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→