text{e}^\text{x}\log\text{x}$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{u}\log\text{u}\Big[\frac{\text{e}^\text{x}}{\text{x}}+\text{e}^\text{x}\log\text{x}\Big]$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{e}^{\text{x}^{\text{e}^{\text{x}}}}\times\text{x}^{\text{e}^\text{x}}\Big[\frac{\text{e}^\text{x}}{\text{x}}+\text{e}^\text{x}\log\text{x}\Big]\ .....(\text{A})$
[Using equation (ii) and (iii)]
Now, $\text{v}=\text{x}^{\text{e}^{\text{e}^\text{x}}}\ .....(\text{iv})$
Taking log on both sides,
$\log\text{v}=\log\text{x}^{\text{e}^{\text{e}^\text{x}}}$
$\Rightarrow\log\text{v}=\text{e}^{\text{e}^\text{x}}\log\text{x}$
$\Rightarrow\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\text{e}^{\text{e}^\text{x}}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{e}^\text{x}}\big)$
$\Rightarrow\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\text{e}^{\text{e}^\text{x}}\big(\frac{1}{\text{x}}\big)+\log\text{xe}^{\text{e}^\text{x}}\frac{\text{d}}{\text{dx}}(\text{e}^\text{x})$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\text{v}\Big[\text{e}^{\text{e}^\text{x}}\big(\frac{1}{\text{x}}\big)+\log\text{xe}^{\text{e}^\text{x}}\text{e}^\text{x}\Big]$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\text{e}^{\text{e}^{\text{e}^\text{x}}}\times\text{e}^{\text{e}^\text{x}}\Big[\frac{1}{\text{x}}+\text{e}^\text{x}\log\text{x}\Big]\ .....(\text{B})$
[Using equation (4)]
Now, $\text{w}=\text{e}^{\text{x}^{\text{x}^{\text{e}}}}\ .....(\text{v})$
Taking log on sides,
$\log\text{w}=\log\text{e}^{\text{x}^{\text{x}^{\text{e}}}}$
$\Rightarrow\log\text{w}=\text{x}^{\text{x}^\text{e}}\log\text{e}$
$\Rightarrow\log\text{w}=\text{x}^{\text{x}^{\text{e}}}\ .....(\text{vi})$
Taking log on both sides,
$\log\log\text{w}=\log\text{x}^{\text{x}^{\text{e}}}$
$\Rightarrow\log\log\text{w}=\text{x}^{\text{e}}\log\text{x}$
$\Rightarrow\frac{1}{\log\text{w}}\frac{\text{d}}{\text{dx}} (\log\text{w})=\text{x}^\text{e}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x}^\text{e})$
$\Rightarrow\frac{1}{\log\text{w}}\big(\frac{1}{\text{w}}\big)\frac{\text{dw}}{\text{dx}}=\text{x}^{\text{e}}\big(\frac{1}{\text{x}}\big)\log\text{xex}^{\text{e}-1}$
$\Rightarrow\frac{\text{dw}}{\text{dx}}=\text{w}\log\text{w}\big[\text{x}^{\text{e}-1}+\text{e}\log\text{xx}^{\text{e}-1}\big]$
$\Rightarrow\frac{\text{dw}}{\text{dx}}=\text{e}^{\text{x}^{\text{x}^\text{e}}}\text{x}^{\text{x}^\text{e}}\text{x}^{\text{e}-1}(1+\text{e}\log\text{x})\ .....(\text{C})$
Using equation (A), (B) and (C) in equation (i), we get
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}^{\text{e}^{\text{x}}}}\times\text{x}^{\text{e}^\text{x}}\Big[\frac{\text{e}^\text{x}}{\text{x}}+\text{e}^\text{x}\log\text{x}\Big] + \text{e}^{\text{e}^{\text{e}^\text{x}}}\times\text{e}^{\text{e}^\text{x}}\Big[\frac{1}{\text{x}}+\text{e}^\text{x}\log\text{x}\Big] + \text{e}^{\text{x}^{\text{x}^\text{e}}}\text{x}^{\text{x}^\text{e}}\text{x}^{\text{e}-1}(1+\text{e}\log\text{x})$
View full question & answer→Question 1525 Marks
If $\text{y}=\sin\Big[2\tan^{-1}\Big\{\sqrt{\frac{1-\text{x}}{1+\text{x}}}\Big\}\Big],$ find $\frac{\text{dy}}{\text{dx}}.$
AnswerHere, $\text{y}=\sin\Big[2\tan^{-1}\Big[\sqrt{\frac{1-\text{x}}{1+\text{x}}}\Big]\Big]$
Put, $\text{x}=\cos2\theta, \text{So},$
$\text{y}=\sin\Big[2\tan^{-1}\sqrt{\frac{1-\cos2\theta}{1+\cos2\theta}}\Big]$
$=\sin\Big[2\tan^{-1}\sqrt{\frac{2\sin^2\theta}{2\cos^2\theta}}\Big]$
$=\sin\big[2\tan^{-1}\sqrt{\tan^2\theta}\big]$
$=\sin\big[2\tan^{-1}(\tan\theta)\big]$
$=\sin{2\theta}$
$=\sin\Big[2\times\frac{1}{2}\cos^{-1}\text{x}\Big]\ \big[\text{Since, x}=\cos2\theta\big]$
$=\sin\big(\sin^{-1}\sqrt{1-\text{x}^2}\big)$
$\text{y}=\sqrt{1-\text{x}^2}$
Differentiating with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2\sqrt{1-\text{x}^2}}\frac{\text{d}}{\text{dx}}(1-\text{x}^2)$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2\sqrt{1-\text{x}^2}}(-2\text{x})$
$\frac{\text{dy}}{\text{dx}}=\frac{-\text{x}}{\sqrt{1-\text{x}^2}}$
View full question & answer→Question 1535 Marks
Differentiate the following functions with respect to x:
$\frac{\text{e}^\text{x}\sin\text{x}}{(\text{x}^2+2)^3}$
AnswerConsider $\text{y}=\frac{\text{e}^\text{x}\sin\text{x}}{(\text{x}^2+2)^3}$
Differentiating it with respect to x and applying the chain and product rule, we get
$\frac{\text{dy}}{\text{dx}}=\frac{(\text{x}^2+2)^3\frac{\text{d}}{\text{dx}}(\text{e}^\text{x}\sin\text{x})-\text{e}^\text{x}\sin\text{x}\frac{\text{d}}{\text{dx}}(\text{x}^2+2)^3}{\big[(\text{x}^2+2)^3\big]^2}$
$=\frac{(\text{x}^2+2)^3[\text{e}^\text{x}\cos\text{x}+\sin\text{xe}^\text{x}]-\text{e}^\text{x}\sin\text{x}3(\text{x}^2+2)^2(2\text{x})}{(\text{x}^2+2)^6}$
$=\frac{(\text{x}^2+2)^3[\text{e}^\text{x}\cos\text{x}+\sin\text{xe}^\text{x}]-6\text{xe}^\text{x}\sin\text{x}(\text{x}^2+2)^2}{(\text{x}^2+2)^6}$
$=\frac{(\text{x}^2+2)^2\big[(\text{x}^2+2)(\text{e}^\text{x}\cos\text{x}+\sin\text{xe}^\text{x})-6\text{xe}^\text{x}\sin\text{x}}{(\text{x}^2+2)^6}$
$=\frac{\text{x}^2\text{e}^\text{x}\cos\text{x}+\text{x}^2\sin\text{xe}^\text{x}+2\text{e}^\text{x}\cos\text{x}+2\sin\text{xe}^\text{x}-6\text{xe}^\text{x}\sin\text{x}}{(\text{x}^2+2)^4}$
$=\frac{\text{e}^\text{x}\sin\text{x}}{(\text{x}^2+2)^3}+\frac{\text{e}^\text{x}\cos\text{x}}{(\text{x}^2+2)^3}-\frac{6\text{xe}^\text{x}\sin\text{x}}{(\text{x}^2+2)^4}$
Therefore,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^\text{x}\sin\text{x}}{(\text{x}^2+2)^3}+\frac{\text{e}^\text{x}\cos\text{x}}{(\text{x}^2+2)^3}-\frac{6\text{xe}^\text{x}\sin\text{x}}{(\text{x}^2+2)^4}$
View full question & answer→Question 1545 Marks
Differentiate the following functions with respect to x:
$\sqrt{\frac{\text{a}^2-\text{x}^2}{\text{a}^2+\text{x}^2}}$
AnswerLet $\text{y}=\sqrt{\frac{\text{a}^2-\text{x}^2}{\text{a}^2+\text{x}^2}}$
$\Rightarrow\text{y}=\Big({\frac{\text{a}^2-\text{x}^2}{\text{a}^2+\text{x}^2}}\Big)^\frac{1}{2}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big({\frac{\text{a}^2-\text{x}^2}{\text{a}^2+\text{x}^2}}\Big)^\frac{1}{2}$
$=\frac{1}{2}\Big({\frac{\text{a}^2-\text{x}^2}{\text{a}^2+\text{x}^2}}\Big)^{\frac{1}{2}-1}\times\frac{\text{d}}{\text{dx}}\Big(\frac{\text{a}^2-\text{x}^2}{\text{a}^2+\text{x}^2}\Big)$
[Using chain rule]
$=\frac{1}{2}\Big({\frac{\text{a}^2-\text{x}^2}{\text{a}^2+\text{x}^2}}\Big)^{\frac{1}{2}}\times\Bigg\{\frac{\big(\text{a}^2+\text{x}^2\big)\frac{\text{a}}{\text{dx}}\big(\text{a}^2-\text{x}^2\big)-\big(\text{a}^2-\text{x}^2\big)\frac{\text{d}}{\text{dx}}(\text{a}^2+\text{x}^2)}{\big(\text{a}^2+\text{x}^2\big)}\Bigg\}$
[Using chain rule]
$=\frac{1}{2}\Big({\frac{\text{a}^2-\text{x}^2}{\text{a}^2+\text{x}^2}}\Big)^{\frac{1}{2}}\Bigg\{\frac{-2\text{x}\big(\text{a}^2+\text{x}^2\big)-2\text{x}\big(\text{a}^2-\text{x}^2\big)}{\big(\text{a}^2+\text{a}^2\big)^2}\Bigg\}$
$=\frac{1}{2}\Big({\frac{\text{a}^2-\text{x}^2}{\text{a}^2+\text{x}^2}}\Big)^{\frac{1}{2}}\bigg\{\frac{-2\text{xa}^2-2\text{x}^3-2\text{xa}^2+2\text{x}^3}{\big(\text{a}^2+\text{a}^2\big)^2}\bigg\}$
$ =\frac{1}{2}\Big({\frac{\text{a}^2-\text{x}^2}{\text{a}^2+\text{x}^2}}\Big)^{\frac{1}{2}}\Bigg(\frac{-4\text{xa}^2}{\big(\text{a}^2+\text{x}^2\big)^3}\Bigg)$
$=\frac{-2\text{xa}^2}{\sqrt{\text{a}^2-\text{x}^2}\big(\text{a}^2+\text{x}^2\big)^{\frac{3}{2}}}$
So,
$\frac{\text{d}}{\text{dx}}\bigg(\sqrt{\frac{\text{a}^2-\text{x}^2}{\text{a}^2+\text{x}^2}}\bigg)=\frac{-2\text{xa}^2}{\sqrt{\text{a}^2-\text{x}^2}\big(\text{a}^2+\text{x}^2\big)^{\frac{3}{2}}}$
View full question & answer→Question 1555 Marks
Differentiate the following functions with respect to x:
$\text{e}^{\text{x}\log\text{x}}$
AnswerLet $\text{y}=\text{e}^{\text{x}\log\text{x}}$
$\Rightarrow\ \text{y}=\text{e}^{\log\text{x}^\text{x}} \ \big[\text{Since}, \log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
$\Rightarrow\text{y}=\text{x}^{\text{x}}\ .....(\text{i})\ \big[\text{Since, e}^{\log\text{a}}=\text{a}\big]$
Taking log on both the sides,
$\log\text{y}=\log\text{x}^\text{x}$
$\log\text{y}=\text{x}\log\text{x}$
Differentiating with respect to x, using product rule,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x})$
$=\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}(1)$
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=1+\log\text{x}$
$\frac{\text{dy}}{\text{dx}}=\text{y}[1+\log\text{x}]$
$\frac{\text{dy}}{\text{dx}}=\text{x}^\text{x}(1+\log\text{x})$
[Using equation (i)]
View full question & answer→Question 1565 Marks
Find $\frac{\text{dy}}{\text{dx}},$ when
$\text{x}=\text{a}(\theta+\sin\theta)$ and $\text{y}=\text{a}(1-\cos\theta)$
AnswerHere, $\text{x}=\text{a}(\theta+\sin\theta)$Differentiating it with respect to $\theta$
$\frac{\text{dx}}{\text{d}\theta}=\text{a}(1+\cos\theta).....(\text{i}) $
And, $\text{y}=\text{a}(1-\cos\theta)$
Differentiating it with respect to $\theta$,
$ \frac{\text{dy}}{\text{d}\theta}=\text{a}(\theta+\sin\theta)$
and
$\frac{\text{dy}}{\text{d}\theta}=\text{a}\sin\theta...(\text{ii}) $
Using equation (i) and (ii),
$=\frac{\text{a}\sin\theta}{\text{a}(1-\cos\theta)} $
$=\frac{\frac{2\sin\theta}{2}\frac{\cos\theta}{2}}{\frac{2\sin^{2}\theta}{2}}, \begin{Bmatrix} \text{Since, }1-\cos\theta=\frac{2\sin^{2\theta}}{2}\\\frac{2\sin\theta}{2}\frac{\cos\theta}{2}=\sin\theta \end{Bmatrix}$
$=\frac{\text{dy}}{\text{dx}}=\frac{\tan}{2}$
View full question & answer→Question 1575 Marks
If $\text{u}=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$ and $\text{v}=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big),$ where -1 < x < 1 then write the value of $\frac{\text{dy}}{\text{dx}}.$
Answer$\text{u}=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$ and $\text{v}=\tan^{-1}\Big(\frac{2\text{x}}{1+\text{x}^3}\Big)$
We know, $\frac{\text{du}}{\text{dx}}=\frac{2}{1+\text{x}^2}$
Using the chain rule of differentiation,
$\frac{\text{dv}}{\text{dx}}=\frac{1}{1+\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)^2}\cdot\frac{(1+\text{x}^2)\cdot(2\text{x})'-(1+\text{x}^2)'\cdot(2\text{x})}{(1+\text{x}^2)^2}$
$=\frac{(1+\text{x}^2)^2}{(1+\text{x}^2)^2+(2\text{x})^2}\cdot\frac{2(1+\text{x}^2)-(2\text{x})(2\text{x})}{(1+\text{x}^2)^2}$
$=\frac{2(1-\text{x}^2)}{(1+\text{x}^2)^2+(2\text{x})^2}$
Using Chain Rule of Differentiation,
$\frac{\text{du}}{\text{dv}}=\frac{\text{du}}{\text{dx}}\cdot\frac{\text{dx}}{\text{dv}}$
$=\frac{2}{1+\text{x}^2}\cdot\frac{(1+\text{x}^2)^2+(2\text{x})^2}{2(1-\text{x})^2}$
$=\frac{(1+\text{x}^2)^2+(2\text{x})^2}{(1+\text{x}^2)(1-\text{x}^2)}$
Dividing numerator and denominator by $(1 + x^2)^2,$
$\frac{\text{du}}{\text{dv}}=\frac{1+\big(\frac{2\text{x}}{1+\text{x}^2}\big)}{\frac{1-\text{x}^2}{1+\text{x}^2}}$
$=\frac{1+\sin^2\text{u}}{\cos\text{u}}$
$=\sec\text{u}(1+\tan\text{u})$
View full question & answer→Question 1585 Marks
Differentiate the following functions with respect to x:
$\text{x}^{\frac{1}{\text{x}}}$
AnswerLet $\text{y}=\text{x}^{\frac{1}{\text{x}}}\ .....(\text{i})$
Taking log on both sides,
$\log\text{y}=\log\text{x}^{\frac{1}{\text{x}}}$
$\Rightarrow\log\text{y}=\frac{1}{\text{x}}\log\text{x}\ \big[\because\log\text{a}^{\text{b}}=\text{b}\log\text{a}\big]$
Differentiating with respect to x,
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}\big(\text{x}^{-1}\big)$
[Using product rule]
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}\times\frac{1}{\text{x}}(\log\text{x})\times\Big(-\frac{1}{\text{x}^2}\Big)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}^2}-\frac{\log\text{x}}{\text{x}^2}$
$\Rightarrow \frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{(1-\log\text{x})}{\text{x}^2}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\text{x}^\frac{1}{\text{x}}\Big[\frac{1-\log\text{x}}{\text{x}}\Big]$
View full question & answer→Question 1595 Marks
Differentiate the following functions with respect to x:
$\cos^{-1}\Big\{\frac{\cos\text{x}+\sin\text{x}}{\sqrt{2}}\Big\},-\frac{\pi}{4}<\text{x}<\frac{\pi}{4}$
AnswerLet $\text{y}=\cos^{-1}\Big\{\frac{\cos\text{x}+\sin\text{x}}{\sqrt{2}}\Big\}$
$\text{y}=\cos^{-1}\Big\{\cos\text{x}\Big(\frac{1}{\sqrt{2}}\Big)+\sin\text{x}\Big(\frac{1}{\sqrt{2}}\Big)\Big\}$
$=\cos^{-1}\Big\{\cos\text{x}\cos\Big(\frac{\pi}{4}\Big)+\sin\text{x}\sin\text{x}\Big(\frac{\pi}{4}\Big)\Big\}$
$\text{y}=\cos^{-1}\Big[\cos\Big(\text{x}-\frac{\pi}{4}\Big)\Big]$
Here, $-\frac{\pi}{4}<\text{x}<\frac{\pi}{4}$
$\Rightarrow\ \Big(-\frac{\pi}{4}-\frac{\pi}{4}\Big)<\Big(\text{x}-\frac{\pi}{4}\Big)<\Big(\frac{\pi}{4}-\frac{\pi}{4}\Big)$
$\Rightarrow-\frac{\pi}{2}<\Big(\text{x}-\frac{\pi}{4}\Big)<0$
So, from equation (i),
$\text{y}=-\Big(\text{x}-\frac{\pi}{4}\Big)$
$\Big[\text{Since}, \cos^{-1}(\cos\theta)=-\theta,\text{ if }\theta\in [-\pi, 0]\Big]$
$\text{y}=-\text{x}+\frac{\pi}{4}$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=-1$
View full question & answer→Question 1605 Marks
Find $\frac{\text{dy}}{\text{dx}},$ when
$\text{x}=\text{a}(\cos\theta+\theta\sin\theta)$ and $\text{y}=\text{a}(\sin\theta-\theta\sin\theta-\theta\cos\theta)$
AnswerWe have, $\text{x}=\text{a}(\cos\theta+\theta\sin\theta)$ and $\text{y}=\text{a}(\sin\theta-\theta\sin\theta-\theta\cos\theta)$
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{a}\Big[\frac{\text{d}}{\text{d}\theta}\cos\theta+\frac{\text{d}}{\text{d}\theta}(\theta\sin\theta)\Big] $ and
$\frac{\text{dy}}{\text{d}\theta}=\text{a}\Big[\frac{\text{d}}{\text{d}\theta}\sin\theta-\frac{\text{d}}{\text{d}\theta}(\theta\cos\theta)\Big] $
$ \Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{a}\Big[-\sin\theta+\theta\frac{\text{d}}{\text{d}\theta}(\sin\theta)+\sin\theta\frac{\text{d}}{\text{d}\theta}(\theta)\Big]$ and
$\frac{\text{dy}}{\text{d}\theta}=\text{a}\Big[\cos\theta-\left\{\theta\frac{\text{d}}{\text{d}\theta}(\cos\theta)+\cos\theta\frac{\text{d}}{\text{d}\theta}(\theta)\right\}\Big]$
$ \Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{a}\big[-\sin\theta+\theta\cos\theta\big]$ and
$\frac{\text{dy}}{\text{d}\theta}=\text{a}\big[\cos\theta+\theta\sin\theta-\cos\theta\big]$
$ \Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{a}\theta\cos\theta$ and $\frac{\text{dy}}{\text{d}\theta}=\text{a}\theta\sin\theta$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}=\frac{\text{a}\theta\sin\theta}{\text{a}\theta\cos\theta}=\tan\theta$
View full question & answer→Question 1615 Marks
Find $\frac{\text{dy}}{\text{dx}}$ in the following cases:
$(x + y)^2 = 2axy$
AnswerWe Have,$ (x + y)^2 = 2axy$
Differentiating with respect to x, we get,
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{x}+\text{y}\big)^2=\frac{\text{d}}{\text{dx}}\big(2\text{axy}\big)$
$\Rightarrow2(\text{x}+\text{y})\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})=2\text{a}\Big[\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\frac{\text{d}}{\text{dx}}(\text{x})\Big]$
$\Rightarrow2(\text{x}+\text{y})\Big[1+\frac{\text{dy}}{\text{dx}}\Big]=2\text{a}\Big[\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}(1)\Big]$
$\Rightarrow2(\text{x}+\text{y})+2(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}=2\text{a}\text{x}\frac{\text{dy}}{\text{dx}}+2\text{ay}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\big[2(\text{x}+\text{y})-2\text{a}\text{x}\big]=2\text{ay}-2(\text{x}+\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2[\text{ay}-\text{x}-\text{y}]}{2[\text{x}+\text{y}-\text{a}\text{x}]}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\Big(\frac{\text{ay}-\text{x}-\text{y}}{\text{x}+\text{y}-\text{a}\text{x}}\Big)$
View full question & answer→Question 1625 Marks
Find $\frac{\text{dy}}{\text{dx}}$ in the following cases:
$4\text{x}+3\text{y}=\log\big(4\text{x}-3\text{y}\big)$
AnswerWe have, $4\text{x}+3\text{y}=\log\big(4\text{x}-3\text{y}\big)$
Differentiating with respect to x, we get,
$\frac{\text{d}}{\text{dx}}\big(4\text{x}\big)+\frac{\text{d}}{\text{dx}}\big(3\text{y}\big)=\frac{\text{d}}{\text{dx}}\big\{\log\big(4\text{x}-3\text{y}\big)\big\}$
$\Rightarrow4+3\frac{\text{dy}}{\text{dx}}=\frac{1}{\big(4\text{x}-3\text{y}\big)}\frac{\text{d}}{\text{dx}}\big(4\text{x}-3\text{y}\big)$
$\Rightarrow4+3\frac{\text{dy}}{\text{dx}}=\frac{1}{\big(4\text{x}-3\text{y}\big)}\Big(4-3\frac{\text{dy}}{\text{dx}}\Big)$
$\Rightarrow3\frac{\text{dy}}{\text{dx}}+\frac{3}{\big(4\text{x}-3\text{y}\big)}\frac{\text{dy}}{\text{dx}}=\frac{4}{\big(4\text{x}-3\text{y}\big)}-4$
$\Rightarrow3\frac{\text{dy}}{\text{dx}}\Big\{1+\frac{1}{(4\text{x}-3\text{y})}\Big\}=4\Big\{\frac{1}{(4\text{x}-3\text{y})}-1\Big\}$
$\Rightarrow3\frac{\text{dy}}{\text{dx}}\Big\{\frac{4\text{x}-3\text{y}+1}{(4\text{x}+3\text{y})}\Big\}=4\Big\{\frac{1-4\text{x}-3\text{y}}{4\text{x}-3\text{y}}\Big\}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{4}{3}\Big\{\frac{1-4\text{x}+3\text{y}}{(4\text{x}-3\text{y})}\Big\}\Big(\frac{4\text{x}-3\text{y}}{4\text{x}-3\text{y}+1}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{4}{3}\Big(\frac{1-4\text{x}+3\text{y}}{4\text{x}-3\text{y}+1}\Big)$
View full question & answer→Question 1635 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=\text{x}^{\text{x}}+(\sin\text{x})^\text{x}$
AnswerWe have, $\text{y}=\text{x}^{\text{x}}+(\sin\text{x})^\text{x}$
$\Rightarrow\text{y}=\text{e}^{\log\text{x}^\text{x}}+\text{e}^{\log(\sin\text{x})^\text{x}}$
$\Rightarrow\text{y}=\text{e}^{\text{x}\log\text{x}}+\text{e}^{\text{x}\log\sin\text{x}}$
Differentiating with respect to x using chain rule and product rule,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}\log\text{x}}\big)+\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}\log\sin\text{x}}\big)$
$=\text{e}^{\text{x}\log\text{x}}\frac{\text{d}}{\text{dx}}(\text{x}\log\text{x})+\text{e}^{\text{x}\log\sin\text{x}}\frac{\text{d}}{\text{dx}}(\text{x}\log\sin\text{x})$
$=\text{e}^{\text{x}\log\text{x}}\Big[\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big] \\ +\text{e}^{\log(\sin\text{x})^\text{x}}\Big[\text{x}\frac{\text{d}}{\text{dx}}(\log\sin\text{x})+\log\sin\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big]$
$=\text{x}^{\text{x}}\Big[\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}(1)\Big] \\ +(\sin\text{x})^\text{x}\Big[\text{x}\Big(\frac{1}{\sin\text{x}}\Big)\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}\big]$
$=\text{x}^{\text{x}}\big[1+\log\text{x}\big]+(\sin\text{x})^\text{x}\Big[\text{x}\Big(\frac{1}{\sin\text{x}}\Big)(\cos\text{x})+\log\sin\text{x}\Big]$
$=\text{x}^{\text{x}}\big[1+\log\text{x}\big]+(\sin\text{x})^\text{x}\big[\text{x}\cot\text{x}+\log\sin\text{x}\big]$
View full question & answer→Question 1645 Marks
Differentiate the following functions with respect to x:
$\log(\text{cosec x}-\cot\text{x})$
AnswerConsider $\text{y}=\log(\text{cosec x}-\cot\text{x})$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\log\big(\text{cosec x}-\cot\text{x}\big)$
$=\frac{1}{(\text{cosec x}-\cot\text{x})}\times\big(-\text{cosec x}\cot\text{x}+\text{cosec}^2\text{x}\big)$
[Using chain rule]
$=\frac{1}{(\text{cosec x}-\cot\text{x})}\times\big(-\text{cosec x}\cot\text{x}+\text{cosec}^2\text{x}\big)$
$=\frac{\text{cosec x}(\text{cosec x}-\cos\text{x})}{(\text{cosec x}-\cot\text{x})}$
$=\text{cosec x}$
Hence, the solution is, $\frac{\text{d}}{\text{dx}}\big(\log(\text{cosec x}-\cot\text{x})\big)=\text{cosec x}$
View full question & answer→Question 1655 Marks
Differentiate the following functions with respect to x:
$\text{x}\sin2\text{x}+5^{\text{x}}+\text{k}^\text{k}+(\tan^2\text{x})^3$
AnswerLet $\text{y}=\text{x}\sin2\text{x}+5^{\text{x}}+\text{k}^\text{k}+(\tan^2\text{x})^3$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[\text{x}\sin2\text{x}+5^{\text{x}}+\text{k}^\text{k}+(\tan^2\text{x})^3\big]$
$=\frac{\text{d}}{\text{dx}}(\text{x}\sin2\text{x})+\frac{\text{d}}{\text{dx}}(5^\text{x})+\frac{\text{d}}{\text{dx}}(\text{k}^\text{k})+\frac{\text{d}}{\text{dx}}\big(\tan^6\text{x}\big)$
$=\Big[\text{x}\frac{\text{d}}{\text{dx}}(\sin2\text{x})+\sin2\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big] \\ +5^\text{x}\log5+0+6\tan^5\text{x}\frac{\text{d}}{\text{dx}}(\tan\text{x})$
[Using product rule and chain rule]
$=\Big[\text{x}\cos2\text{x}\frac{\text{d}}{\text{dx}}(2\text{x})+\sin2\text{x}\Big]+5^\text{x}\log5+6\tan^5\text{x}\sec^2\text{x}$
$=2\text{x}\cos2\text{x}+\sin2\text{x}+5^\text{x}\log5+6\tan^5\text{x}\sec^2\text{x}$
So,
$\frac{\text{d}}{\text{dx}}\big(\text{x}\sin2\text{x}+5^\text{x}+\text{k}^\text{k}+(\tan^2\text{x})^3\big) \\ =2\text{x}\cos2\text{x}+\sin2\text{x}+5^\text{x}\log5+6\tan^5\text{x}\sec^2\text{x}$
View full question & answer→Question 1665 Marks
If $\text{y}=1+\frac{\alpha}{\big(\frac{1}{\text{x}}-\alpha\big)}+\frac{\frac{\beta}{\text{x}}}{\big(\frac{1}{\text{x}}-\alpha\big)\big(\frac{1}{\text{x}}-\beta\big)}+\frac{\frac{\gamma}{\text{x}^2}}{\big(\frac{1}{\text{x}}-\alpha\big)\big(\frac{1}{\text{x}}-\beta\big)\big(\frac{1}{\text{x}}-\gamma\big)},$ find $\frac{\text{dy}}{\text{dx}}$
Answer$\text{y}=1+\frac{\alpha}{\big(\frac{1}{\text{x}}-\alpha\big)}+\frac{\frac{\beta}{\text{x}}}{\big(\frac{1}{\text{x}}-\alpha\big)\big(\frac{1}{\text{x}}-\beta\big)}+\frac{\frac{\gamma}{\text{x}^2}}{\big(\frac{1}{\text{x}}-\alpha\big)\big(\frac{1}{\text{x}}-\beta\big)\big(\frac{1}{\text{x}}-\gamma\big)}$
Using the theorem,
If $\text{y}=1+\frac{\text{ax}^2}{(\text{x}-\text{a})(\text{x}-\text{b})(\text{x}-\text{c})}+\frac{\text{bx}}{(\text{x}-\text{b})(\text{x}-\text{c})}+\frac{\text{c}}{(\text{x}-\text{x})}$ then,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}\Big\{\frac{\text{a}}{\text{a}-\text{x}}+\frac{\text{b}}{\text{b}-\text{x}}+\frac{\text{c}}{\text{c}-\text{x}}\Big\}$
Here, we have $\frac{1}{\text{x}}$ instead of x,
So, using above theorem we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\alpha}{\big(\frac{1}{\text{a}}-\alpha\big)}+\frac{\beta}{\big(\frac{1}{\text{x}}-\beta\big)}+\frac{\gamma}{\big(\frac{1}{\text{x}}-\gamma\big)}$
View full question & answer→Question 1675 Marks
Differentiate the following functions with respect to x:
$\text{e}^{\text{ax}}\sec\text{x}\tan2\text{x}$
AnswerLet $\text{y}=\text{e}^{\text{ax}}\sec\text{x}\tan2\text{x}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{e}^{\text{ax}}\sec\text{x}\tan2\text{x})$
$=\text{e}^{\text{ax}}\frac{\text{d}}{\text{dx}}\big\{\sec\text{x}\tan2\text{x}\big\}+\sec\text{x}\tan2\text{x}\frac{\text{d}}{\text{dx}}\big\{\text{e}^{\text{ax}}\big\}$
$=\text{e}^{\text{ax}}\big[\text{sec}\text{x}\tan\text{x}\tan2\text{x}+2\sec^2 2\text{x}\sec\text{x}\big]+\text{ae}^{\text{ax}}\sec\text{a}\tan^{2\text{x}}$
$=\text{ae}^{\text{ax}}\sec\text{x}\tan2\text{x}+\text{e}^{\text{ax }}\sec\text{x}\tan\text{x}\tan2\text{x}+2\text{e}^{\text{ax}}\sec\text{x}\sec^2 2\text{x}$
$=\text{e}^{\text{ax}}\sec\text{x}\big\{\text{a}\tan2\text{x}+\tan\text{x}\tan2\text{x}+2\sec^22\text{x}\big\}$
So,
$\frac{\text{d}}{\text{dx}}(\text{e}^{\text{ax}}\sec\text{x}\tan2\text{x})=\text{e}^{\text{ax}}\sec\text{x}\big\{\text{a}\tan2\text{x}+\tan\text{x}\tan2\text{x}+2\sec^22\text{x}\big\}$
View full question & answer→Question 1685 Marks
Differentiate the following functions with respect to x:
$\sin^{-1}\Big\{\sqrt{\frac{1-\text{x}}{2}}\Big\},0<\text{x}<1$
AnswerLet $\text{y}=\sin^{-1}\Big\{\sqrt{\frac{1-\text{x}}{2}}\Big\}$
Put $\text{x}=\cos2\theta$
$\text{y}=\sin^{-1}\Big\{\sqrt{\frac{1+\cos2\theta}{2}}\Big\}$
$=\sin^{-1}\Big\{\sqrt{\frac{2\sin^2\theta}{2}}\Big\}$
$\text{y}=\sin^{-1}(\sin\theta)\ .....(\text{i})$
Here, $0<\text{x}<1$
$\Rightarrow\ 0<\cos2\theta<1$
$\Rightarrow\ 0<2\theta<\frac{\pi}{2}$
$\Rightarrow\ 0<\theta<\frac{\pi}{4}$
So, from equation (i),
$\text{y}=\theta$
$\Big[\text{Since, } \sin^{-1}(\sin\theta)=\theta\text{ if }\theta \in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\text{y}=\frac{1}{2}\cos^{-1}\text{x}\ \big[\text{Since x}=\cos2\theta\big]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=-\frac{1}{2\sqrt{1-\text{x}^2}}$
View full question & answer→Question 1695 Marks
Differentiate the following functions with respect to x:
$\text{x}^{\sin{\text{x}}}$
AnswerLet $\text{y}=\text{x}^{\sin{\text{x}}}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=\log\text{x}^{\sin{\text{x}}}$
$\log\text{y}=\sin\text{x}\log\text{x}\ \big[\text{Since,}\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating with respect to x,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\sin\text{x}\frac{\text{d}}{\text{dx}}\log\text{x}+\log\text{x}\frac{\text{d}}{\text{dx}}\sin\text{x}$
[Using product rule]
$\frac{1}{\text{y}}\frac{\text{dt}}{\text{dx}}=\sin\text{x}\big(\frac{1}{\text{x}}\big)+\log\text{x}(\cos\text{x})$
$\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{\sin\text{x}}{\text{x}}+(\log\text{x})(\cos\text{x})\Big]$
Put the value of y,
$\frac{\text{dy}}{\text{dx}}=\text{x}^{\sin\text{x}}\Big[\frac{\sin\text{x}}{\text{x}}+(\log\text{x})(\cos\text{x})\Big]$
View full question & answer→Question 1705 Marks
Differentiate $\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)$ with respect to $\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big),$ if 0 < x < 1.
AnswerLet $\text{u}=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)$
Put $\text{x}=\tan\theta,\text{so}$
$\text{u}=\tan^{-1}\Big(\frac{2\tan\theta}{1-\tan^2\theta}\Big)$
$\text{u}=\tan^{-1}(\tan2\theta)\ .....(\text{i})$
Let $\text{v}=\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$
$=\cos^{-1}\Big(\frac{1-\tan^2\theta}{1+\tan^2\theta}\Big)$
$\text{v}=\cos^{-1}(\cos2\theta)\ .....(\text{ii})$
Hrer, 0 < x < 1
$\Rightarrow0< \tan\theta<1$
$\Rightarrow0<\theta<\frac{\pi}{4}$
So, from equation (i),
$\text{u}=2\theta\Big[\text{Since,}\tan^{-1}(\tan\theta)=\theta,\text{if }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\text{u}=2\tan^{-1}\text{x }[\text{Since,x}=\tan\theta]$
Differentiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=\frac{2}{1+\text{x}^2}\ .....(\text{iii})$
From equation (ii),
$\text{v}=\theta \big[\text{Since,} \cos^{-1}(\cos\theta)=\theta,\text{if }\theta\in[0,\pi]\big]$
$\text{v}=2\tan^{-1}\text{x }[\text{Since},\text{x}=\tan\theta]$
Differentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{2}{1+\text{x}^2}\ .....(\text{iv})$
Dividing equation (iii) by (iv)
$\frac{\frac{\text{}du}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{2}{1+\text{x}^2}\times\frac{1+\text{x}^2}{2}$
$\frac{\text{du}}{\text{dv}}=1$
View full question & answer→Question 1715 Marks
If $\text{y}=\cos^{-1}(2\text{x})+2\cos^{-1}\sqrt{1-4\text{x}^2}, -\frac{1}{2}<\text{x}<0,$ find $\frac{\text{dy}}{\text{dx}}.$
AnswerHere, $\text{y}=\cos^{-1}(2\text{x})+2\cos^{-1}\sqrt{1-4\text{x}^2}$
Put $2\text{x}=\cos\theta, \text{So},$
$\text{y}=\cos^{-1}(\cos\theta)+2\cos^{-1}\sqrt{1-\cos^2\theta}$
$=\cos^2(\cos\theta)+2\cos^{-1}(\sin\theta)$
$\text{y}=\cos^{-1}(\cos\theta)+2\cos^{-1}\Big(\cos\Big(\frac{\pi}{2}-\theta\Big)\Big)\ .....(\text{i})$
Now, $-\frac{1}{2}<\text{x}<0$
$\Rightarrow -1<2\text{x}<0$
$\Rightarrow -1<\cos\theta<0$
$\Rightarrow \frac{\pi}{2}<\theta<\pi$
And
$\Rightarrow -\frac{\pi}{2}>-\theta>-\pi$
$\Rightarrow \Big(\frac{\pi}{2}-\frac{\pi}{2}\Big)>\Big(\frac{\pi}{2}-\theta\Big)>\Big(\frac{\pi}{2}-\pi\Big)$
$\Rightarrow 0>\Big(\frac{\pi}{2}-\theta\Big)>-\frac{\pi}{2}$
So, from equation (i),
$\text{y}=\theta+2\Big[-\Big(\frac{\pi}{2}-\theta\Big)\Big]$
$\begin{bmatrix} \text{Since}, \cos^{-1}\cos(\theta)=\theta, \text{if }\theta\in[0,\pi] \\ \cos^{-1}\cos(\theta)=-\theta, \text{if }\theta\in[-\pi,0] \end{bmatrix}$
$\text{y}=\theta-2\times\frac{\pi}{2}+2\theta$
$\text{y}=-\pi+3\theta$
$\text{y}=-\pi+3\cos^{-1}(2\text{x})\ \big[\text{Since}, 2\text{x}=\cos\theta\big]$
Differentiating it with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=0+3\Big(\frac{1}{\sqrt{1-(2\text{x})^2}}\Big)\frac{\text{d}}{\text{dx}}(2\text{x})$
$=\frac{-3}{\sqrt{1-4\text{x}^2}}(2)$
$\frac{\text{dy}}{\text{dx}}=-\frac{6}{\sqrt{1-4\text{x}^2}}$
View full question & answer→Question 1725 Marks
If $\text{y}=\frac{\text{x}}{\text{x}+2},$ show that $\text{x}\frac{\text{dy}}{\text{dx}}=(1-\text{y})\text{y}$
AnswerWe have, $\text{y}=\frac{\text{x}}{\text{x}+2}$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}}{\text{x}+2}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(\text{x}+2)\frac{\text{d}}{\text{dx}}(\text{x})-\text{x}\frac{\text{d}}{\text{dx}}(\text{x}+2)}{(\text{x}+2)^2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}+2-\text{x}}{(\text{x}+2)^2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}+2}{(\text{x}+2)^2}-\frac{\text{x}}{(\text{x}+2)^2}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}+2}-\frac{\text{xy}^2}{\text{x}^2} \Big[\because\ \text{x}+2=\frac{\text{x}}{\text{y}}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}-\frac{\text{y}^2}{\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}\text{y}(1-\text{y})$
$\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}=(1-\text{y})\text{y}$
Hence, proved.
View full question & answer→Question 1735 Marks
Differentiate the following functions with respect to x:
$(1+\cos\text{x})^\text{x}$
AnswerLet $\text{y}=(1+\cos\text{x})^\text{x}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=\log(1+\cos\text{x})^\text{x}$
$\log\text{y}=\text{x}\log(1-\cos\text{x})$
Differentiating with respect to x,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}\log(1+\cos\text{x})+\log(1+\cos\text{x})\frac{\text{d}}{\text{dx}}(\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{1}{(1+\cos\text{x})}\frac{\text{d}}{\text{dx}}(1+\cos\text{x})+\log(1+\cos\text{x})(1)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{(1+\cos\text{x})}(0-\sin\text{x})+\log(1+\cos\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\log(1+\cos\text{x})-\frac{\text{x}\sin\text{x}}{(1+\cos\text{x})}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\log(1+\cos\text{x})-\frac{\text{x}\sin\text{x}}{(1+\cos\text{x})}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=(1+\cos\text{x})^\text{x}\Big[\log(1+\cos\text{x})-\frac{\text{x}\sin\text{x}}{(1+\cos\text{x})}\Big]$
[Using equation (i)]
View full question & answer→Question 1745 Marks
If $\text{y}=\log\big\{\sqrt{\text{x}-1}-\sqrt{\text{x}+1}\big\},$ show that $\frac{\text{dy}}{\text{dt}}=\frac{-1}{2\sqrt{\text{x}^2-1}}.$
AnswerHere $\text{y}=\log\big\{\sqrt{\text{x}-1}-\sqrt{\text{x}+1}\big\}$
Differentiating it with respect to x and applying the chain and product rule, we get
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}} \log\big\{\sqrt{\text{x}-1}-\sqrt{\text{x}+1}\big\}$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\big\{\sqrt{\text{x}-1}-\sqrt{\text{x}+1}\big\}}\frac{\text{d}}{\text{dx}}(\sqrt{\text{x}-1}-\sqrt{\text{x}+1})$
$=\frac{1}{\big\{\sqrt{\text{x}-1}-\sqrt{\text{x}+1}\big\}}\Big[\frac{\text{d}}{\text{dx}}\sqrt{\text{x}-1}-\frac{\text{d}}{\text{dx}}\sqrt{\text{x}+1}\Big]$
$=\frac{1}{\big\{\sqrt{\text{x}-1}-\sqrt{\text{x}+1}\big\}}\Big[\frac{1}{2}(\text{x}-1)^{\frac{1}{2}}-\frac{1}{2}(\text{x}+1)^{\frac{1}{2}}\Big]$
$=\frac{1}{2}\frac{1}{\big\{\sqrt{\text{x}-1}-\sqrt{\text{x}+1}\big\}}\Big[\frac{1}{\sqrt{\text{x}-1}}-\frac{1}{\sqrt{\text{x}+1}}\Big]$
$=\frac{1}{2}\frac{1}{\big\{\sqrt{\text{x}-1}-\sqrt{\text{x}+1}\big\}}\Bigg(\frac{-\big\{\sqrt{\text{x}-1}-\sqrt{\text{x}-1}\big\}}{\big(\sqrt{\text{x}-1}\big)\big(\sqrt{\text{x}+1}\big)}\Bigg)$
$=\frac{1}{2}\bigg(\frac{1}{\big(\sqrt{\text{x}-1}\big)\big(\sqrt{\text{x}+1}\big)}\bigg)$
$\frac{\text{dy}}{\text{dx}}=\frac{-1}{2\sqrt{\text{x}^2-1}}$
Therefore,
$\frac{\text{dy}}{\text{dx}}=\frac{-1}{2\sqrt{\text{x}^2-1}}$
View full question & answer→Question 1755 Marks
If $\text{f(x)}=\log\Big\{\frac{\text{u(x)}}{\text{v(x)}}\Big\},\text{u}(1)=\text{v}(1)$ and u'(1) = v'(1) = 2, then find the value of f(1).
AnswerWe have, $\text{f(x)}=\log\Big\{\frac{\text{u(x)}}{\text{v(x)}}\Big\}$
And,
$\text{u}(1)=\text{v}(1),\text{u}'(1)=\text{v}'(1)=2\ .....(\text{i})$
$\Rightarrow\text{f}'\text{(x)}=\frac{\text{d}}{\text{dx}}\Big[\text{f(x)}=\log\Big\{\frac{\text{u(x)}}{\text{v(x)}}\Big\}\Big]$
$\Rightarrow\text{f}'\text{(x)}=\frac{1}{\Big[\frac{\text{u(x)}}{\text{v(x)}}\Big]}\times\frac{\text{d}}{\text{dx}}\Big[\frac{\text{u(x)}}{\text{v(x)}}\Big]$
$\Rightarrow\text{f}'\text{(x)}=\frac{\text{v(x)}}{\text{u(x)}}\times\bigg[\frac{\text{v(x)}\frac{\text{d}}{\text{dx}}\{\text{u(x)}\}-\text{u(x)}\frac{\text{d}}{\text{dx}}\{\text{v(x)}\}}{\{\text{v(x)}\}^2}\bigg]$
$\Rightarrow\text{f}'\text{(x)}=\frac{\text{v(x)}}{\text{u(x)}}\times\Big[\frac{\text{u(x)}\times\text{u}'\text{(x)}-\text{u(x)}\times\text{v}'\text{(x)}}{\{\text{v(x)}\}^2}\Big]$
Putting x = 1, we get,
$\text{f}'(1)=\frac{\text{v}(1)}{\text{u}(1)}\times\Big[\frac{\text{u}(1)\times\text{u}'(1)-\text{u}1\times\text{v}'(1)}{\{\text{v}(1)\}^2}\Big]$
$\Rightarrow\text{f}'(1)=1\times\Big[\frac{\text{u}(1)\times2-\text{u}(1)\times 2}{\{\text{u}(1)\}^2}\Big]$
[Using eqn (1)]
$\Rightarrow\text{f}'(1)=\Big[\frac{0}{\{\text{u}(1)\}^2}\Big]$
$\Rightarrow\text{f}'(1)=0$
View full question & answer→Question 1765 Marks
Differentiate the following functions with respect to x:
$10^{\log\sin\text{x}}$
AnswerLet $\text{y}=10^{\log\sin\text{x}}\ .....(\text{i})$
Taking log on both sides,
$\log\text{y}=\log10^{\log\sin\text{x}}$
$\Rightarrow\log\text{y}=\log\sin\text{x}\log10$
Differentiating with respect to x,
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\log10\frac{\text{d}}{\text{dx}}\log\sin\text{x}$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\log10\frac{1}{\sin\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\log10\Big(\frac{1}{\sin\text{x}}\Big)(\cos\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}\big[\log10\times\cot\text{x}\big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=10^{\log\sin\text{x}}\times\log10\times\cot\text{x}$
[Using equation (i)]
View full question & answer→Question 1775 Marks
If $\text{y}=\text{x}\sin\text{y},$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}(1-\text{x}\cos\text{y})}$
AnswerWe have, $\text{y}=\text{x}\sin\text{y}\ .....(\text{i})$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x}\sin\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}(\sin\text{y})+\sin\text{y}\frac{\text{d}}{\text{dx}}(\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}\cos\text{y}\frac{\text{dy}}{\text{dx}}+\sin\text{y}(1)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}-\text{x}\cos\text{y}\frac{\text{dy}}{\text{dx}}=\sin\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}(1-\text{x}\cos\text{y})=\sin\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sin\text{y}}{(1-\text{x}\cos\text{y})}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}(1-\text{x}\cos\text{y})}\Big[\because\sin\text{y}=\frac{\text{y}}{\text{x}}\Big]$
View full question & answer→Question 1785 Marks
If $\text{y}=\text{e}^{\text{x}}\cos\text{x},$ Prvoe that $\frac{\text{dy}}{\text{dx}}=\sqrt{2}\text{e}^\text{x}.\cos\Big(\text{x}+\frac{\pi}{4}\Big)$
AnswerGiven, $\text{y}=\text{e}^{\text{x}}\cos\text{x}$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^\text{x}\cos\text{x}\big)$
$=\text{e}^\text{x}\frac{\text{d}}{\text{dx}}\cos\text{x}+\cos\text{x}\frac{\text{d}}{\text{dx}}\text{e}^{\text{x}}$ [Using product rule]
$=\text{e}^\text{e}(-\sin\text{x})+\text{e}^\text{x}\cos\text{x}$
$=\text{e}^\text{x}(\cos\text{x}-\sin\text{x})$
$=\sqrt{2}\text{e}^\text{x}\Big(\frac{\cos\text{x}}{\sqrt{2}}-\frac{\sin\text{x}}{\sqrt{2}}\Big)$ $\big[$Multiplying and dividing by $\sqrt{2}\big]$
$=\sqrt{2}\text{e}^\text{x}\Big(\cos\frac{\pi}{4}\cos\text{x}-\sin\frac{\pi}{4}\sin\text{x}\Big)$
$\frac{\text{dy}}{\text{dx}}=\sqrt{2}\text{e}^\text{x}\cos\Big(\text{x}+\frac{\pi}{4}\Big)$
View full question & answer→Question 1795 Marks
Differentiate the following functions with respect to x:
$3\text{e}^{-3\text{x}}\log(1+\text{x})$
AnswerConsider $\text{y}=3\text{e}^{-3\text{x}}\log(1+\text{x})$
Differentiating it with respect to x and applying the chain and product rule, we get
$\frac{\text{dy}}{\text{dx}}=3\frac{\text{d}}{\text{dx}}\big[3\text{e}^{-3\text{x}}\log(1+\text{x})\big]$
$\frac{\text{dy}}{\text{dt}}=3\Big(\text{e}^{-3\text{x}}\frac{1}{1+\text{x}}+\log(1+\text{x})\big(-3\text{e}^{-3\text{x}}\big)\Big)$
$=3\Big(\frac{\text{e}^{-3\text{x}}}{1+\text{x}}-3\log(1+\text{x})\Big)$
The solution is,
$=3\text{e}^{-3\text{e}}\Big(\frac{1}{1+\text{x}}-3\log(1-\text{x})\Big)$
View full question & answer→Question 1805 Marks
Differentiate the following functions with respect to x:
$(\log\text{x})^{\cos\text{x}}$
AnswerLet $\text{y}=(\log\text{x})^{\cos\text{x}}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=(\log\text{x})^{\cos\text{x}}$
$\Rightarrow\log\text{y}=\cos\text{x}\log(\log\text{x})$
Differentiating with respect to x,
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\cos\text{x}\frac{\text{d}}{\text{dx}}\log(\log\text{x})+\log(\log\text{x})\frac{\text{d}}{\text{dx}}(\cos\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\cos\text{x}}{\log\text{x}}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log(\log\text{x})\times(-\sin\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\cos\text{x}}{\log\text{x}}\times\big(\frac{1}{\text{x}}\big)-\sin\text{x}\log(\log\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{\cos\text{x}}{\text{x}\log\text{x}}-\sin\text{x}\log(\log\text{x})\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=(\log\text{x}^{\cos\text{x}})\Big[\frac{\cos\text{x}}{\text{x}\log\text{x}}-\sin\text{x}\log(\log\text{x})\Big]$
[Using equation (i)]
View full question & answer→Question 1815 Marks
Differentiate the following functions with respect to x:
$\text{x}^{\text{x}\cos\text{x}}+\frac{\text{x}^2+1}{\text{x}^2-1}$
AnswerLet $\text{y}=\text{x}^{\text{x}\cos\text{x}}+\frac{\text{x}^2+1}{\text{x}^2-1}$
Also, let $\text{u}=\text{x}^{\text{x}\cos\text{x}}\text{ and v}=\frac{\text{x}^2+1}{\text{x}^2-1}$
$\therefore\ \text{y}=\text{u}+\text{v}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{du}}{\text{dx}}+\frac{\text{dv}}{\text{dx}}\ .....(\text{i})$
$\text{u}=\text{x}^{\text{x}\cos\text{x}}$
$\Rightarrow\ \log\text{u}=\log(\text{x}^{\text{x}\cos\text{x}})$
$\Rightarrow\log\text{u}=\text{x}\cos\text{x}\log\text{x}$
Diffrerentiating both sides with respect to x, we obtain
$\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x}).\cos\text{x}\log\text{x}+\text{x}.\frac{\text{d}}{\text{dx}}(\cos\text{x}).\log\text{x}+\text{x}\cos\text{x}.\frac{\text{d}}{\text{dx}}(\log\text{x})$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{u}\Big[1.\cos\text{x}.\log\text{x}+\text{x}.(-\sin\text{x})\log\text{x}+\text{x}\cos\text{x}.\frac{1}{\text{x}}\Big]$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{x}^{\text{x}\cos\text{x}}(\cos\text{x}\log\text{x}-\text{x}\sin\text{x}\log\text{x}+\cos\text{x})$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{x}^{\text{x}\cos\text{x}}\big[\cos\text{x}(1+\cos\text{x})-\text{x}\sin\text{x}\log\text{x}\big]\ .....(\text{ii})$
$\text{v}=\frac{\text{x}^2+1}{\text{x}^2-1}$
$\Rightarrow\log\text{v}=\log(\text{x}^2+1)-\log(\text{x}^2-1)$
Differentiating both sides with respect to x, we obtain
$\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\frac{2\text{x}}{\text{x}^2+1}-\frac{2\text{x}}{\text{x}^2-1}$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\text{v}\Big[\frac{2\text{x}(\text{x}^2-1)-2\text{x}(\text{x}^2+1)}{(\text{x}^2+1)(\text{x}^2-1)}\Big]$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{\text{x}^2+1}{\text{x}^2-1}\times\Big[\frac{-4\text{x}}{(\text{x}^2+1)(\text{x}^2-1)}\Big]$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{-4\text{x}}{(\text{x}^2-1)^2}\ .....(\text{iii})$
From (1), (2) and (3), we obtain
$\frac{\text{dy}}{\text{dx}}=\text{x}^{\text{x}\cos\text{x}}\big[\cos\text{x}(1+\log\text{x})-\text{x}\sin\text{x}\log\text{x}\big]-\frac{4\text{x}}{(\text{x}^2-1)^2}$
View full question & answer→Question 1825 Marks
Differentiate the following functions with respect to x:
$\sin^{-1}\big\{\sqrt{1-\text{x}^2}\big\},0<\text{x}<1$
AnswerLet $\text{y}=\sin^{-1}\big\{\sqrt{1-\text{x}^2}\big\}$
Put $\text{x}=\cos2\theta$
$\text{y}=\sin^{-1}\big\{\sqrt{1-\cos^2\theta}\big\}$
$\text{y}=\sin^{-1}(\sin\theta)\ .....(\text{i})$
Here, $0<\text{x}<1$
$\Rightarrow\ 0<\cos2\theta<1$
$\Rightarrow\ 0<2\theta<\frac{\pi}{2}$
From equation (i),
$\text{y}=\theta$
$\Big[\text{Since, } \sin^{-1}(\sin\theta)=\theta\text{ if }\theta \in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\text{y}=\cos^{-1}\text{x}\ \big[\text{Since x}=\cos\theta\big]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=-\frac{1}{\sqrt{1-\text{x}^2}}$
View full question & answer→Question 1835 Marks
If $\text{y}=(\tan\text{x})^{(\tan\text{x})^{(\tan\text{x})^{....\infty}}},$ prove that $\frac{\text{dy}}{\text{dx}}=2\text{ at x}=\frac{\pi}{4}$
AnswerWe have, $\text{y}=(\tan\text{x})^{(\tan\text{x})^{(\tan\text{x})^{....\infty}}}$
$\Rightarrow\text{y}=(\tan\text{x})^{\text{y}}$
Taking log on both sides,
$\log\text{y}=\log(\tan\text{x})^\text{y}$
$\Rightarrow\log\text{y}=\text{y}\log\tan\text{x}$
Differentaiting with respect to x using chain rule,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{y}\frac{\text{d}}{\text{dx}}\big\{\log\tan\text{x}\big\}+\log\tan\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\tan\text{x}}\frac{\text{d}}{\text{dx}}(\tan\text{x})+\log\tan\frac{\text{d}}{\text{dx}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\Big(\frac{1}{\text{y}}-\log\tan\text{x}\Big)=\frac{\text{y}}{\tan\text{y}}\sec^2\text{x}$
Now, $\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=\frac{\pi}{4}}=\frac{\text{y}\sec^3\big(\frac{\pi}{4}\big)}{\tan\big(\frac{\pi}{4}\big)}\times\frac{\text{y}}{1-\text{y}\log\tan\big(\frac{\pi}{4}\big)}$
$\Rightarrow\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=\frac{\pi}{4}}=\frac{\text{y}^2\big(\sqrt{2}\big)^2}{1(1-\text{y}\log\tan1)}$
$\Rightarrow \Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=\frac{\pi}{4}}=\frac{2(1)^2}{(1-0)}\Bigg[\because(\text{y})_{\frac{\pi}{4}}=\big(\tan\frac{\pi}{4}\big)^{\big(\tan\frac{\pi}{4}\big)^{\big(\tan\frac{\pi}{4}\big)^{\ .....\infty}}}=1\Bigg]$
$\Rightarrow \Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=\frac{\pi}{4}}=2$
View full question & answer→Question 1845 Marks
Differentiate the following functions with respect to x:
$\big(\sin^{-1}\text{x}^4\big)^4$
AnswerConsider $\text{y}=\big(\sin^{-1}\text{x}^4\big)^4$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\sin^{-1}\text{x}^4\big)^4$
$=4\big(\sin^{-1}\text{x}^4\big)\frac{\text{d}}{\text{dx}}\big(\sin^{-1}\text{x}^4\big)$
[Using chain rule]
$=4\big(\sin^{-1}\text{x}^4\big)^3\frac{1}{\sqrt{1-\big(\text{x}^4\big)^2}}\frac{\text{d}}{\text{dx}}\big(\text{x}^4\big)$
$=4\big(\sin^{-1}\text{x}^4\big)^3\frac{4\text{x}^3}{\sqrt{1-\text{x}^8}}$
$=\frac{16\text{x}^3\big(\sin^{-1}\text{x}^4\big)^3}{\sqrt{1-\text{x}^8}}$
Hence, the solution is, $\frac{\text{d}}{\text{dx}}\big(\sin^{-1}\text{x}^4\big)=\frac{16\text{x}^3\big(\sin^{-1}\text{x}^4\big)^3}{\sqrt{1-\text{x}^8}}$
View full question & answer→Question 1855 Marks
Differentiate the following functions with respect to x:
$\frac{2^\text{x}\cos\text{x}}{(\text{x}^2+3)^2}$
AnswerLet $\text{y}=\frac{\text{2}^\text{x}\cos\text{x}}{(\text{x}^2+3)^3}$
Differentiate it with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big[\frac{\text{2}^\text{x}\cos\text{x}}{(\text{x}^2+3)^3}\Big]$
$=\bigg[\frac{(\text{x}^2+3)^2\frac{\text{d}}{\text{dx}}(2^\text{x}\cos\text{x})-(2^\text{x}\cos\text{x})\frac{\text{d}}{\text{dx}}(\text{x}^2+3)^2}{\big[(\text{x}^2+3)\big]^2}\bigg]$
[Using quotient rule]
$=\Bigg[\frac{(\text{x}^2+3)^2\Big\{2^\text{x}\frac{\text{d}}{\text{dx}}\cos\text{x}+\cos\text{x}\frac{\text{d}}{\text{dx}}2^\text{x}\Big\}-(2^\text{x}+3)2(\text{x}^2+3)\frac{\text{d}}{\text{dx}}(\text{x}^2+3)}{(\text{x}^2+3)^4}\Bigg]$
[Using Product rule and chain rule]
$=\bigg[\frac{(\text{x}^2+3)^2\big\{-2^\text{x}\sin\text{x}+\cos\text{x}2^\text{x}\log_\text{e}2\big\}-2(2^\text{x}\cos\text{x})(\text{x}^2+3)(2\text{x})}{(\text{x}^2+3)^4}\bigg]$
$=\bigg[\frac{2^\text{x}(\text{x}^2+3)\big\{(\text{x}^2+3)(\cos\text{x}\log_\text{e}2-\sin\text{x})-4\text{x}\cos\text{x}\big\}}{(\text{x}^2+3)^4}\bigg]$
$=\frac{2^\text{x}}{(\text{x}^2+3)^2}\bigg[\cos\text{x}\log_\text{e}2-\sin\text{x}-\frac{4\text{x}\cos\text{x}}{(\text{x}^2+3)}\bigg]$
So,
$\frac{\text{d}}{\text{dx}}\Big[\frac{2^\text{x}\cos\text{x}}{(\text{x}^2+3)^2}\Big]=\frac{2^\text{x}}{(\text{x}^2+3)^2}\bigg[\cos\text{x}\log_\text{e}2-\sin\text{x}-\frac{4\text{x}\cos\text{x}}{(\text{x}^2+3)}\bigg]$
View full question & answer→Question 1865 Marks
Differentiate the following functions with respect to x:
$\text{x}^{\sin^{-1}\text{x}}$
AnswerLet $\text{y}=\text{x}^{\sin^{-1}\text{x}}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=\log\text{x}^{\sin^{-1}\text{x}}$
$\log\text{y}=\sin^{-1}\text{x}\log\text{x}$
Differentiating it with respect to x,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\sin^{-1}\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+(\log\text{x})\frac{\text{d}}{\text{dx}}(\sin^{-1}\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\sin^{-1}\text{x}\Big(\frac{1}{\text{x}}\Big)+(\log\text{x})\Big(\frac{1}{\sqrt{1-\text{x}^2}}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{\sin^{-1}\text{x}}{\text{x}}+\frac{\log\text{x}}{\sqrt{1-\text{x}^2}}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}^{\sin^{-1}\text{x}}\Big[\frac{\sin^{-1}\text{x}}{\text{x}}+\frac{\log\text{x}}{\sqrt{1-\text{x}^2}}\Big]$
[Using equation (i)]
View full question & answer→Question 1875 Marks
If $xy^2 = 1$, prove that $2\frac{\text{dy}}{\text{dx}}+\text{y}^3=0$
AnswerWe have $xy^2 = 1 .....(i)$
Differentiating with respect to x, we get,
$\frac{\text{d}}{\text{dx}}(\text{xy}^2)=\frac{\text{d}}{\text{dx}}(1)$
$\Rightarrow\text{x}\frac{\text{d}}{\text{dx}}(\text{y}^2)+\text{y}^2\frac{\text{d}}{\text{dx}}(\text{x})=0$
$\Rightarrow\text{x}(2\text{y})\frac{\text{d}}{\text{dx}}+\text{y}^2(1)=0$
$\Rightarrow2\text{xy}\frac{\text{dy}}{\text{dx}}=-\text{y}^2$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-\text{y}^2}{2\text{xy}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-\text{y}}{2\text{x}}$
Put $\text{x}=\frac{1}{\text{y}^2}$ from equation (i)
$ \Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-\text{y}}{2\Big(\frac{1}{\text{y}^2}\Big)}$
$\Rightarrow2\frac{\text{dy}}{\text{dx}}=-\text{y}^3$
$\Rightarrow2\frac{\text{dy}}{\text{dx}}+\text{y}^3=0$
View full question & answer→Question 1885 Marks
If $\text{y}=\tan^{-1}\Big(\frac{1-\text{x}}{1+\text{x}}\Big),$, find $\frac{\text{dy}}{\text{dx}}.$
AnswerHere,
$\text{y}=\tan^{-1}\Big(\frac{1-\text{x}}{1+\text{x}}\Big)$
Differentiating it with respect to x using chain rule and quotinet rule,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{1+\big(\frac{1-\text{x}}{1+\text{x}}\big)^2}\frac{\text{d}}{\text{dx}}\Big(\frac{1-\text{x}}{1+\text{x}}\Big)$
$=\frac{(1+\text{x})^2}{(1+\text{x}^2+2\text{x}+1+\text{x}^2-2\text{x})}\bigg[\frac{(1+\text{x})\frac{\text{d}}{\text{dx}}(1-\text{x})-(1-\text{x})\frac{\text{d}}{\text{dx}}(1+\text{x})}{(1+\text{x})^2}\bigg]$
$=\frac{(1+\text{x})^2}{2\text{x}^2+2}\Big[\frac{(1+\text{x})(-1)-(1-\text{x})(1)}{(1+\text{x})^2}\Big]$
$=\frac{(1+\text{x})^2}{2(\text{x}^2+1)}\Big(\frac{-\text{x}-1-1+\text{x}}{(1+\text{x})^2}\Big)$
$=\frac{(1+\text{x})^2}{2(\text{x}^2+1)}\times\frac{-2}{(1+\text{x})^2}$
$\frac{\text{dy}}{\text{dx}}=-\frac{1}{\text{x}^2+1}$
View full question & answer→Question 1895 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\bigg[\frac{\text{x}^\frac{1}{3}+\text{a}^{\frac{1}{3}}}{1-(\text{ax})^\frac{1}{3}}\bigg]$
AnswerLet $\text{y}=\tan^{-1}\bigg[\frac{\text{x}^\frac{1}{3}+\text{a}^{\frac{1}{3}}}{1-(\text{ax})^\frac{1}{3}}\bigg]$
$\Rightarrow\text{y}=\tan^{-1}\big(\text{x}^\frac{1}{3}\big)+\tan^{-1}\big(\text{a}^\frac{1}{3}\big)$
$\Big[\text{Since},\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)\Big]$
Differentiate it with respect to x usign chain rule,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{1+\big(\text{x}^\frac{1}{3}\big)^2}\times\frac{\text{d}}{\text{dx}}\big(\text{x}^\frac{1}{2}\big)+0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\bigg(\frac{1}{2}\times\text{x}^{\frac{1}{3}-1}\bigg)}{1+\text{x}^\frac{2}{3}}$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{1}{3\text{x}^\frac{2}{3}\Big(1+\text{x}^\frac{2}{3}\Big)}$
View full question & answer→Question 1905 Marks
Differentiate the following functions with respect to x:
$\frac{\text{e}^\text{x}\log\text{x}}{\text{x}^2}$
AnswerLet $\text{y}=\frac{\text{e}^\text{x}\log\text{x}}{\text{x}^2}$
Differentiate with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}^2\frac{\text{d}}{\text{dx}}(\text{e}^\text{x}\log\text{x})-(\text{e}^\text{x}\log\text{x})\frac{\text{d}}{\text{dx}}\text{x}^2}{\big(\text{x}^2\big)^2}$
[Using quotient rule]
$=\frac{\text{x}^2\Big\{\text{e}^\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{e}^\text{x})\Big\}-\text{e}^\text{x}\log\text{x}\times2\text{x}}{\text{x}^4}$
[Using product rule]
$=\frac{\text{x}^2\Big[\frac{\text{e}^\text{x}}{\text{x}}+\text{e}^\text{x}\log\text{x}\Big]-2\text{xe}^\text{x}\log\text{x}}{\text{x}^4}$
$=\frac{\frac{\text{x}^2\text{e}^\text{z}(1+\text{x}\log\text{x})}{\text{x}}-2\text{xe}^\text{z}\log\text{x}}{\text{x}^4}$
$=\frac{\text{xe}^\text{x}[1+\text{x}\log\text{x}-2\log\text{x}]}{\text{x}^4}$
$=\frac{\text{xe}^\text{x}}{\text{x}^3}\Big[\frac{1}{\text{x}}+\frac{\text{x}\log\text{x}}{\text{x}}-\frac{2\log\text{x}}{\text{x}}\Big]$
$=\text{e}^\text{x}\text{x}^{-2}\Big[\frac{1}{\text{x}}+\log\text{x}-\frac{2}{\text{x}}\log\text{x}\Big]$
So,
$\frac{\text{d}}{\text{dx}}\Big[\frac{\text{e}^\text{x}\log\text{x}}{\text{x}^2}\Big]=\text{e}^\text{x}\text{x}^{-2}\Big[\frac{1}{\text{x}}+\log\text{x}-\frac{2}{\text{x}}\log\text{x}\Big]$
View full question & answer→Question 1915 Marks
Differentiate the following functions with respect to x:
$\sqrt{\frac{1-\text{x}^2}{1+\text{x}^2}}$
AnswerLet $\text{y}=\sqrt{\frac{1-\text{x}^2}{1+\text{x}^2}}$
$\text{y}=\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)^\frac{1}{2}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)^\frac{1}{2}$
$=\frac{1}{2}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)^{\frac{1}{2}-1}\frac{\text{d}}{\text{dx}}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$
[Using chain rule]
$=\frac{1}{2}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)^{\frac{-1}{2}}\bigg[\frac{(1+\text{x}^2)\frac{\text{d}}{\text{dx}}(1-\text{x}^2)-(1-\text{x}^2)\frac{\text{d}}{\text{dx}}(1+\text{x}^2)}{\big(1+\text{x}^2\big)^2}\bigg]$
[Using chain rule]
$=\frac{1}{2}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)^{\frac{1}{2}}\bigg[\frac{(1+\text{x}^2)(-2\text{x})-(1-\text{x}^2)(2\text{x})}{\big(1+\text{x}^2\big)^2}\bigg]$
$=\frac{1}{2}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)^{\frac{1}{2}}\bigg[\frac{-2\text{x}-2\text{x}^3-2\text{x}+2\text{x}^3}{\big(1+\text{x}^2\big)^2}\bigg]$
$=\frac{1}{2}\frac{-4\text{x}}{\sqrt{1-\text{x}^2}\big(1+\text{x}^2)^\frac{3}{2}}$
So,
$\frac{\text{d}}{\text{dx}}\bigg(\sqrt{\frac{1-\text{x}^2}{1+\text{x}^2}}\bigg)=\frac{-4\text{x}}{\sqrt{1-\text{x}^2}\big(1+\text{x}^2)^\frac{3}{2}}$
View full question & answer→Question 1925 Marks
Find $\frac{\text{dy}}{\text{dx}}$ in the following cases:
$y^3 - 3xy^2 = x^3 + 3x^2y$
AnswerHere, $y^3 - 3xy^2 = x^3 + 3x^2y$ Differentiating with respect to x,
$\Rightarrow\frac{\text{d}}{\text{dy}}(\text{y}^3)-\frac{\text{d}}{\text{dx}}(3\text{xy}^2)=\frac{\text{d}}{\text{dx}}(\text{x}^3)+\frac{\text{d}}{\text{dx}}(3\text{x}^2\text{y})$
$\Rightarrow 3\text{y}^2\frac{\text{dy}}{\text{dx}}-3\Big[\text{x}\frac{\text{d}}{\text{dx}}\text{y}^2\frac{\text{d}}{\text{dx}}(\text{x})\Big]=3\text{x}^2+3\Big[\text{x}^2\frac{\text{d}}{\text{dx}}(\text{y})+\text{y}\frac{\text{d}}{\text{dx}}(\text{x}^2)\Big]$
$\Rightarrow 3\text{y}^2\frac{\text{dy}}{\text{dx}}-3\Big[\text{x}(2\text{y})\frac{\text{dy}}{\text{dx}}+\text{y}^2\Big]=3\text{x}^2+3\Big[\text{x}^2\frac{\text{dy}}{\text{dx}}+\text{y}(2\text{x})\Big]$
$\Rightarrow 3\text{y}^2\frac{\text{dy}}{\text{dx}}-6\text{xy}\frac{\text{dy}}{\text{dx}}-3\text{y}^2+3\text{x}^2+3\text{x}^2\frac{\text{dy}}{\text{dx}}+6\text{xy}$
$\Rightarrow 3\text{y}^2\frac{\text{dy}}{\text{dx}}-6\text{xy}\frac{\text{dy}}{\text{dx}}-3\text{x}^2\frac{\text{dy}}{\text{dx}}=3\text{x}^2+6\text{xy}+3\text{y}^2$
$=3\frac{\text{dy}}{\text{dx}}(\text{y}^2-2\text{xy}-\text{x}^2)=3(\text{x}^2+2\text{xy}+\text{y}^2)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{3(\text{x}+\text{y})^2}{3(\text{y}^2-2\text{xy}-\text{x}^2)}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(\text{x}+\text{y})^2}{\text{y}^2-2\text{xy}-\text{x}^2)}$
View full question & answer→Question 1935 Marks
Differentiate the following functions with respect to x:
$\sin^{-1}\big(2\text{x}^2-1\big),0<\text{x}<1$
AnswerLet $\text{y}=\sin^{-1}\big\{2\text{x}^2-1\big\}$
Let $\text{x}=\cos\theta$
$\text{y}=\sin^{-1}\big(2\cos^2\theta-1\big)$
$=\sin^{-1}(\cos2\theta)$
$\text{y}=\sin^{-1}\Big\{\sin\Big(\frac{\pi}{2}-2\theta\Big)\Big\}\ .....(\text{i})$
Here, $0<\text{x}<1$
$\Rightarrow\ 0<\cos\theta<1$
$\Rightarrow\ 0<2\theta<\frac{\pi}{2}$
$\Rightarrow\ 0> -2\theta>-\pi$
$\Rightarrow\ \frac{\pi}{2}>\Big(\frac{\pi}{2}-2\theta\Big)>-\frac{\pi}{2}$
So, from equation (i),
$\text{y}=\frac{\pi}{2}-2\theta$
$\Big[\text{Sicne}, \sin^{-1}(\cos\theta)=\theta,\text{ if }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\text{y}=\frac{\pi}{2}-2\cos^{-1}\text{x}\ \big[\text{Since x}=\cos\theta\big]$
$\frac{\text{dy}}{\text{dx}}=0-2\frac{\text{d}}{\text{dx}}\big(\cos^{-1}\text{x}\big)$
$=-2\Big(-\frac{1}{\sqrt{1-\text{x}^2}}\Big)$
$\frac{\text{dy}}{\text{dx}}=\frac{2}{\sqrt{1-\text{x}^2}}$
View full question & answer→Question 1945 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{\text{a}+\text{bx}}{\text{b}-\text{ax}}\Big)$
AnswerLet $\text{y}=\tan^{-1}\Big(\frac{\text{a}+\text{bx}}{\text{b}-\text{ax}}\Big)$
$=\tan^{-1}\bigg(\frac{\frac{\text{a}+\text{bx}}{\text{b}}}{\frac{\text{b}-\text{ax}}{\text{b}}}\bigg)$
$=\tan^{-1}\bigg(\frac{\frac{\text{a}}{\text{b}}+\frac{\text{bx}}{\text{b}}}{\frac{\text{b}}{\text{a}}-\frac{\text{ax}}{\text{b}}}\bigg)$
$=\tan^{-1}\bigg(\frac{\frac{\text{a}}{\text{b}}+\text{x}}{1-\big(\frac{\text{a}}{\text{b}}\big)\text{x}}\bigg)$
$\text{y}=\tan^{-1}\big(\frac{\text{a}}{\text{b}}\big)+\tan^{-1}\text{x}$
$\Big[\text{Since},\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)\Big]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=0+\frac{1}{1+\text{x}^2}$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{1+\text{x}^2}$
View full question & answer→Question 1955 Marks
Differentiate the following functions with respect to x:
$\log(\text{x}+\sqrt{\text{x}^2+1})$
AnswerLet $\text{y}=\log(\text{x}+\sqrt{\text{x}^2+1})$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dt}}=\frac{\text{d}}{\text{dy}}\log\big(\text{x}+\sqrt{\text{x}^2+1}\big)$
$=\frac{1}{\text{x}+\sqrt{\text{x}^2+1}}\frac{\text{d}}{\text{dx}}\Big(\text{x}+\big(\text{x}^2+1\big)^\frac{1}{2}\Big)$
[Using chain rule]
$=\frac{1}{\text{x}+\sqrt{\text{x}^2+1}}\Big[1+\frac{1}{2}\big(\text{x}^2+1\big)^{\frac{1}{2}-1}\frac{\text{d}}{\text{dx}}\big(\text{x}^2+1\big)\Big]$
$=\frac{1}{\text{x}+\sqrt{\text{x}^2+1}}\Big[1+\frac{1}{2\sqrt{\text{x}^2+1}}\times2\text{x}\Big]$
$=\frac{1}{\text{x}+\sqrt{\text{x}^2+1}}\Big[\frac{\sqrt{\text{x}^2+1}+\text{x}}{\sqrt{\text{x}^2+1}}\Big]$
$=\frac{1}{\sqrt{\text{x}^2+1}}$
So,
$\frac{\text{d}}{\text{dx}}\Big(\log\big(\text{x}+\sqrt{\text{x}^2+1}\big)\Big)=\frac{1}{\sqrt{\text{x}^2+1}}$
View full question & answer→Question 1965 Marks
Differentiate the following functions with respect to x:
$\log_\text{x}3$
AnswerLet, $\text{y}=\log_\text{x}3$
$\Rightarrow\ \text{y}=\frac{\log3}{\log\text{x}}\ \Big[\because\ \log_\text{a}\text{b}=\frac{\log\text{b}}{\log\text{a}}\Big]$
Differentiate it with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\frac{\log3}{\log\text{x}}\Big)$
$=\log3\frac{\text{d}}{\text{dx}}(\log\text{x})^{-1}$
$=\log3\times\Big[-1(\log\text{x})^{-2}\Big]\frac{\text{d}}{\text{dx}}(\log\text{x})$
[Using chain rule]
$=-\frac{\log 3}{(\log\text{x})^2}\times\frac{1}{\text{x}}$
$=-\Big(\frac{\log 3}{\log\text{x}}\Big)^2\times\frac{1}{\text{x}}\times\frac{1}{\log3}$
$=-\frac{1}{\text{x}\log3(\log_3\text{x})^2} \Big[\because \frac{\log\text{b}}{\log\text{a}}=\log_\text{a}\text{b}\Big]$
So,
$\frac{\text{d}}{\text{dx}}(\log_\text{x}3)=-\frac{1}{\text{x}\log3(\log_3\text{x})^2}$
View full question & answer→Question 1975 Marks
If $xy = e^{x-y}$, find $\frac{\text{dy}}{\text{dx}}$
AnswerThe given function is $xy = e^{x-y}$
Taking logarithm on both the sides, we obtain
$\log(\text{xy})=\log\big(\text{e}^{\text{x}-\text{y}})$
$\Rightarrow\log\text{x}+\log\text{y}=(\text{x}-\text{y})\log\text{e}$
$\Rightarrow\log\text{x}=\log\text{y}=(\text{x}-\text{y})\times1$
$\Rightarrow\log\text{x}=\log\text{y}=\text{x}-\text{y}$
Differentiating both sides with respect to x, we obtain
$\frac{\text{d}}{\text{dx}}(\log\text{x})+\frac{\text{d}}{\text{dx}}(\log\text{y})=\frac{\text{d}}{\text{dx}}(\text{x})-\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{1}{\text{x}}+\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=1-\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\big(1+\frac{1}{\text{y}}\big)\frac{\text{dy}}{\text{dx}}=1-\frac{1}{\text{x}}$
$\Rightarrow\big(\frac{\text{y}+1}{\text{y}}\big)\frac{\text{dy}}{\text{dx}}=\frac{\text{x}-1}{\text{x}}$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\text{y}(\text{x}-1)}{\text{x}(\text{y}+1)}$
View full question & answer→Question 1985 Marks
Find $\frac{\text{dy}}{\text{dx}}$ in the following cases:
$(x^2 + y^2)^2 = xy$
AnswerGiven, $(x^2 + y^2)^2 = xy$
Differentiating with respect to x,
$\frac{\text{d}}{\text{dx}}\Big(\big(\text{x}^2+\text{y}^2\big)^2\Big)=\frac{\text{d}}{\text{dx}}(\text{xy})$
$\Rightarrow2(\text{x}^2+\text{y}^2\big)\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{y}^2\big)=\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\frac{\text{d}}{\text{dx}}\big(\text{x}\big)$
[Using chain rule]
$\Rightarrow2(\text{x}^2+\text{y}^2\big)\Big(2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}\Big)=\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}(1)$
$\Rightarrow4\text{x}(\text{x}^2+\text{y}^2\big)+4\text{y}\big(\text{x}^2+\text{y}^2\big)\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}$
$\Rightarrow4\text{y}\big(\text{x}^2+\text{y}^2\big)\frac{\text{dy}}{\text{dx}}-\text{x}\frac{\text{dy}}{\text{dx}}=\text{y}-4\text{x}(\text{x}^2+\text{y}^2\big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\big[4\text{y}\text{x}^2+4\text{y}^3-\text{x}\big]=\text{y}-4\text{x}^3-4\text{x}\text{y}^2$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\Big(\frac{\text{y}-4\text{x}^3-4\text{x}\text{y}^2}{4\text{y}\text{x}^2+4\text{y}^3-\text{x}}\Big)$
View full question & answer→Question 1995 Marks
If $\text{y}=\cos^{-1}\Big\{\frac{2\text{x}-3\sqrt{1-\text{x}^2}}{\sqrt{13}}\Big\},$ find $\frac{\text{dy}}{\text{dx}}.$
AnswerHere, $\text{y}=\cos^{-1}\Big\{\frac{2\text{x}-3\sqrt{1-\text{x}^2}}{\sqrt{13}}\Big\}$
Let $\text{x}=\cos\theta, \text{So},$
$\text{y}=\cos^{-1}\Big\{\frac{2\cos\theta-3\sqrt{1-\cos^2\theta}}{\sqrt{13}}\Big\}$
$=\cos^{-1}\Big\{\frac{2}{\sqrt{13}}\cos\theta-\frac{3}{13}\sin\theta\Big\}$
Let $\cos\phi=\frac{2}{\sqrt{13}}$
$\Rightarrow\sin\phi=\sqrt{1-\cos^2\phi}$
$=\sqrt{1-\Big(\frac{2}{\sqrt{13}}\Big)^2}$
$=\sqrt{\frac{13-4}{13}}$
$=\sqrt{\frac{9}{13}}$
$\sin\phi=\frac{3}{\sqrt{13}}$
So,
$\text{y}=\cos^{-1}\big\{\cos\phi\cos\theta-\sin\phi\sin\theta\big\}$
$=\cos^{-1}\big[\cos(\theta+\phi)\big]$
$\text{y}=\phi+\theta$
$\text{y}=\cos^{-1}\Big(\frac{2}{\sqrt{13}}\Big)+\cos^{-1}\text{x}$
$\Big[\text{Since, x}=\cos\theta,\cos\phi=\frac{2}{\sqrt{13}}\Big]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=0+\Big(-\frac{1}{\sqrt{1-\text{x}^2}}\Big)$
$\frac{\text{dy}}{\text{dx}}=-\frac{1}{\sqrt{1-\text{x}^2}}$
View full question & answer→Question 2005 Marks
Differentiate the following functions with respect to x:
$\frac{\text{x}^2(1-\text{x}^2)}{\cos2\text{x}}$
AnswerConsider $\text{y}=\frac{\text{x}^2+2}{\sqrt{\cos\text{x}}}$
Differentiate it with respect to x and applying the chain and product rule, we get
$\frac{\text{dy}}{\text{dx}}=\frac{\cos2\text{x}\frac{\text{d}}{\text{dx}}\text{x}^2(1-\text{x}^2)^3-\text{x}^2(1-\text{x}^2)^3\frac{\text{d}}{\text{dx}}\cos2\text{x}}{\cos^2 2\text{x}}$
$=\frac{\cos2\text{x}\Big[\text{x}^2\frac{\text{d}}{\text{dx}}(1-\text{x}^2)^3+(1-\text{x}^2)^3\frac{\text{d}}{\text{dx}}\text{x}^2-\text{x}^2(1-\text{x}^2)^3(-2\sin2\text{x})\Big]}{\cos^2 2\text{x}}$
$=\frac{\cos2\text{x}\Big[-6\text{x}^2(1-\text{x}^2)^2+(1-\text{x}^2)^32\text{x}+2\text{x}^2(1-\text{x}^2)^3\sin2\text{x}\Big]}{\cos^2 2\text{x}}$
$=\frac{2\text{x}(1-\text{x}^2)^2}{\cos2\text{x}}-\frac{6\text{x}^3(1-\text{x}^2)^2}{\cos2\text{x}}+\frac{2\text{x}^2(1-\text{x}^2)^3\sin2\text{x}}{\cos^2 2\text{x}}$
$=2\text{x}(1-\text{x}^2)\sec2\text{x}\big\{1-4\text{x}^2+\text{x}(1-\text{x}^2)\tan2\text{x}\big\}$
Therefore,
$\frac{\text{dy}}{\text{dx}}=2\text{x}(1-\text{x}^2)\sec2\text{x}\big\{1-4\text{x}^2+\text{x}(1-\text{x}^2)\tan2\text{x}\big\}$
View full question & answer→Question 2015 Marks
If $e^x + x^y = e^{x+y}$, prove that $\frac{\text{dy}}{\text{dx}}+\text{e}^{\text{y}-\text{x}}=0$
AnswerHere,
$e^x + e^y = e^{x+y} ......(i)$
Differentiating both the sides using chain rule,
$\frac{\text{d}}{\text{dx}}(\text{e}^\text{x})+\frac{\text{d}}{\text{dx}}(\text{e}^\text{y})=\frac{\text{d}}{\text{dx}}(\text{e}^{\text{x}+\text{y}})$
$\text{e}^{\text{x}}+\text{e}^{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})$
$\text{e}^{\text{x}}+\text{e}^{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}\Big[1+\frac{\text{d}}{\text{dx}}\Big]$
$\text{e}^{\text{y}}\frac{\text{dy}}{\text{dx}}-\text{e}^{\text{x}+\text{y}}\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}-\text{e}^\text{x}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^{\text{x}+\text{y}}-\text{e}^\text{x}}{\text{e}^\text{y}-\text{e}^{\text{x}+\text{y}}}$
$=\Big(\frac{\text{e}^\text{x}+\text{e}^\text{x}-\text{e}^\text{x}}{\text{e}^\text{y}-\text{e}^\text{x}-\text{e}^\text{y}}\Big)$
[Using equation (i)]
$\frac{\text{dy}}{\text{dx}}=-\text{e}^{\text{y}-\text{x}}$
$\frac{\text{dy}}{\text{dx}}+\text{e}^{\text{y}-\text{x}}=0$
View full question & answer→Question 2025 Marks
Differentiate $(\log\text{x})^\text{x}$ with respect to x.
AnswerLet $\text{u}=(\log1+\text{x})^\text{x}$
Taking log on both sides,
$\log\text{u}=\log(\log\text{x})^\text{x}$
$\Rightarrow\log\text{u}=\text{x}\log(\log\text{x})$
$\Rightarrow\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}\big\{\log(\log\text{x})\big\}+\log(\log\text{x})\frac{\text{d}}{\text{dx}}(\text{x})$
$\Rightarrow\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\text{x}\Big(\frac{1}{\log\text{x}}\Big)\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\log\text{x}(1)$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{u}\Big[\frac{\text{x}}{\log\text{x}}\big(\frac{1}{\text{x}}\big)+\log\log\text{x}\Big]$
$\Rightarrow\frac{\text{du}}{\text{dx}}=(\log\text{x})^\text{x}\Big[\frac{1}{\log\text{x}}+\log\log\text{x}\Big]\ .....(\text{i})$
Again, let $\text{v}=\log\text{x}$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{1}{\text{x}}\ .....(\text{ii})$
Dividing equation (i) by (ii), we get
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{(\log\text{x})^\text{x}\Big[\frac{1}{\log\text{x}}+\log\log\text{x}\Big]}{\frac{1}{\text{x}}}$
$\Rightarrow\frac{\text{du}}{\text{dv}}=\frac{(\log\text{x})^\text{x}\Big[\frac{1+\log\text{x}(\log\log\text{x})}{\log\text{x}}\Big]}{\frac{1}{\text{x}}}$
$\Rightarrow\frac{\text{du}}{\text{dv}}=\text{x}(\log\text{x})^{\text{x}{-1}}(1+\log\text{x}\times\log\log\text{x})$
View full question & answer→Question 2035 Marks
If $\text{x}-\text{e}^{\frac{\text{x}}{\text{y}}},$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\text{x}-\text{y}}{\text{x}\log\text{x}}$
Answer$\text{x}-\text{e}^{\frac{\text{x}}{\text{y}}}$
Taking logarithm on both sides, we get
$\log\text{x}=\frac{\text{x}}{\text{y}}$
$\Rightarrow\text{y}\log\text{x}=\text{x}$
$\Rightarrow\log\text{x}\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}}=1$
$\Rightarrow\log\text{x}\frac{\text{dy}}{\text{dx}}=1-\frac{\text{y}}{\text{x}}$
$\Rightarrow\log\text{x}\frac{\text{dy}}{\text{dx}}=\frac{\text{x}-\text{y}}{\text{x}}$
$\Rightarrow \text{x}\log\text{x}\frac{\text{dy}}{\text{dx}}=\text{x}-\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}-\text{y}}{\text{x}\log\text{x}}$
View full question & answer→Question 2045 Marks
Differentiate $\sin^{-1}\Big(2\text{x}\sqrt{1-\text{x}^2}\Big)$ with respect to $\sec^{-1}\Big(\frac{1}{\sqrt{1+\text{x}^2}}\Big),$ if:
$\text{x}\in\Big(0,\frac{1}{\sqrt{2}}\Big)$
AnswerLet $\text{u}=\sin^{-1}\Big(2\text{x}\sqrt{1-\text{x}^2}\Big)$
Put $\text{x}=\sin\theta$
$\Rightarrow\text{u}=\sin^{-1}\Big(2\sin\theta\sqrt{1-\sin^2\theta}\Big)$
$\Rightarrow\text{u}=\sin^{-1}(2\sin\theta\cos\theta)$
$\Rightarrow\text{u}=\sin^{-1}(\sin2\theta)\ .....(\text{i})$
And,
Let $\text{v}=\sec^{-1}\Big(\frac{1}{\sqrt{1-\text{x}^2}}\Big)$
$\Rightarrow\text{v}=\sec^{-1}\Big(\frac{1}{\sqrt{1-\sin^2\theta}}\Big)$
$\Rightarrow\text{v}=\sec^{-1}\Big(\frac{1}{\cos\theta}\Big)$
$\Rightarrow\text{v}=\sec^{-1}(\sec\theta)$
$\Rightarrow\text{v}=\cos^{-1}\bigg(\frac{1}{\frac{1}{\cos\theta}}\bigg)\ \Big[\text{Since},\sec^{-1}\text{x}=\cos^{-1}\big(\frac{1}{\text{x}}\big)\Big]$
$\Rightarrow\text{v}=\cos^{-1}(\cos\theta)\ .....(\text{ii})$
Here,
$\text{x}\in\Big(0,\frac{1}{\sqrt{2}}\Big)$
$\Rightarrow\sin\theta\in\Big(0,\frac{1}{\sqrt{2}}\Big)$
$\Rightarrow\theta\in\Big(0,\frac{\pi}{4}\Big)$
So, from equation (i),
$\text{u}=2\theta\Big[\text{Since},\sin^{-1}(\sin\theta)=\theta,\text{ if }\theta\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\Big]$
Let, $\text{u}=2\sin^{-1}\text{x}\big[\text{Since},\text{x}=\sin\theta\big]$
Differentiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=2\Big(\frac{1}{\sqrt{1-\text{x}^2}}\Big)$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\frac{2}{\sqrt{1-\text{x}^2}}\ .....(\text{iii})$
And, from equation (ii),
$\text{v}=\theta\big[\text{Since},\cos^{-1}(\cos\theta)=\theta,\text{ if }\theta\in[0,\pi]\big]$
$\Rightarrow\text{v}=\sin^{-1}\text{x}\big[\text{Since},\text{x}=\sin\theta\big]$
Differentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}\ .....(\text{iv})$
Dividing equation (iii) by (iv),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{2}{\sqrt{1-\text{x}^2}}\times\frac{\sqrt{1-\text{x}^2}}{1}$
$\therefore\frac{\text{du}}{\text{dv}}=2$
View full question & answer→Question 2055 Marks
Differentiate $\sin^{-1}\sqrt{1-\text{x}^2}$ with respect to $\cos^{-1}\text{x},$ if
$\text{x}\in(0, 1)$
AnswerLet $\text{u}=\sin^{-1}\sqrt{1-\text{x}^2}$
Put $\text{x}=\cos\theta$
$\Rightarrow\text{u}=\sin^{-1}\sqrt{1-\cos^2\theta}$
$\Rightarrow\text{u}=\sin^{-1}(\sin\theta)\ .....(\text{i})$
And, $\text{v}=\cos^{-1}\text{x}\ .....(\text{ii})$
Now, $\text{x}\in(0,1)$
$\Rightarrow\cos\theta\in(0,1)$
$\Rightarrow\theta\in\Big(0,\frac{\pi}{2}\Big)$
So, from equation (i),
$\text{u}=\theta$
$\Big[\text{Since},\sin^{-1}(\sin\theta)=\theta\text{ if }\theta\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\Big]$
$\Rightarrow\text{u}=\cos^{-1}\text{x}\big[\text{Since},\cos\theta=\text{x}\big]$
Differentiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=\frac{-1}{\sqrt{1-\text{x}^2}}\ .....(\text{iii})$
From equation (ii),
$\text{v}=\cos^{-1}\text{x}$
Differentaiting it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{-1}{\sqrt{1-\text{x}^2}}\ .....(\text{iv})$
Dividing equation (iii) by (iv),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{-1}{\sqrt{1-\text{x}^2}}\times\frac{\sqrt{1-\text{x}^2}}{-1}$
$\therefore\frac{\text{du}}{\text{dx}}=1$
View full question & answer→Question 2065 Marks
Differentiate $\tan^{-1}\Big(\frac{\cos\text{x}}{1+\sin\text{x}}\Big)$ with respect to $\sec^{-1}\text{x}$
AnswerLet, $\text{u}=\tan^{-1}\Big(\frac{\cos\text{x}}{1+\sin\text{x}}\Big)$
$\Rightarrow\text{u}=\tan^{-1}\Big[\tan\Big(\frac{\pi}{4}-\frac{\text{x}}{2}\Big)\Big]$
$\Rightarrow\text{u}=\frac{\pi}{4}-\frac{\text{x}}{2}$
Differentiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=0-\big(\frac{1}{2}\big)$
$\frac{\text{du}}{\text{dx}}=-\frac{1}{2}\ .....(\text{i})$
Let, $\text{v}=\sec^{-1}\text{x}$
Differentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{1}{\text{x}\sqrt{\text{x}^2-1}}\ .....\text{(ii)}$
Differentiating equation (i) by (ii),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=-\frac{1}{2}\times\frac{\text{x}\sqrt{\text{x}^2-1}}{1}$
$\frac{\text{du}}{\text{dv}}=\frac{-\text{x}\sqrt{\text{x}^2-1}}{2}$
View full question & answer→Question 2075 Marks
If $\text{e}^{\text{x}}+\text{e}^{\text{y}}=\text{e}^{\text{x}+\text{y}},$ prove that $\frac{\text{dy}}{\text{dx}}=-\frac{\text{e}^{\text{x}}(\text{e}^\text{y}-1)}{\text{e}^{\text{y}}(\text{e}^{\text{x}}-1)}$ or $\frac{\text{dy}}{\text{dx}}+\text{e}^{\text{y}-\text{x}}=0$
Answer$\text{e}^\text{x}+\text{e}^\text{y}=\text{e}^{\text{x}+\text{y}}$
$\Rightarrow\text{e}^\text{x}+\text{e}^\text{y}\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}\Big(1+\frac{\text{dy}}{\text{dx}}\Big)$
$\Rightarrow\text{e}^\text{x}+\text{e}^{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}+\text{e}^{\text{x}+\text{y}}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\text{e}^\text{y}\frac{\text{dy}}{\text{dx}}-\text{e}^{\text{x}+\text{y}}\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}-\text{e}^{\text{x}}$
$\Rightarrow\frac{\text{dx}}{\text{dy}}(\text{e}^\text{y}-\text{e}^{\text{x}+\text{y}})=\text{e}^{\text{x}+\text{y}}-\text{e}^{\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^{\text{x}+\text{y}}-\text{e}^{\text{x}}}{\text{x}^\text{y}-\text{e}^{\text{x}+\text{y}}}$
$=\frac{\text{e}^\text{x}(\text{e}^\text{y}-1)}{\text{e}^\text{y}({1-\text{e}}^\text{x})}$
$=-\frac{\text{e}^\text{x}(\text{e}^\text{y}-1)}{\text{e}^\text{y}(\text{e}^\text{x}-1)}$
View full question & answer→Question 2085 Marks
If $\text{y}=\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}},$ prove that $2\text{x}\frac{\text{dy}}{\text{dx}}=\sqrt{\text{x}}-\frac{1}{\sqrt{\text{x}}}$
Answer$\text{y}=\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}\Big)$
$=\frac{\text{d}}{\text{dx}}\big(\sqrt{\text{x}}\big)+\frac{\text{d}}{\text{dx}}\Big(\text{x}^{-1\frac{1}{2}}\Big)$
$=\frac{1}{2\sqrt{\text{x}}}+\Big(-\frac{1}{2}\times\text{x}^{-\frac{1}{2}-1}\Big)$
$=\frac{2}{2\sqrt{\text{x}}}-\frac{1}{2\sqrt[\text{x}]{\text{x}}}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}-1}{2\text{x}\sqrt{\text{x}}}$
$\Rightarrow 2\text{x}\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\sqrt{\text{x}}}-\frac{1}{\sqrt{\text{x}}}$
Hence, the solution is, $2\text{x}\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\sqrt{\text{x}}}-\frac{1}{\sqrt{\text{x}}}$
View full question & answer→Question 2095 Marks
If $\cos\text{y}=\text{x}\cos(\text{a}+\text{y}),$ with $\cos\text{a}\neq\pm1,$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\cos^2(\text{a}+\text{y})}{\sin\text{a}}$
AnswerWe have, $\cos\text{y}=\text{x}\cos(\text{a}+\text{y})$
Differentiating with respect to x, we get,
$\frac{\text{d}}{\text{dx}}(\cos\text{y})=\frac{\text{d}}{\text{dx}}\big\{\text{x}\cos(\text{a}+\text{y})\big\}$
$\Rightarrow -\sin\text{y}\frac{\text{dy}}{\text{dx}}=\cos(\text{a}+\text{y})\frac{\text{d}}{\text{dx}}(\text{x})+\text{x}\frac{\text{d}}{\text{dx}}\cos(\text{a}+\text{y})$
$\Rightarrow -\sin\frac{\text{dy}}{\text{dx}}=\cos(\text{a}+\text{y})+\text{x}\big[-\sin(\text{a}+\text{y})\big]\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\big[\text{x}\sin(\text{a}+\text{y})-\sin\text{y}\big]\frac{\text{dy}}{\text{dx}}=\cos(\text{a}+\text{y})$
$\Rightarrow\Big[\frac{\cos\text{y}}{\cos(\text{a}+\text{y})}\sin(\text{a}+\text{y})-\sin\text{y}\Big]\frac{\text{dy}}{\text{dx}}=\cos(\text{a}+\text{y}) \\ \Big[\because \cos\text{y}=\text{x}\cos(\text{a}+\text{y})\Rightarrow\text{x}=\frac{\cos\text{y}}{\cos(\text{a}+\text{y})}\Big] $
$\Rightarrow\big[\cos\text{y}\sin(\text{a}+\text{y})-\sin\text{y}\cos(\text{a}+\text{y})\big]\frac{\text{dy}}{\text{dx}}=\cos^2(\text{a}+\text{y})$
$\Rightarrow \sin(\text{a}+\text{y}-\text{y})\frac{\text{dy}}{\text{dx}}=\cos^2(\text{a}+\text{y})$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\frac{\cos^2(\text{a}+\text{y})}{\sin\text{a}}$
View full question & answer→Question 2105 Marks
If $\text{y}=\frac{\text{x}\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}},$ prove that $(1-\text{x}^2)\frac{\text{dy}}{\text{dx}}=\text{x}+\frac{\text{y}}{\text{x}}$
AnswerGivne, $\text{y}=\frac{\text{x}\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}$
Differentiate with respect to x,
$\frac{\text{d}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}\Big)$
$=\bigg[\frac{\sqrt{1-\text{x}^2}\frac{\text{d}}{\text{dx}}(\text{x}\sin^{-1}\text{x})-(\text{x}\sin^{-1}\text{x})\frac{\text{d}}{\text{dx}}(\sqrt{1-\text{x}^2})}{(\sqrt{1-\text{x}^2})^2}\bigg]$
[Using quotient rule, product rule, chain rule]
$=\begin{bmatrix}\frac{\sqrt{1-\text{x}^2}\Big\{\text{x}\frac{\text{d}}{\text{dx}}\sin^{-1}\text{x}+\sin^{-1}\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big\}-\big(\text{x}\sin^{-1}\text{x}\big)\frac{1}{2\sqrt{1-\text{x}^2}}\frac{\text{d}}{\text{dx}}\big(1-\text{x}^2\big)}{\big(1-\text{x}^2\big)} \end{bmatrix}$
$=\begin{bmatrix}\frac{\sqrt{1-\text{x}^2}\Big\{\frac{\text{x}}{\sqrt{1-\text{x}^2}}+\sin^{-1}\text{x}\Big\}-\frac{\text{x}\sin{-1}\text{x}(-2\text{x})}{2\sqrt{1-\text{x}^2}}}{\big(1-\text{x}^2\big)} \end{bmatrix}$
$=\begin{bmatrix}\frac{\text{x}+\sqrt{1-\text{x}^2}\sin^{-1}\text{x}+\frac{\text{x}^2\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}}{\big(1-\text{x}^2\big)} \end{bmatrix}$
$\Rightarrow\big(1-\text{x}^2\big)\frac{\text{dy}}{\text{dx}}=\text{x}+\frac{\sqrt{1-\text{x}^2}\sin^{-1}}{1}+\frac{\text{x}^2\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}$
$\Rightarrow\big(1-\text{x}^2\big)\frac{\text{dy}}{\text{dx}}=\text{x}+\bigg(\frac{(1-\text{x}^2)\sin^{-1}\text{x}+\text{x}^2\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}\bigg)$
$\Rightarrow\big(1-\text{x}^2\big)\frac{\text{dy}}{\text{dx}}=\text{x}+\bigg(\frac{\sin^{-1}\text{x}-\text{x}^2\sin^{-1}\text{x}+\text{x}^2\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}\bigg)$
$\Rightarrow\big(1-\text{x}^2\big)\frac{\text{dy}}{\text{dx}}=\text{x}+\bigg(\frac{\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}\bigg)$
$\Rightarrow(1-\text{x}^2)\frac{\text{dy}}{\text{dx}}=\text{x}+\frac{\text{y}}{\text{x}}\ \Big\{\text{Since, given y}=\frac{\text{x}\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}\Big\}$
View full question & answer→Question 2115 Marks
Differentiate the following functions with respect to x:
$\frac{\text{x}^2+2}{\sqrt{\cos\text{x}}}$
AnswerLet $\text{y}=\frac{\text{x}^2+2}{\sqrt{\cos\text{x}}}$
Differentiate it with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\sqrt{\cos\text{x}}\frac{\text{d}}{\text{dx}}(\text{x}^2+2)-(\text{x}^2+2)\frac{\text{d}}{\text{dx}}(\sqrt{\cos\text{x}})}{(\sqrt{\cos\text{x}})^2}$
[Using quotient rule and chain rule]
$=\frac{2\text{x}\sqrt{\cos\text{x}}-(\text{x}^2+2)\Big(-\frac{1}{2}\frac{\sin\text{x}}{\sqrt{\cos\text{x}}}\Big)}{\cos\text{x}}$
$=\frac{2\text{x}\sqrt{\cos\text{x}}+\frac{(\text{x}^2+2)\sin\text{x}}{2\sqrt{\cos\text{x}}}}{\cos\text{x}}$
$=\frac{4\text{x}\cos\text{x}+(\text{x}^2+2)\sin\text{x}}{2(\cos\text{x})^\frac{3}{2}}$
$=\frac{2\text{x}}{\sqrt{\cos\text{x}}}+\frac{1}{2}\frac{(\text{x}^2+2)\sin\text{x}}{(\cos\text{x})^\frac{3}{2}}$
$=\frac{1}{\sqrt{\cos\text{x}}}\Big\{2\text{x}+\frac{1}{2}\frac{(\text{x}^2+2)\sin\text{x}}{\cos\text{x}}\Big\}$
$=\frac{1}{\sqrt{\cos\text{x}}}\Big\{2\text{x}+\frac{(\text{x}^2+2)\tan\text{x}}{2}\Big\}$
So,
$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^2+2}{\sqrt{\cos\text{x}}}\Big)=\frac{1}{\sqrt{\cos\text{x}}}\Big\{2\text{x}+\frac{(\text{x}^2+2)\sin\text{x}}{2}\Big\}$
View full question & answer→Question 2125 Marks
Prove that $\frac{\text{dy}}{\text{dx}}\Big\{\frac{\text{x}}{2}\sqrt{\text{a}^2-\text{x}^2}+\frac{\text{a}^2}{2}\sin^{-1}\frac{\text{x}}{\text{a}}\Big\}=\sqrt{\text{a}^2-\text{x}^2}$
Answer$\frac{\text{dy}}{\text{dx}}\Big\{\frac{\text{x}}{2}\sqrt{\text{a}^2-\text{x}^2}+\frac{\text{a}^2}{2}\sin^{-1}\frac{\text{x}}{\text{a}}\Big\}=\sqrt{\text{a}^2-\text{x}^2}$
$\text{L.H.S}=\frac{\text{dy}}{\text{dx}}\Big\{\frac{\text{x}}{2}\sqrt{\text{a}^2-\text{x}^2}+\frac{\text{a}^2}{2}\sin^{-1}\frac{\text{x}}{\text{a}}\Big\}$
$=\frac{\text{dy}}{\text{dx}}\Big(\frac{\text{x}}{2}\sqrt{\text{a}^2-\text{x}^2}\Big)+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{a}^2}{2}\sin^{-1}\frac{\text{x}}{\text{a}}\Big)$
$=\frac{1}{2}\Big[\text{x}\frac{\text{d}}{\text{dx}}\sqrt{\text{a}^2-\text{x}^2}+\sqrt{\text{a}^2-\text{x}^2}\frac{\text{d}}{\text{dx}}(\text{x})\Big] \\ +\frac{\text{a}^2}{2}\times\frac{1}{\sqrt{1-\Big(\frac{\text{x}}{\text{x}}\Big)^2}}\times\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}}{\text{a}}\Big)$
[Using product rule, chain rule]
$=\frac{1}{2}\bigg[\text{x}\times\frac{1}{2\sqrt{\text{a}^2-\text{x}^2}}\frac{\text{d}}{\text{dx}}\big(\text{a}^2-\text{x}^2\big)+\sqrt{\text{a}^2-\text{x}^2}\Big] \\ +\Big(\frac{\text{a}^2}{2}\Big)\times\frac{1}{\sqrt{\frac{\text{a}^2-\text{x}^2}{\text{a}^2}}}\times\Big(\frac{1}{\text{a}}\Big)$
$=\frac{1}{2}\Big[\frac{\text{x}(-2\text{x})}{2\sqrt{\text{a}^2-\text{x}^2}}+\sqrt{\text{a}^2-\text{x}^2}\Big]+\Big(\frac{\text{a}^2}{2}\Big)\times\frac{1}{\sqrt{\text{a}^2-\text{x}^2}}\times\Big(\frac{1}{\text{a}}\Big)$
$=\frac{1}{2}\bigg[\frac{-2\text{x}^2+2\big(\text{a}^2-\text{x}^2\big)}{2\sqrt{\text{a}^2-\text{x}^2}}\bigg]+\frac{\text{a}^2}{2\sqrt{\text{a}^2-\text{x}^2}}$
$=\frac{1}{2}\bigg[\frac{2\big(\text{a}^2-2\text{x}^2\big)}{2\sqrt{\text{a}^2-\text{x}^2}}\bigg]+\frac{\text{a}^2}{2\sqrt{\text{a}^2-\text{x}^2}}$
$=\frac{\text{a}^2-2\text{x}^2}{2\sqrt{\text{a}^2-\text{x}^2}}+\frac{\text{a}^2}{2\sqrt{\text{a}^2-\text{x}^2}}$
$=\frac{\text{a}^2-2\text{x}^2+\text{a}^2}{2\sqrt{\text{a}^2-\text{x}^2}}$
$=\frac{2\text{a}^2-2\text{x}^2}{2\sqrt{\text{a}^2-\text{x}^2}}$
$=\frac{2\big(\text{a}^2-\text{a}^2\big)}{2\sqrt{\text{a}^2-\text{x}^2}}$
$=\frac{(\text{a}^2-\text{x}^2)}{\sqrt{\text{a}^2-\text{x}^2}}$
$=\sqrt{\text{a}^2-\text{x}^2}$
$=\text{R.H.S}$
View full question & answer→Question 2135 Marks
Find $\frac{\text{dy}}{\text{dx}}$ in the following cases:
$\text{x}^{\frac{2}{3}}+\text{y}^{\frac{2}{3}}=\text{a}^{\frac{2}{3}}$
AnswerWe have, $\text{x}^{\frac{2}{3}}+\text{y}^{\frac{2}{3}}=\text{a}^{\frac{2}{3}}$
Differentiating it with respect to x, we get,
$\frac{\text{d}}{\text{dx}}\Big(\text{x}^{\frac{2}{3}}\Big)+\frac{\text{d}}{\text{dx}}\Big(\text{y}^{\frac{2}{3}}\Big)=\frac{\text{d}}{\text{dx}}\Big(\text{a}^{\frac{2}{3}}\Big)$
$\Rightarrow\frac{2}{3}\big(\text{x}\big)^{\frac{2}{3}-1}+\frac{2}{3}\big(\text{y}\big)^{\frac{2}{3}-1}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{2}{3}\big(\text{x}\big)^{\frac{-1}{3}}+\frac{2}{3}\big(\text{y}\big)^{\frac{-1}{3}}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{2}{3}\big(\text{y}\big)^{\frac{-1}{3}}\frac{\text{dy}}{\text{dx}}=-\frac{2}{3}\big(\text{x}\big)^{\frac{-1}{3}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{2}{3}\big(\text{x}\big)^{\frac{-1}{3}}\times\frac{3}{2\text{y}^{\frac{-1}{3}}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{x}^{\frac{-1}{3}}}{\text{y}^{\frac{-1}{3}}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{x}^{\frac{1}{3}}}{\text{y}^{\frac{1}{3}}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\big(\frac{\text{x}}{\text{y}}\big)^{\frac{1}{3}}$
View full question & answer→Question 2145 Marks
Differentiate the following functions with respect to x:
$\cos^{-1}\big\{2\text{x}\sqrt{1-\text{x}^2}\big\},\frac{1}{\sqrt{2}}<\text{x}<1$
AnswerLet $\text{y}=\cos^{-1}\Big\{2\text{x}\sqrt{1-\text{x}^2}\Big\}$
Put $\text{x}=\cos\theta$
$\text{y}=\cos^{-1}\Big\{2\cos\sqrt{1-\cos^2\theta}\Big\}$
$=\cos^{-1}\big\{2\cos\theta\sin\theta\big\}$
$\text{y}=\cos^{-1}\big\{\sin2\theta\big\}$
$\big[\text{Since}, \sin2\theta=2\sin\theta\cos\theta,\sin^2\theta+\cos^2\theta=1\big]$
$\text{y}=\cos^{-1}\Big[\cos\Big(\frac{\pi}{2}-\theta\Big)\Big]\ .....(\text{i})$
Now,
$\frac{1}{\sqrt{2}}<\text{x}<1$
$\Rightarrow\frac{1}{\sqrt{2}}<\cos\theta<1$
$\Rightarrow 0<\theta<\frac{\pi}{4}$
$\Rightarrow 0<2\theta<\frac{\pi}{2}$
$\Rightarrow 0>-2\theta>-\frac{\pi}{2}$
$\Rightarrow\frac{\pi}{2}>\Big(\frac{\pi}{2}-2\theta\Big)>0$
Hence, from equation (i),
$\text{y}=\frac{\pi}{2}-2\theta$
$\big[\text{Sicne}, \cos^{-1}(\cos\theta)=\theta,\text{ if }\theta\in[0,\pi]\big]$
$\text{y}=\frac{\pi}{2}-2\cos^{-1}\text{x}\ \big[\text{Since x}=\cos\theta\big]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\frac{\pi}{2}\Big)-2\frac{\text{d}}{\text{dx}}\big(\cos^{-1}\text{x}\big)$
$=0-2\Big(\frac{-1}{\sqrt{1-\text{x}^2}}\Big)$
$\frac{\text{dy}}{\text{dx}}=\frac{2}{\sqrt{1-\text{x}^2}}$
View full question & answer→Question 2155 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=\text{e}^{3\text{x}}\sin4\text{x}\times2^\text{x}$
AnswerWe have, $\text{y}=\text{e}^{3\text{x}}\sin4\text{x}\times2^\text{x}\ .....(\text{i})$
Taking log on both sides,
$\log\text{y}=\log\text{e}^{3\text{x}}+\log\sin4\text{x}+\log2^\text{x}$
$\Rightarrow\log\text{y}=3\text{x}\log\text{e}+\log\sin4\text{x}+\text{x}\log2$
$\Rightarrow\log\text{y}=3\text{x}+\log\sin4\text{x}+\text{x}\log2$
Differentiating with resepect to x,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(3\text{x})+\frac{\text{d}}{\text{dx}}(\sin4\text{x})+\frac{\text{d}}{\text{dx}}(\text{x}\log2)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=3+\frac{1}{\sin4\text{x}}\frac{\text{d}}{\text{dx}}(\sin4\text{x})+\log2(1)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=3+\frac{1}{\sin4\text{x}}(\cos4\text{x})\frac{\text{d}}{\text{dx}}(4\text{x})+\log2$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=3+\cot4\text{x}(4)+\log2$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=3+4\cot4\text{x}+\log2$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}\big[3+4\cot4\text{x}+\log2\big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{e}^{3\text{x}}\sin4\text{x}2^\text{x}\big[3+4\cot4\text{x}+\log2\big]$
[Using equation (i)]
View full question & answer→Question 2165 Marks
Differentiate the following functions with respect to x:
$\sin(2\sin^{-1}\text{x})$
AnswerLet, $\text{y}=\sin(2\sin^{-1}\text{x})$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\sin(2\sin^{-1}\text{x})\Big)$
$=\cos\big(2\sin^{-1}\text{x}\big)\frac{\text{d}}{\text{dx}}\big(2\sin^{-1}\text{x}\big)$
[Using chain rule]
$=\cos\big(2\sin^{-1}\text{x}\big)\times2\frac{1}{\sqrt{1-\text{x}^2}}$
$=\frac{2\cos\big(2\sin^{-1}\text{x}\big)}{\sqrt{1-\text{x}^2}}$
So,
$\frac{\text{d}}{\text{dx}}\Big(\sin\big(2\sin^{-1}\text{x}\big)\Big)=\frac{2\cos\big(2\sin^{-1}\text{x}\big)}{\sqrt{1-\text{x}^2}}$
View full question & answer→Question 2175 Marks
If $\text{x}=\text{a}\Big(\frac{1+\text{t}^2}{1-\text{t}^2}\Big)\text{ and y}=\frac{2\text{t}}{1-\text{t}^2},$ find $\frac{\text{dy}}{\text{dx}}$
AnswerWe have, $\text{x}=\text{a}\Big(\frac{1+\text{t}^{2}}{1-\text{t}^{2}}\Big)$ $\Rightarrow\frac{\text{dx}}{\text{dt}}=\text{a}\bigg[\frac{(1-\text{t}^{2})\frac{\text{d}}{\text{dt}}(1+\text{t}^{2})-(1+\text{t}^{2})\frac{\text{d}}{\text{dt}}(1-\text{t}^{2})}{(1-\text{t}^{2})}\bigg]$[Using quotient rule]
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\text{a}\bigg[\frac{(1-\text{t}^{2})(2\text{t)}-(1+\text{t}^{2})(-2\text{t})}{(1-\text{t}^{2})^{2}}\bigg]$ $\Rightarrow\frac{\text{dx}}{\text{dt}}=\text{a}\bigg[\frac{2\text{t}-2\text{t}^{3}+2\text{t}+2\text{t}^{3}}{(1-\text{t}^{2})^{2}}\bigg]$ $\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{4\text{a}\text{t}}{(1-\text{t}^{2})^{2}}\ .....(\text{i})$ and, $\text{y}=\frac{2\text{t}}{1-\text{t}^{2}}$ $\Rightarrow\frac{\text{dx}}{\text{dt}}=2\bigg[\frac{(1-\text{t}^{2})\frac{\text{d}}{\text{dt}}(\text{t})-\text{t}\frac{\text{d}}{\text{dt}}(1-\text{t}^{2})}{(1-\text{t}^{2})^{2}}\bigg]$ $\Rightarrow\frac{\text{dx}}{\text{dt}}=2\bigg[\frac{(1-\text{t}^{2})(1)-\text{t}(-2\text{t})}{(1-\text{t}^{2})^{2}}\bigg]$ $\Rightarrow\frac{\text{dx}}{\text{dt}}=2\bigg[\frac{(1-\text{t}^{2})+2\text{t}^{2}}{(1-\text{t}^{2})^{2}}\bigg]$ $\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{2(1-\text{t}^{2})}{(1-\text{t}^{2})^{2}}\ .....(\text{ii})$ Dividing equation (ii) by (i), $\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{2(1+\text{t}^{2})}{(1-\text{t}^{2})^{2}}\times\frac{(1-\text{t}^{2})^{2}}{4\text{a}\text{t}}$ $\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(1+\text{t}^{2})}{2\text{a}\text{t}}$
View full question & answer→Question 2185 Marks
If $\text{y}=\text{x}\sin^{-1}\text{x}+\sqrt{1-\text{x}^2},$ prove that $\frac{\text{dy}}{\text{dx}}=\sin^{-1}\text{x}$
AnswerWe have, $\text{y}=\text{x}\sin^{-1}\text{x}+\sqrt{1-\text{x}^2}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big[\text{x}\sin^{-1}\text{x}+\sqrt{1-\text{x}^2}\Big]$
$=\frac{\text{d}}{\text{dx}}\big(\text{x}\sin^{-1}\text{x}\big)+\frac{\text{d}}{\text{dx}}\big(\sqrt{1-\text{x}^2}\big)$
$=\Big[\text{x}\frac{\text{d}}{\text{dx}}\sin^{-1}\text{x}+\sin^{-1}\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big]+\frac{1}{2\sqrt{1-\text{x}^2}}\frac{\text{d}}{\text{dx}}(1-\text{x}^2)$
$=\Big[\frac{\text{x}}{\sqrt{1-\text{x}^2}}+\sin^{-1}\text{x}\Big]-\frac{2\text{x}}{2\sqrt{1-\text{x}^2}}$
$=\frac{\text{x}}{\sqrt{1-\text{x}^2}}+\sin^{-1}\text{x}-\frac{\text{x}}{\sqrt{1-\text{x}^2}}$
$=\sin^{-1}\text{x}$
View full question & answer→Question 2195 Marks
Differentiate the following functions with respect to x:
$\text{e}^{3\text{x}}\cos2\text{x}$
AnswerConsider $\text{y}=\text{e}^{3\text{x}}\cos2\text{x}$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\text{e}^{3\text{x}}\cos2\text{x}$
$=\text{e}^{3\text{x}}\times\frac{\text{d}}{\text{dx}}(\cos2\text{x})+\cos2\text{x}\frac{\text{d}}{\text{dx}}(\text{e}^{3\text{x}})$
[Using chain rule]
$=\text{e}^{3\text{x}}\times(-\sin2\text{x})\frac{\text{d}}{\text{dx}}(2\text{x})+\cos2\text{xe}^{3\text{x}}\frac{\text{d}}{\text{dx}}(3\text{x})$
[Using chain rule]
$=-2\text{e}^{3\text{x}}\sin2\text{x}+3\text{e}^{3\text{x}}\cos2\text{x}$
$=\text{e}^{3\text{x}}(3\cos2\text{x}-2\sin2\text{x})$
Hence, the solution is, $\frac{\text{d}}{\text{dx}}(\text{e}^{3\text{x}}\cos2\text{x})=\text{e}^{3\text{x}}(3\cos2\text{x}-2\sin2\text{x})$
View full question & answer→Question 2205 Marks
Differentiate the following functions with respect to x:
$\text{e}^{\tan^{-1}\sqrt{\text{x}}}$
AnswerLet, $\text{y}=\text{e}^{\tan^{-1}\sqrt{\text{x}}}$
Differentiate it with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\tan^{-1}\sqrt{\text{x}}}\big)$
$=\text{e}^{\tan^{-1}\sqrt{\text{x}}}\frac{\text{d}}{\text{dx}}\big(\tan^{-1}\sqrt{\text{x}}\big)$
[Using chain rule]
$=\text{e}^{\tan^{-1}\sqrt{\text{x}}}\times\frac{1}{1+(\sqrt{\text{x}})^2}\frac{\text{d}}{\text{dx}}\big(\sqrt{\text{x}}\big)$
$=\frac{\text{e}^{\tan^{-1}\sqrt{\text{x}}}}{1+\text{x}}\times\frac{1}{2\sqrt{\text{x}}}$
$=\frac{\text{e}^{\tan^{-1}\sqrt{\text{x}}}}{2\sqrt{\text{x}}(1+\text{x})}$
So,
$\frac{\text{d}}{\text{dx}}\Big(\text{e}^{\tan^{-1}\sqrt{\text{x}}}\Big)=\frac{\text{e}^{\tan^{-1}\sqrt{\text{x}}}}{2\sqrt{\text{x}}(1+\text{x})}$
View full question & answer→Question 2215 Marks
If $(\cos\text{x})^\text{y}=(\cos\text{y})^\text{x},$ find $\frac{\text{dy}}{\text{dx}}$
Answer$(\cos\text{x})^\text{y}=(\cos\text{y})^\text{x}$
Taking log on both sides we get
$\text{y}\log\cos\text{x}=\text{x}\log\cos\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\log\cos\text{x}-\text{y}\tan\text{x}=\log\cos\text{y}-\text{x}\tan\text{y}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\log\cos\text{x}+\text{x}\tan\text{y}\frac{\text{dy}}{\text{dx}}=\log\cos\text{y}+\text{y}\tan\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}(\log\cos\text{x}+\text{x}\tan\text{y})=\log\cos\text{y}+\text{y}\tan\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\log\cos\text{y}+\text{y}\tan\text{x}}{\log\cos\text{x}+\text{x}\tan\text{y}}$
View full question & answer→Question 2225 Marks
If $\text{y}=\sin^{-1}\big(6\text{x}\sqrt{1-9\text{x}^2}\big), -\frac{1}{3\sqrt{2}}<\text{x}<\frac{1}{3\sqrt{2}},$ then find $\frac{\text{dy}}{\text{dx}}.$
AnswerWe have, $\text{y}=\sin^{-1}\big(6\text{x}\sqrt{1-9\text{x}^2}\big), -\frac{1}{3\sqrt{2}}<\text{x}<\frac{1}{3\sqrt{2}}$
So, $\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[\sin^{-1}\big(6\text{x}\sqrt{1-9\text{x}^2}\big)\big]$
$=\frac{\text{d}}{\text{dx}}\Big[\sin^{-1}\big(6\text{x}\sqrt{1-9\text{x}^2}\big)\Big]$
$=\frac{1}{\sqrt{1-\big(6\text{x}\sqrt{1-9\text{x}^2}\big)^2}}\times\frac{\text{d}}{\text{dx}}\big(6\text{x}\sqrt{1-9\text{x}^2}\big)$
$=\frac{1}{\sqrt{1-[36\text{x}^2(1-9\text{x}^2)]}}\times\Big(6\text{x}\frac{\text{d}}{\text{dx}}\sqrt{1-9\text{x}^2}+\sqrt{1-9\text{x}^2}\frac{\text{d}}{\text{dx}}(6\text{x})\Big)$
$=\frac{1}{\sqrt{1-36\text{x}^2-324\text{x}^4}}\times\Big(6\text{x}\times\frac{1}{2\sqrt{1-9\text{x}^2}}\frac{\text{d}}{\text{dx}}(1-9\text{x}^2)+\sqrt{1-9\text{x}^2}(6)\Big)$
$=\frac{1}{\sqrt{1-36\text{x}^2-324\text{x}^4}}\times\Big(6\text{x}\times\frac{1}{2\sqrt{1-9\text{x}^2}}(-18\text{x}^2)+6\sqrt{1-9\text{x}^2}\Big)$
$=\frac{1}{\sqrt{1-36\text{x}^2-324\text{x}^4}}\times\Big(\frac{-54\text{x}^2}{\sqrt{1-9\text{x}^2}}+6\sqrt{1-9\text{x}^2}\Big)$
$=\frac{1}{\sqrt{1-36\text{x}^2-324\text{x}^4}}\times\Big(\frac{-54\text{x}^2+6\sqrt{1-9\text{x}^2}}{\sqrt{1-9\text{x}^2}}\Big)$
$=\frac{-54\text{x}^2+6-54\text{x}}{\sqrt{1-9\text{x}^2}\sqrt{1-36\text{x}^2-324\text{x}^4}}$
$=\frac{6-108\text{x}}{\sqrt{1-9\text{x}^2}\sqrt{1-36\text{x}^2-324\text{x}^4}}$
View full question & answer→Question 2235 Marks
If $\text{y}=\sqrt{\text{a}^2-\text{x}^2},$ prove that $\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}=0$
AnswerDIfferentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\sqrt{\text{a}^2-\text{x}^2}\big)$
$=\frac{1}{2\sqrt{\text{a}^2-\text{x}^2}}\frac{\text{d}}{\text{dx}}\big(\text{a}^2-\text{x}^2\big)$
[Using chain rule]
$=\frac{1}{2\sqrt{\text{a}^2-\text{x}^2}}(-2\text{x})$
$=\frac{-\text{x}}{\sqrt{\text{a}^2-\text{x}^2}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-\text{x}}{\text{y}}$
$\big[\text{Since},\sqrt{\text{a}^2-\text{x}^2}=\text{y}\big]$
$\Rightarrow\text{y}\frac{\text{dy}}{\text{dx}}=-\text{x}$
Hence, the solution is, $\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}=0$
View full question & answer→Question 2245 Marks
If $\text{x}-\text{e}^{\tan\text{x}}+\sqrt{\frac{\text{x}^2+1}{2}},$ find $\frac{\text{dy}}{\text{dx}}$
Answer$\text{y}=\text{x}^{\tan\text{x}}+\sqrt{\frac{\text{x}^2+1}{2}}$
$\text{y}=\text{e}^{\tan\text{x}\log\text{x}}+\text{e}^{\frac{1}{2}\log\big(\frac{\text{x}^2+1}{2}\big)}$
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\tan\text{x}\log\text{x}}\frac{\text{d}}{\text{dx}}(\tan\text{x}\log\text{x})+\text{e}^{\frac{1}{2}\log\big(\frac{\text{x}^2+1}{2}\big)}\frac{\text{d}}{\text{dx}}\Big(\frac{1}{2}\log\Big(\frac{\text{x}^2+1}{2}\Big)\Big)$
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\tan\text{x}}\Big[\frac{\tan\text{x}}{\text{x}}+\sec^3\text{x}\log\text{x}\Big]+\sqrt{\frac{\text{x}^2+1}{2}}\Big(\frac{1}{2}\times\frac{2}{\text{x}^2+1}\times(\text{x})\Big)$
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\tan\text{x}}\Big[\frac{\tan\text{x}}{\text{x}}+\sec^3\text{x}\log\text{x}\Big]+\sqrt{\frac{\text{x}^2+1}{2}}\Big(\frac{\text{x}}{\text{x}^2+1}\Big)$
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\tan\text{x}}\Big[\frac{\tan\text{x}}{\text{x}}+\sec^3\text{x}\log\text{x}\Big]+\frac{\text{x}}{\sqrt{2(\text{x}^2+1)}}$
View full question & answer→Question 2255 Marks
If $\text{xy}=1,$ prove that $\frac{\text{dy}}{\text{dx}}+\text{y}^2=0$
AnswerHere, xy = 1 .....(i)
Differentiating with respect to x,
$\frac{\text{d}}{\text{dx}}(\text{xy})=\frac{\text{d}}{\text{dx}}(1)$
$\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\frac{\text{d}}{\text{dx}}(\text{x})=0$
[Using product rule]
$ \Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}(1)=0$
$\Big[\text{Put x}=\frac{1}{\text{y}}\text{ from equation (i)}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}}{\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}}{\frac{1}{\text{y}}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\text{y}^2$
$\Rightarrow\frac{\text{dy}}{\text{dx}}+\text{y}^2=0$
View full question & answer→Question 2265 Marks
Differentiate the following functions with respect to x:
$\sin^{-1}\Big\{\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}}\Big\}$
AnswerLet $\text{y}=\sin^{-1}\Big\{\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}}\Big\}$
Put $\text{x}=\text{a}\tan\theta$
$\Rightarrow\text{y}=\sin^{-1}\Big\{\frac{\text{a}\tan\theta}{\sqrt{\text{a}^2\tan^2\theta+\text{a}^2}}\Big\}$
$\Rightarrow\text{y}=\sin^{-1}\bigg\{\frac{\text{a}\tan\theta}{\sqrt{\text{a}^2(\tan^2\theta+1)}}\bigg\}$
$\Rightarrow\text{y}=\sin^{-1}\Big(\frac{\text{a}\tan\theta}{\text{a}\sec\theta}\Big)$
$\Rightarrow\text{y}=\sin^{-1}(\sin\theta)$
$\Rightarrow\text{y}=\theta$
$\Rightarrow\text{y}=\tan^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)\big[\text{Since, x} = \text{a}\tan\theta\big]$
Differentiating it with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{1+\big(\frac{\text{x}}{\text{a}}\big)^2}\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}}{\text{a}}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{a}^2}{\text{a}^2+\text{x}^2}\times\big(\frac{1}{\text{a}}\big)$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\text{a}}{\text{a}^2+\text{x}^2}$
View full question & answer→Question 2275 Marks
Differentiate the following functions with respect to x:
$\text{x}^{\cos^{-1}\text{x}}$
AnswerLet $\text{y}=\text{x}^{\cos^{-1}\text{x}}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=\log\text{x}^{\cos^{-1}\text{x}}$
$\Rightarrow\log\text{y}=\cos^{-1}\text{x}\log\text{x}$
Differentiating with respect to x,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\cos^{-1}\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}\cos^{-1}\text{x}$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\cos^{-1}\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}\Big(\frac{-1}{\sqrt{1-\text{x}^2}}\Big)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\cos^{-1}}{\text{x}}-\frac{\log\text{x}}{\sqrt{1-\text{x}^2}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{\cos^{-1}\text{x}}{\text{x}}-\frac{\log\text{x}}{\sqrt{1-\text{x}^2}}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}^{\cos^{-1}\text{x}}\Big[\frac{\cos^{-1}\text{x}}{\text{x}}-\frac{\log\text{x}}{\sqrt{1-\text{x}^2}}\Big]$
[Using equation (i)]
View full question & answer→Question 2285 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big\{\frac{\text{x}}{1+\sqrt{1-\text{x}^3}}\Big\},-1<\text{x}<1$
AnswerLet $\text{y}=\tan^{-1}\Big\{\frac{\text{x}}{1+\sqrt{1-\text{x}^3}}\Big\}$
Put $\text{x}=\sin\theta$
$\text{y}=\tan^{-1}\Big\{\frac{\sin\theta}{1+\sqrt{1-\sin^2\theta}}\Big\}$
$\text{y}=\tan^{-1}\Big(\frac{\sin\theta}{1+\cos\theta}\Big)$
$\text{y}=\tan^{-1}\bigg\{\frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}}\bigg\}$
$\text{y}=\tan^{-1}\Big(\tan\frac{\theta}{2}\Big)\ .....(\text{i})$
Here, $-1<\text{x}<1$
$\Rightarrow\ -1<\sin\theta<1$
$\Rightarrow -\frac{\pi}{2}<\theta<\frac{\pi}{2}$
$\Rightarrow -\frac{\pi}{4}<\frac{\theta}{2}<\frac{\pi}{4}$
So, from equation (i),
$\text{y}=\frac{\theta}{2}\ \Big[\text{Since}, \tan^{-1}(\tan\theta)=\theta,\text{ if }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\text{y}=\frac{1}{2}\sin^{-1}\text{x}\big[\text{Since, x}=\sin\theta\big]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2\sqrt{1-\text{x}^2}}$
View full question & answer→Question 2295 Marks
Differentiate $\tan^{-1}\Big(\frac{\sqrt{1+\text{x}^2}-1}{\text{x}}\Big)$ with respect to $\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big),$ if $-1<\text{x}<1,\text{x}\neq0.$
AnswerLet $\text{u}=\tan^{-1}\Big(\frac{\sqrt{1+\text{x}^2}-1}{\text{x}}\Big)$
Put $\text{x}=\tan\theta, \text{So}$
$\text{u}=\tan^{-1}\Big(\frac{\sqrt{1+\tan^2\theta}-1}{\tan\theta}\Big)$
$=\tan^{-1}\Big(\frac{\sec\theta-1}{\tan\theta}\Big)$
$=\tan^{-1}\Big(\frac{1-\cos\theta}{\sin\theta}\Big)$
$=\tan^{-1}\bigg(\frac{\frac{2\sin^2\theta}{2}}{\frac{2\sin\theta}{2}\frac{\cos\theta}{2}}\bigg)$
$=\tan^{-1}\bigg(\frac{\frac{\sin\theta}{2}}{\frac{\cos\theta}{2}}\bigg)$
$\text{u}=\tan^{-1}\Big(\frac{\tan\theta}{2}\Big)\ .....(\text{i})$
And,
Let, $\text{v}=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$
$=\sin^{-1}\Big(\frac{2\tan\theta}{1+\tan^2\theta}\Big)$
$\text{v}=\sin^{-1}(\sin2\theta)\ .....(\text{ii})$
Here,
$-1<\text{x}<1$
$\Rightarrow -1<\tan\theta<1$
$\Rightarrow-\frac{\pi}{4}<\theta<\frac{\pi}{4}\ .....(\text{A})$
So, from equation (i),
$\text{u}=\frac{\theta}{2}\Big[\text{Since},\tan^{-1}(\tan\theta)=\theta\text{ if }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\text{u}=\frac{1}{2}\tan^{-1}\text{x}\big[\text{Since},\text{x}=\tan\theta\big]$
Differentiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=\frac{1}{2}\Big(\frac{1}{1+\text{x}^2}\Big)$
$\frac{\text{du}}{\text{dx}}=\frac{1}{2}\frac{1}{(1+\text{x}^2)}\ .....(\text{iii})$
Now, from equation (ii) and (A)
$\text{v}=2\theta\Big[\text{Since},\sin^{-1}(\sin\theta)=\theta,\text{ if }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\text{v}=2\tan^{-1}\text{x}\big[\text{Since},\text{x}=\tan\theta\big]$
Differentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=2\Big(\frac{1}{1+\text{x}^2}\Big)\ .....(\text{iv})$
Dividing equation (iii) by (iv),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{1}{2(1+\text{x}^2)}\times\frac{1+\text{x}^2}{2}$
$\frac{\text{du}}{\text{dv}}=\frac{1}{4}$
View full question & answer→Question 2305 Marks
Find $\frac{\text{dy}}{\text{dx}},$ when
$\text{x}=\text{ae}^{\theta}(\sin\theta-\cos\theta),\text{y}=\text{ae}^\theta(\sin\theta+\cos\theta)$
AnswerWe have, $\text{x}=\text{ae}^{\theta}(\sin\theta-\cos\theta)$ and $\text{y}=\text{ae}^{\theta}(\sin\theta+\cos\theta)$
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{a}\Big[\text{e}^\theta\frac{\text{d}}{\text{d}\theta}(\sin\theta-\cos\theta)+(\sin\theta-\cos\theta)\frac{\text{d}}{\text{d}\theta}(\text{e}^\theta)\Big]$ and
$\frac{\text{dy}}{\text{d}\theta}=\text{a}\Big[\text{e}^\theta\frac{\text{d}}{\text{d}\theta}(\sin\theta+\cos\theta)+(\sin\theta+\cos\theta)\frac{\text{d}}{\text{d}\theta}(\text{e}^\theta)\Big]$
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{a}\Big[\text{e}^\theta(\cos\theta+\sin\theta)+(\sin\theta-\cos\theta)(\text{e}^\theta)\Big]$ and $\frac{\text{dy}}{\text{d}\theta}=\text{a}\Big[\text{e}^\theta(\cos\theta-\sin\theta)+(\sin\theta+\cos\theta)(\text{e}^\theta)\Big]$
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{a}\big[2\text{e}^\theta(\sin\theta)\big]$ and $\frac{\text{dy}}{\text{d}\theta}=\text{a}\big[2\text{e}^\theta(\cos\theta)\big]$
$\therefore\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}=\frac{\text{a}(2\text{e}^\theta\cos\theta)}{\text{a}(2\text{e}^\theta\sin\theta)}=\cot\theta$
View full question & answer→Question 2315 Marks
Find $\frac{\text{dy}}{\text{dx}},$ when
$\text{x}=\frac{\text{e}^\text{t}+\text{e}^{-\text{t}}}{2}\text{ and y}=\frac{\text{e}^\text{t}-\text{e}^\text{-t}}{2}$
AnswerWe have, $ \text{x}=\frac{\text{e}^{\text{t}}+\text{e}^{\text{-t}}}{2}$ and $ \text{y}=\frac{\text{e}^{\text{t}}+\text{e}^{\text{-t}}}{2}$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{1}{2}\bigg[\frac{\text{d}}{\text{dt}}(\text{e}^{\text{t}})+\frac{\text{d}}{\text{dt}}(\text{e}^{\text{-t}})\bigg]$ and $\frac{\text{dy}}{\text{dt}}=\frac{1}{2}\bigg[\frac{\text{d}}{\text{dt}}(\text{e}^{\text{t}})-\frac{\text{d}}{\text{dt}}(\text{e}^{\text{-t}})\bigg]$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{1}{2}\bigg[\text{e}^{\text{t}}+\text{e}^{\text{-t} \frac{\text{d}}{\text{dt}}}(\text{-t})\bigg]$ and $\frac{\text{dy}}{\text{dt}}=\frac{1}{2}\bigg[\text{e}^{\text{t}}-\text{e}^{\text{-t}}\frac{\text{d}}{\text{dt}}({\text{e}^{\text{-t}}})\bigg]$
$\Rightarrow\frac{1}{2}(\text{e}^{\text{t}}-\text{e}^\text{-t})=\text{y}$ and $\frac{\text{dy}}{\text{dt}}=\frac{1}{2}(\text{e}^\text{t}+\text{e}^{\text{-t}})=\text{x}$
$\therefore\frac{\text{dy}}{\text{dt}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{\text{x}}{\text{y}}$
View full question & answer→Question 2325 Marks
If $\text{xy}=4,$ prove that $\text{x}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}^2\Big)=3\text{y}$
AnswerWe have, $\text{xy}=4$
$\Rightarrow\text{y}=\frac{4}{\text{x}}$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\frac{4}{\text{x}}\big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=4\frac{\text{d}}{\text{dx}}\big(\text{x}^{-1}\big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=4(-1\times\text{x}^{-1-1})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=4\Big(-\frac{1}{\text{x}^2}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-4}{\text{x}^2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{4}{\big(\frac{4}{\text{y}}\big)^2}\ \Big[\because\text{x}=\frac{4}{\text{y}}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{4\text{y}^2}{16}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}^2}{4}$
$\Rightarrow4\frac{\text{dy}}{\text{dx}}=-\text{y}^2$
$\Rightarrow4\frac{\text{dy}}{\text{dx}}+4\text{y}^2=3\text{y}^2$
$\Rightarrow4\Big(\frac{\text{dy}}{\text{dx}}+\text{y}^2\Big)=3\text{y}^2$
Dividing both side by x,
$\Rightarrow\frac{4}{\text{x}}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}^2\Big)=\frac{3\text{y}^2}{\text{x}}$
$\Rightarrow\text{y}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}^2\Big)=\frac{3\text{y}^2}{\text{y}}$
$\Rightarrow\text{x}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}^2\Big)=\frac{3\text{y}^2}{\text{y}}$
$\Rightarrow\text{x}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}^2\Big)=3\text{y}$
View full question & answer→Question 2335 Marks
If $\text{y}=\log\sqrt{\frac{1+\tan\text{x}}{1-\tan\text{x}}},$ prove that $\frac{\text{dy}}{\text{dx}}=\sec2\text{x}$
AnswerWe have, $\text{y}=\sqrt{\frac{1+\text{e}^\text{x}}{1-\text{e}^\text{x}}}$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\sqrt{\frac{1+\text{e}^\text{x}}{1-\text{e}^\text{x}}}\Big)$
$=\frac{1}{\sqrt[2]{\Big(\frac{1+\text{e}^\text{x}}{1-\text{e}^\text{x}}\Big)}}\times\frac{\text{d}}{\text{dx}}\Big(\frac{1+\text{e}^\text{x}}{1-\text{e}^\text{x}}\Big)$
[Using chain rule]
$=\frac{1}{2}\times\sqrt{\frac{1-\text{e}^\text{x}}{1+\text{e}^\text{x}}}\Bigg[\frac{\big(1-\text{e}^\text{x}\frac{\text{d}}{\text{dx}}(1+\text{e}^\text{x})-(1+\text{e}^\text{x})\frac{\text{d}}{\text{dx}}(1-\text{e}^\text{x})\big)}{(1-\text{e}^\text{x})^2}\Bigg]$
$=\frac{1}{2}\sqrt{\frac{1-\text{e}^\text{x}}{1+\text{e}^\text{x}}}\bigg[\frac{(1-\text{e}^\text{x})\text{e}^\text{x}+(1+\text{e}^\text{x})\text{e}^\text{x}}{(1-\text{e}^\text{x})^2}\bigg]$
$=\frac{1}{2}\sqrt{\frac{1-\text{e}^\text{x}}{1+\text{e}^\text{x}}}\Big[\frac{2\text{e}^\text{x}}{(1-\text{e}^\text{x})^2}\Big]$
$=\frac{\text{e}^\text{x}}{\sqrt{(1+\text{e}^\text{x})\sqrt{(1+\text{e}^\text{x})}}}\frac{1}{(1-\text{e}^\text{x})}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^\text{x}}{(1-\text{e}^\text{x})\sqrt{1-\text{e}^{2\text{x}}}}$
View full question & answer→Question 2345 Marks
If $x^m + y^n = 1,$ Prove that $\frac{\text{dy}}{\text{dx}}=-\frac{\text{my}}{\text{nx}}$
AnswerWe have, $x^m + y^n = 1$
Taking log on both side,
$\log(\text{x}^\text{m}\text{y}^{\text{n}})=\log(1)$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}(\text{m}\log\text{x})+\frac{\text{d}}{\text{dx}}(\text{n}\log\text{y})=\frac{\text{d}}{\text{dx}}\big\{\log(1)\big\}$
$\Rightarrow\frac{\text{m}}{\text{n}}+\frac{\text{n}}{\text{y}}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{m}}{\text{x}}\times\frac{\text{y}}{\text{n}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{my}}{\text{nx}}$
View full question & answer→Question 2355 Marks
Differentiate the following functions with respect to x:
$(\sin\text{x})^{\cos\text{x}}$
AnswerLet $\text{y}=(\sin\text{x})^{\cos\text{x}}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=(\sin\text{x})^{\cos\text{x}}$
$\Rightarrow\log\text{y}=\cos\text{x}\log(\sin\text{x})$
Differentiating with respect to x,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\cos\text{x}\frac{\text{d}}{\text{dx}}(\log\sin\text{x})+\log\sin\text{x}\frac{\text{d}}{\text{dx}}(\cos\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\cos\text{x}\frac{1}{\sin\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}(-\sin\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\cos\text{x}}{\sin\text{x}}(\cos\text{x})-\sin\text{x}\log\sin\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}[\cos\text{x}\cot\text{x}-\sin\text{x}\log\sin\text{x}]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=(\sin\text{x})^{\cos\text{x}}[\cos\text{x}\cot\text{x}-\sin\text{x}\log\sin\text{x}]$
View full question & answer→Question 2365 Marks
Differentiate the following functions with respect to x:
$\text{x}^{\tan^{-1}\text{x}}$
AnswerLet $\text{y}=\text{x}^{\tan^{-1}\text{x}}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=\log\text{x}^{\tan^{-1}\text{x}}$
$\log\text{y}=\tan^{-1}\text{x}\log\text{x}\ \big[\text{Since},\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating it with respect to x using product rule,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\tan^{-1}\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\tan^{-1}\text{x})$
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\tan^{-1}\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}\Big(\frac{1}{1+\text{x}^2}\Big)$
$\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{\tan^{-1}\text{x}}{\text{x}}+\frac{\log\text{x}}{1+\text{x}^2}\Big]$
$\frac{\text{dy}}{\text{dx}}=\text{x}^{\tan^{-1}\text{x}}\Big[\frac{\tan^{-1}\text{x}}{\text{x}}+\frac{\log\text{x}}{1+\text{x}^2}\Big]$
[Using equation (i)]
View full question & answer→Question 2375 Marks
Differentiate $\sin^{-1}\Big(4\text{x}\sqrt{1-4\text{x}^2}\Big)$ with respect to $\sqrt{1-4\text{x}^2},$ if:
$\text{x}\in\Big(-\frac{1}{2\sqrt{2}},\frac{1}{2}\Big)$
AnswerLet $\text{u}=\sin^{-1}\Big(4\text{x}\sqrt{1-4\text{x}^2}\Big)$
Put $2\text{x}=\cos\theta \text{ So},$
$\Rightarrow\text{u}=\sin^{-1}\Big(2\times\cos\theta\sqrt{1-\cos^2\theta}\Big)$
$\Rightarrow\text{u}=\sin^{-1}(2\cos\theta\sin\theta)$
$\Rightarrow\text{u}=\sin^{-1}(\sin2\theta)\ .....(\text{i})$
Let, $\text{v}=\sqrt{1-4\text{x}^2}\ .....(\text{ii})$
Here,
$\text{x}\in\Big(-\frac{1}{2\sqrt{2}},\frac{1}{2}\Big)$
$\Rightarrow2\text{x}\in\Big(-\frac{1}{\sqrt{2}},1\Big)$
$\Rightarrow\cos\theta\in\Big(\frac{1}{\sqrt{2}},1\Big)$
$\Rightarrow\theta\in\Big(0,\frac{\pi}{4}\Big)$
So, from equation (i),
$\text{u}=2\theta$
$\Big[\text{Since},\sin^{-1}(\sin\theta)=\theta,\text{ if }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\Rightarrow\text{u}=2\cos^{-1}(2\text{x})\big[\text{Since},2\text{x}=\cos\theta\big]$
Differentiating it with respect to x using chain rule,
$\frac{\text{dv}}{\text{dx}}=2\bigg(\frac{-1}{\sqrt{1-(2\text{x})^2}}\bigg)\frac{\text{d}}{\text{dx}}(2\text{x})$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\Big(\frac{-2}{\sqrt{1-4\text{x}^2}}(2)\Big)$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{-4}{\sqrt{1-4\text{x}^2}}\ .....(\text{v})$
Dividing equation (v) by (iv)
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{-4}{\sqrt{1-4\text{x}^2}}\times\frac{\sqrt{1-4\text{x}^2}}{-4\text{x}}$
$\frac{\text{du}}{\text{dv}}=\frac{1}{\text{x}}$
View full question & answer→Question 2385 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big\{\frac{\text{x}}{\sqrt{\text{a}^2-\text{x}^2}}\Big\},-\text{a}<\text{x}<\text{a}$
AnswerLet $\text{y}=\tan^{-1}\Big\{\frac{\text{x}}{\sqrt{\text{a}^2-\text{x}^2}}\Big\}$ Put $\text{x}=\text{a}\sin\theta$ $\text{y}=\tan^{-1}\Big\{\frac{\text{a}\sin\theta}{\sqrt{\text{a}^2-\text{a}^2\sin^2\theta}}\Big\}$ $\text{y}=\tan^{-1}\bigg\{\frac{\text{a}\sin\theta}{\sqrt{\text{a}^2(1-\sin^2\theta)}}\bigg\}$ $\text{y}=\tan^{-1}\Big\{\frac{\text{a}\sin\theta}{\text{a}\cos\theta}\Big\}$ $\text{y}=\tan^{-1}(\tan\theta)\ .....(\text{i})$ Here, $-\text{a}<\text{x}<\text{a}$ $\Rightarrow-1<\frac{\text{x}}{\text{a}}<1$ $\Rightarrow -\frac{\pi}{2}<\theta<\frac{\pi}{2}$ From equation (i), $\text{y}=\theta$ $\Big[\text{Since},\tan^{-1}(\tan\theta)=\theta,\text{ if }\theta \in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$ $\text{y}=\sin^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)\big[\text{Since x}=\text{a }\sin \theta\big]$ Differentiating it with respect to x, $\frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{1-\big(\frac{\text{x}}{\text{a}}\big)^2}}\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}}{\text{a}}\Big)$ $=\frac{\text{a}}{\sqrt{\text{a}^2-\text{x}^2}}\times\big(\frac{1}{\text{a}}\big)$$\frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{\text{a}^2-\text{x}^2}}$
View full question & answer→Question 2395 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{5\text{x}}{1+6\text{x}^3}\Big), -\frac{1}{\sqrt{6}}<\text{x}<\frac{1}{\sqrt{6}}$
AnswerLet $\text{y}=\tan^{-1}\Big(\frac{5\text{x}}{1+6\text{x}^3}\Big)$
$=\tan^{-1}\Big(\frac{3\text{x}+2\text{x}}{1-(3\text{x})(2\text{x})}\Big)$
$\text{y}=\tan^{-1}(3\text{x})+\tan^{-1}(2\text{x})$
$\Big[\text{Since},\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)\Big]$
Differentiating it with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{1+(3\text{x})^2}\frac{\text{d}}{\text{dx}}(3\text{x})+\frac{1}{1+(2\text{x})^2}\frac{\text{d}}{\text{dx}}(2\text{x})$
$=\frac{1}{1+9\text{x}^2}(3)+\frac{1}{1+4\text{x}^2}(2)$
$\frac{\text{dy}}{\text{dx}}=\frac{3}{1+9\text{x}^2}+\frac{2}{1+4\text{x}^2}$
View full question & answer→Question 2405 Marks
If $\text{x}=\text{a}(\theta+\sin\theta),\text{y}=\text{a}(1+\cos\theta),$ find $\frac{\text{dy}}{\text{dx}}.$
AnswerHere,
$\text{x}=\text{a}(\theta+\sin\theta)$
Differentiating it with respect to $\theta$,
$\frac{\text{dx}}{\text{d}\theta}=\text{a}\Big(\frac{\text{d}}{\text{d}\theta}(\theta)+\frac{\text{d}}{\text{d}\theta}(\sin\theta)\Big)$
$\frac{\text{dx}}{\text{d}\theta}=\text{a}(1+\cos\theta)\ .....(\text{i})$
And, $\text{y}=\text{a}(1+\cos\theta)$
Differentiating it with respect to $\theta$,
$\frac{\text{dx}}{\text{d}\theta}=\text{a}(0-\sin\theta)$
$\frac{\text{dx}}{\text{d}\theta}=\text{a}\sin\theta\ .....(\text{ii})$
Dividing equation (ii) by (i),
$\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}=\frac{-\text{a}\sin\theta}{\text{a}(1+\cos\theta)}$
$=\frac{-\frac{2\sin\theta}{2}\frac{\cos\theta}{2}}{\frac{2\cos^2\theta}{2}}$
$\frac{\text{dy}}{\text{dx}}=-\frac{\tan\theta}{2}$
View full question & answer→Question 2415 Marks
If $\text{x}=\cos\text{t}(3-2\cos^2\text{t}),\text{y}\sin\text{t}(3-2\sin^2\text{t})$ find the value of $\frac{\text{dy}}{\text{dx}}\text{ at t}=\frac{\pi}{4}$
Answer$\text{x}=\cos\text{t}(3-2\cos^2\text{t})\text{ and }\text{y}\sin\text{t}(3-2\sin^2\text{t})$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=-\sin\text{t}(3-2\cos^2\text{t})+\cos\text{t}(4\cos\text{t}\sin\text{t})$ and $\frac{\text{dy}}{\text{dx}}=\cos\text{t}(3-2\sin^2\text{t})+\sin\text{t}(-4\sin\text{t}\cos\text{t})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\sin\text{t}+6\sin\text{t}\cos^2\text{t}$ and $ \frac{\text{dy}}{\text{dx}}=3\cos\text{t}-6\sin^2\text{t}\cos\text{t}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\sin\text{t}+(1-2\cos^2\text{t})$ and $\frac{\text{dy}}{\text{dt}}=3\cos\text{t}(1-2\sin^2\text{t})$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=3\sin\text{t}\cos2\text{t}\cos2\text{t}$ and $\frac{\text{dy}}{\text{dt}}=3\cos\text{t}(1-2\sin^2\text{t})$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dx}}}{\frac{\text{dx}}{\text{dt}}}=\frac{3\cos\text{t}\cos2\text{t}}{3\sin\text{t}\cos2\text{t}}=\cot\text{t}$
Now, $\Big(\frac{\text{dy}}{\text{dt}}\Big)_{\text{t}=\frac{\pi}{4}}=\cot\frac{\pi}{4}=1$
View full question & answer→Question 2425 Marks
Find $\frac{\text{dy}}{\text{dx}}$ in the following cases:
$x^5 + y^5 = 5xy$
AnswerWe Heve, $x^5 + y^5 = 5xy$
Differentiating with respect to x, we get,
$\frac{\text{d}}{\text{dx}}\big(\text{x}^5\big)+\frac{\text{d}}{\text{dx}}\big(\text{y}^5\big)=\frac{\text{d}}{\text{dx}}\big(5\text{xy}\big)$
$\Rightarrow5\text{x}^4+5\text{y}^4\frac{\text{dy}}{\text{dx}}=5\Big[\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\frac{\text{dy}}{\text{dx}}\big(\text{x}\big)\Big]$
$\Rightarrow5\text{x}^4+5\text{y}^4\frac{\text{dy}}{\text{dx}}=5\Big[\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\big(1\big)\Big]$
$\Rightarrow5\text{x}^4+5\text{y}^4\frac{\text{dy}}{\text{dx}}=5\text{x}\frac{\text{dy}}{\text{dx}}+5\text{y}$
$\Rightarrow5\text{y}^4\frac{\text{dy}}{\text{dx}}-5\text{x}\frac{\text{dy}}{\text{dx}}=5\text{y}-5\text{x}^4$
$\Rightarrow5\frac{\text{dy}}{\text{dx}}\big(\text{y}^4-\text{x}\big)=5\big(\text{y}-\text{x}^4\big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{5(\text{y}-\text{x}^4)}{5(\text{y}^4-\text{x})}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}-\text{x}^4}{\text{y}^4-\text{x}}$
View full question & answer→Question 2435 Marks
Differentiate $\sin^{-1}\Big\{\frac{2^{\text{x}+1}\times3^\text{x}}{1+(36)^\text{x}}\Big\}$ with respect to x:
AnswerWe have $\text{y}=\sin^{-1}\Big\{\frac{2^{\text{x}+1}\times3^\text{x}}{1+(36)^\text{x}}\Big\}$
$\Rightarrow \text{y}=\sin^{-1}\Big\{\frac{2\times6^\text{x}}{1+6^{2\text{x}}}\Big\}$
Put $6^\text{x}=\tan\theta$
$\Rightarrow \theta=\tan^{-1}(6^\text{x})$
Now, $\text{y}=\sin^{-1}\Big\{\frac{2\tan\theta}{1+\tan^2\theta}\Big\}$
$\Rightarrow \text{y}=\sin^{-1}\big\{\sin2\theta\big\}$
$\Rightarrow \text{y}=2\theta$
$\Rightarrow \text{y}=2\tan^{-1}(6^\text{x})$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=2\times\frac{1}{(6^\text{x})^2}\times6^\text{x}\log6$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{2(\log6)6^\text{x}}{36^\text{x}}$
View full question & answer→Question 2445 Marks
If $\text{y}=\log\sqrt{\text{x}+1}+\sqrt{\text{x}-1},$ show that $\sqrt{\text{x}^2-1}\frac{\text{dy}}{\text{dx}}=\frac{1}{2}\text{y}.$
AnswerConsider $\text{y}=\cos(\log\text{ x})^2$
Differentiating it with respect to x and applying the chain and the product rule, we get
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\sqrt{\text{x}+1}+\frac{\text{d}}{\text{dx}}\sqrt{\text{x}-1}$
$=\frac{1}{2}(\text{x}+1)^\frac{-1}{2}+\frac{1}{2}(\text{x}-1)^\frac{-1}{2}$
$=\frac{1}{2}\Big(\frac{1}{\sqrt{\text{x}+1}}\frac{1}{\sqrt{\text{x}-1}}\Big)$
$=\frac{1}{2}\bigg(\frac{\sqrt{\text{x}-1}+\sqrt{\text{x}+1}}{\big(\sqrt{\text{x}+1}\big)\big(\sqrt{\text{x}-1}\big)}\bigg)$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2}\Big(\frac{\text{y}}{\sqrt{\text{x}^2-1}}\Big)$
So,
$\sqrt{\text{x}^2}-\frac{\text{dy}}{\text{dx}}=\frac{1}{2}\text{y}$
View full question & answer→Question 2455 Marks
Differentiate the following functions with respect to x:
$\cos^{-1}\Big\{\sqrt{\frac{1+\text{x}}{2}}\Big\},-1<\text{x}<1$
AnswerLet $\text{y}=\cos^{-1}\Big\{\sqrt{\frac{1+\text{x}}{2}}\Big\}$
Put $\text{x}=\cos2\theta$
$\text{y}=\cos^{-1}\Big\{\sqrt{\frac{1+\cos2\theta}{2}}\Big\}$
$=\cos^{-1}\Big\{\sqrt{\frac{2\cos^2\theta}{2}}\Big\}$
$\text{y}=\cos^{-1}\{\cos\theta\}$
Here, $-1<\text{x}<1$
$\Rightarrow\ -1<\cos2\theta<1$
$\Rightarrow\ 0<2\theta<\pi$
$\Rightarrow\ 0<\theta<\frac{\pi}{2}$
So, equation (i),
$\text{y}=\theta$
$\big[\text{Since}, \cos^{-1}(\cos\theta)=\theta\text{ if }\theta\in[0,\pi]\big]$
$\text{y}=\frac{1}{2}\cos^{-1}\text{x}\ \big[\text{Since x}=\cos2\theta\big]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=-\frac{1}{2\sqrt{1-\text{x}^2}}$
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