Question 1515 Marks
If $\text{y}=\text{e}^{\text{x}^{\text{e}^\text{x}}}+\text{x}^{\text{e}^{\text{e}^\text{x}}}+\text{e}^{\text{x}^{\text{x}^{\text{e}}}},$ prove that $\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}^{\text{e}^\text{x}}}\times\text{x}^{\text{e}^{\text{x}}}\Big\{\frac{\text{e}^\text{x}}{\text{x}}+\text{e}^\text{x}\log\text{x}\Big\}+\text{e}^{\text{x}^{\text{e}^{\text{x}}}}\times\text{e}^{\text{e}^\text{x}}\Big\{\frac{1}{\text{x}}+\text{e}^\text{x}\times\log\text{x}\Big\}+\text{e}^{\text{x}^{\text{x}^\text{e}}}\text{x}^{\text{x}^{\text{e}}}\times\text{x}^{\text{e}-1}\Big\{\text{x}+\text{e}\log\text{x}\Big\}$
AnswerWe have, $\text{y}=\text{e}^{\text{x}^{\text{e}^\text{x}}}+\text{x}^{\text{e}^{\text{e}^\text{x}}}+\text{e}^{\text{x}^{\text{x}^{\text{e}}}}$
$\Rightarrow\text{y}=\text{u}+\text{v}+\text{w}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{du}}{\text{dx}}+\frac{\text{dv}}{\text{dx}}+\frac{\text{dw}}{\text{dx}}\ .....(\text{i})$
Where $\text{u}=\text{e}^{\text{x}^{\text{e}^{\text{x}}}},\text{v}=\text{x}^{\text{e}^{\text{e}^{\text{x}}}}\text{ and w}=\text{e}^{\text{x}^{\text{x}^{\text{e}}}}$
Now, $\text{u}=\text{e}^{\text{x}^{\text{e}^{\text{x}}}}\ .....(\text{ii})$
Taking log on both sides,
$\log\text{u}=\log\text{e}^{\text{x}^{\text{e}^{\text{x}}}}$
$\Rightarrow\log\text{u}=\text{x}^{\text{e}^\text{x}}\log\text{e}$
$\Rightarrow\log\text{u}=\text{x}^{\text{e}^\text{x}}\ .....(\text{iii})$
Taking $\log$ on both sides,
$\log\log\text{u}=\log\text{x}^{\text{e}^\text{x}}$
$\Rightarrow\log\log\text{u}=\text{e}^\text{x}\log\text{x}$
Differentiating with respect to x,
$\Rightarrow\frac{1}{\log\text{u}}\frac{\text{d}}{\text{dx}}(\log\text{u})=\text{e}^\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{e}^\text{x})$
$\Rightarrow\frac{1}{\log\text{u}}\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\frac{\text{e}^\text{x}}{\text{x}}+\text{e}^\text{x}\log\text{x}$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{u}\log\text{u}\Big[\frac{\text{e}^\text{x}}{\text{x}}+\text{e}^\text{x}\log\text{x}\Big]$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{e}^{\text{x}^{\text{e}^{\text{x}}}}\times\text{x}^{\text{e}^\text{x}}\Big[\frac{\text{e}^\text{x}}{\text{x}}+\text{e}^\text{x}\log\text{x}\Big]\ .....(\text{A})$
[Using equation (ii) and (iii)]
Now, $\text{v}=\text{x}^{\text{e}^{\text{e}^\text{x}}}\ .....(\text{iv})$
Taking log on both sides,
$\log\text{v}=\log\text{x}^{\text{e}^{\text{e}^\text{x}}}$
$\Rightarrow\log\text{v}=\text{e}^{\text{e}^\text{x}}\log\text{x}$
$\Rightarrow\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\text{e}^{\text{e}^\text{x}}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{e}^\text{x}}\big)$
$\Rightarrow\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\text{e}^{\text{e}^\text{x}}\big(\frac{1}{\text{x}}\big)+\log\text{xe}^{\text{e}^\text{x}}\frac{\text{d}}{\text{dx}}(\text{e}^\text{x})$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\text{v}\Big[\text{e}^{\text{e}^\text{x}}\big(\frac{1}{\text{x}}\big)+\log\text{xe}^{\text{e}^\text{x}}\text{e}^\text{x}\Big]$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\text{e}^{\text{e}^{\text{e}^\text{x}}}\times\text{e}^{\text{e}^\text{x}}\Big[\frac{1}{\text{x}}+\text{e}^\text{x}\log\text{x}\Big]\ .....(\text{B})$
[Using equation (4)]
Now, $\text{w}=\text{e}^{\text{x}^{\text{x}^{\text{e}}}}\ .....(\text{v})$
Taking log on sides,
$\log\text{w}=\log\text{e}^{\text{x}^{\text{x}^{\text{e}}}}$
$\Rightarrow\log\text{w}=\text{x}^{\text{x}^\text{e}}\log\text{e}$
$\Rightarrow\log\text{w}=\text{x}^{\text{x}^{\text{e}}}\ .....(\text{vi})$
Taking log on both sides,
$\log\log\text{w}=\log\text{x}^{\text{x}^{\text{e}}}$
$\Rightarrow\log\log\text{w}=\text{x}^{\text{e}}\log\text{x}$
$\Rightarrow\frac{1}{\log\text{w}}\frac{\text{d}}{\text{dx}} (\log\text{w})=\text{x}^\text{e}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x}^\text{e})$
$\Rightarrow\frac{1}{\log\text{w}}\big(\frac{1}{\text{w}}\big)\frac{\text{dw}}{\text{dx}}=\text{x}^{\text{e}}\big(\frac{1}{\text{x}}\big)\log\text{xex}^{\text{e}-1}$
$\Rightarrow\frac{\text{dw}}{\text{dx}}=\text{w}\log\text{w}\big[\text{x}^{\text{e}-1}+\text{e}\log\text{xx}^{\text{e}-1}\big]$
$\Rightarrow\frac{\text{dw}}{\text{dx}}=\text{e}^{\text{x}^{\text{x}^\text{e}}}\text{x}^{\text{x}^\text{e}}\text{x}^{\text{e}-1}(1+\text{e}\log\text{x})\ .....(\text{C})$
Using equation (A), (B) and (C) in equation (i), we get
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}^{\text{e}^{\text{x}}}}\times\text{x}^{\text{e}^\text{x}}\Big[\frac{\text{e}^\text{x}}{\text{x}}+\text{e}^\text{x}\log\text{x}\Big] + \text{e}^{\text{e}^{\text{e}^\text{x}}}\times\text{e}^{\text{e}^\text{x}}\Big[\frac{1}{\text{x}}+\text{e}^\text{x}\log\text{x}\Big] + \text{e}^{\text{x}^{\text{x}^\text{e}}}\text{x}^{\text{x}^\text{e}}\text{x}^{\text{e}-1}(1+\text{e}\log\text{x})$
View full question & answer→Question 1525 Marks
If $\text{y}=\sin\Big[2\tan^{-1}\Big\{\sqrt{\frac{1-\text{x}}{1+\text{x}}}\Big\}\Big],$ find $\frac{\text{dy}}{\text{dx}}.$
AnswerHere, $\text{y}=\sin\Big[2\tan^{-1}\Big[\sqrt{\frac{1-\text{x}}{1+\text{x}}}\Big]\Big]$
Put, $\text{x}=\cos2\theta, \text{So},$
$\text{y}=\sin\Big[2\tan^{-1}\sqrt{\frac{1-\cos2\theta}{1+\cos2\theta}}\Big]$
$=\sin\Big[2\tan^{-1}\sqrt{\frac{2\sin^2\theta}{2\cos^2\theta}}\Big]$
$=\sin\big[2\tan^{-1}\sqrt{\tan^2\theta}\big]$
$=\sin\big[2\tan^{-1}(\tan\theta)\big]$
$=\sin{2\theta}$
$=\sin\Big[2\times\frac{1}{2}\cos^{-1}\text{x}\Big]\ \big[\text{Since, x}=\cos2\theta\big]$
$=\sin\big(\sin^{-1}\sqrt{1-\text{x}^2}\big)$
$\text{y}=\sqrt{1-\text{x}^2}$
Differentiating with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2\sqrt{1-\text{x}^2}}\frac{\text{d}}{\text{dx}}(1-\text{x}^2)$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2\sqrt{1-\text{x}^2}}(-2\text{x})$
$\frac{\text{dy}}{\text{dx}}=\frac{-\text{x}}{\sqrt{1-\text{x}^2}}$
View full question & answer→Question 1535 Marks
Differentiate the following functions with respect to x:
$\frac{\text{e}^\text{x}\sin\text{x}}{(\text{x}^2+2)^3}$
AnswerConsider $\text{y}=\frac{\text{e}^\text{x}\sin\text{x}}{(\text{x}^2+2)^3}$
Differentiating it with respect to x and applying the chain and product rule, we get
$\frac{\text{dy}}{\text{dx}}=\frac{(\text{x}^2+2)^3\frac{\text{d}}{\text{dx}}(\text{e}^\text{x}\sin\text{x})-\text{e}^\text{x}\sin\text{x}\frac{\text{d}}{\text{dx}}(\text{x}^2+2)^3}{\big[(\text{x}^2+2)^3\big]^2}$
$=\frac{(\text{x}^2+2)^3[\text{e}^\text{x}\cos\text{x}+\sin\text{xe}^\text{x}]-\text{e}^\text{x}\sin\text{x}3(\text{x}^2+2)^2(2\text{x})}{(\text{x}^2+2)^6}$
$=\frac{(\text{x}^2+2)^3[\text{e}^\text{x}\cos\text{x}+\sin\text{xe}^\text{x}]-6\text{xe}^\text{x}\sin\text{x}(\text{x}^2+2)^2}{(\text{x}^2+2)^6}$
$=\frac{(\text{x}^2+2)^2\big[(\text{x}^2+2)(\text{e}^\text{x}\cos\text{x}+\sin\text{xe}^\text{x})-6\text{xe}^\text{x}\sin\text{x}}{(\text{x}^2+2)^6}$
$=\frac{\text{x}^2\text{e}^\text{x}\cos\text{x}+\text{x}^2\sin\text{xe}^\text{x}+2\text{e}^\text{x}\cos\text{x}+2\sin\text{xe}^\text{x}-6\text{xe}^\text{x}\sin\text{x}}{(\text{x}^2+2)^4}$
$=\frac{\text{e}^\text{x}\sin\text{x}}{(\text{x}^2+2)^3}+\frac{\text{e}^\text{x}\cos\text{x}}{(\text{x}^2+2)^3}-\frac{6\text{xe}^\text{x}\sin\text{x}}{(\text{x}^2+2)^4}$
Therefore,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^\text{x}\sin\text{x}}{(\text{x}^2+2)^3}+\frac{\text{e}^\text{x}\cos\text{x}}{(\text{x}^2+2)^3}-\frac{6\text{xe}^\text{x}\sin\text{x}}{(\text{x}^2+2)^4}$
View full question & answer→Question 1545 Marks
Differentiate the following functions with respect to x:
$\sqrt{\frac{\text{a}^2-\text{x}^2}{\text{a}^2+\text{x}^2}}$
AnswerLet $\text{y}=\sqrt{\frac{\text{a}^2-\text{x}^2}{\text{a}^2+\text{x}^2}}$
$\Rightarrow\text{y}=\Big({\frac{\text{a}^2-\text{x}^2}{\text{a}^2+\text{x}^2}}\Big)^\frac{1}{2}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big({\frac{\text{a}^2-\text{x}^2}{\text{a}^2+\text{x}^2}}\Big)^\frac{1}{2}$
$=\frac{1}{2}\Big({\frac{\text{a}^2-\text{x}^2}{\text{a}^2+\text{x}^2}}\Big)^{\frac{1}{2}-1}\times\frac{\text{d}}{\text{dx}}\Big(\frac{\text{a}^2-\text{x}^2}{\text{a}^2+\text{x}^2}\Big)$
[Using chain rule]
$=\frac{1}{2}\Big({\frac{\text{a}^2-\text{x}^2}{\text{a}^2+\text{x}^2}}\Big)^{\frac{1}{2}}\times\Bigg\{\frac{\big(\text{a}^2+\text{x}^2\big)\frac{\text{a}}{\text{dx}}\big(\text{a}^2-\text{x}^2\big)-\big(\text{a}^2-\text{x}^2\big)\frac{\text{d}}{\text{dx}}(\text{a}^2+\text{x}^2)}{\big(\text{a}^2+\text{x}^2\big)}\Bigg\}$
[Using chain rule]
$=\frac{1}{2}\Big({\frac{\text{a}^2-\text{x}^2}{\text{a}^2+\text{x}^2}}\Big)^{\frac{1}{2}}\Bigg\{\frac{-2\text{x}\big(\text{a}^2+\text{x}^2\big)-2\text{x}\big(\text{a}^2-\text{x}^2\big)}{\big(\text{a}^2+\text{a}^2\big)^2}\Bigg\}$
$=\frac{1}{2}\Big({\frac{\text{a}^2-\text{x}^2}{\text{a}^2+\text{x}^2}}\Big)^{\frac{1}{2}}\bigg\{\frac{-2\text{xa}^2-2\text{x}^3-2\text{xa}^2+2\text{x}^3}{\big(\text{a}^2+\text{a}^2\big)^2}\bigg\}$
$ =\frac{1}{2}\Big({\frac{\text{a}^2-\text{x}^2}{\text{a}^2+\text{x}^2}}\Big)^{\frac{1}{2}}\Bigg(\frac{-4\text{xa}^2}{\big(\text{a}^2+\text{x}^2\big)^3}\Bigg)$
$=\frac{-2\text{xa}^2}{\sqrt{\text{a}^2-\text{x}^2}\big(\text{a}^2+\text{x}^2\big)^{\frac{3}{2}}}$
So,
$\frac{\text{d}}{\text{dx}}\bigg(\sqrt{\frac{\text{a}^2-\text{x}^2}{\text{a}^2+\text{x}^2}}\bigg)=\frac{-2\text{xa}^2}{\sqrt{\text{a}^2-\text{x}^2}\big(\text{a}^2+\text{x}^2\big)^{\frac{3}{2}}}$
View full question & answer→Question 1555 Marks
Differentiate the following functions with respect to x:
$\text{e}^{\text{x}\log\text{x}}$
AnswerLet $\text{y}=\text{e}^{\text{x}\log\text{x}}$
$\Rightarrow\ \text{y}=\text{e}^{\log\text{x}^\text{x}} \ \big[\text{Since}, \log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
$\Rightarrow\text{y}=\text{x}^{\text{x}}\ .....(\text{i})\ \big[\text{Since, e}^{\log\text{a}}=\text{a}\big]$
Taking log on both the sides,
$\log\text{y}=\log\text{x}^\text{x}$
$\log\text{y}=\text{x}\log\text{x}$
Differentiating with respect to x, using product rule,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x})$
$=\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}(1)$
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=1+\log\text{x}$
$\frac{\text{dy}}{\text{dx}}=\text{y}[1+\log\text{x}]$
$\frac{\text{dy}}{\text{dx}}=\text{x}^\text{x}(1+\log\text{x})$
[Using equation (i)]
View full question & answer→Question 1565 Marks
Find $\frac{\text{dy}}{\text{dx}},$ when
$\text{x}=\text{a}(\theta+\sin\theta)$ and $\text{y}=\text{a}(1-\cos\theta)$
AnswerHere, $\text{x}=\text{a}(\theta+\sin\theta)$Differentiating it with respect to $\theta$
$\frac{\text{dx}}{\text{d}\theta}=\text{a}(1+\cos\theta).....(\text{i}) $
And, $\text{y}=\text{a}(1-\cos\theta)$
Differentiating it with respect to $\theta$,
$ \frac{\text{dy}}{\text{d}\theta}=\text{a}(\theta+\sin\theta)$
and
$\frac{\text{dy}}{\text{d}\theta}=\text{a}\sin\theta...(\text{ii}) $
Using equation (i) and (ii),
$=\frac{\text{a}\sin\theta}{\text{a}(1-\cos\theta)} $
$=\frac{\frac{2\sin\theta}{2}\frac{\cos\theta}{2}}{\frac{2\sin^{2}\theta}{2}}, \begin{Bmatrix} \text{Since, }1-\cos\theta=\frac{2\sin^{2\theta}}{2}\\\frac{2\sin\theta}{2}\frac{\cos\theta}{2}=\sin\theta \end{Bmatrix}$
$=\frac{\text{dy}}{\text{dx}}=\frac{\tan}{2}$
View full question & answer→Question 1575 Marks
If $\text{u}=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$ and $\text{v}=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big),$ where -1 < x < 1 then write the value of $\frac{\text{dy}}{\text{dx}}.$
Answer$\text{u}=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$ and $\text{v}=\tan^{-1}\Big(\frac{2\text{x}}{1+\text{x}^3}\Big)$
We know, $\frac{\text{du}}{\text{dx}}=\frac{2}{1+\text{x}^2}$
Using the chain rule of differentiation,
$\frac{\text{dv}}{\text{dx}}=\frac{1}{1+\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)^2}\cdot\frac{(1+\text{x}^2)\cdot(2\text{x})'-(1+\text{x}^2)'\cdot(2\text{x})}{(1+\text{x}^2)^2}$
$=\frac{(1+\text{x}^2)^2}{(1+\text{x}^2)^2+(2\text{x})^2}\cdot\frac{2(1+\text{x}^2)-(2\text{x})(2\text{x})}{(1+\text{x}^2)^2}$
$=\frac{2(1-\text{x}^2)}{(1+\text{x}^2)^2+(2\text{x})^2}$
Using Chain Rule of Differentiation,
$\frac{\text{du}}{\text{dv}}=\frac{\text{du}}{\text{dx}}\cdot\frac{\text{dx}}{\text{dv}}$
$=\frac{2}{1+\text{x}^2}\cdot\frac{(1+\text{x}^2)^2+(2\text{x})^2}{2(1-\text{x})^2}$
$=\frac{(1+\text{x}^2)^2+(2\text{x})^2}{(1+\text{x}^2)(1-\text{x}^2)}$
Dividing numerator and denominator by $(1 + x^2)^2,$
$\frac{\text{du}}{\text{dv}}=\frac{1+\big(\frac{2\text{x}}{1+\text{x}^2}\big)}{\frac{1-\text{x}^2}{1+\text{x}^2}}$
$=\frac{1+\sin^2\text{u}}{\cos\text{u}}$
$=\sec\text{u}(1+\tan\text{u})$
View full question & answer→Question 1585 Marks
Differentiate the following functions with respect to x:
$\text{x}^{\frac{1}{\text{x}}}$
AnswerLet $\text{y}=\text{x}^{\frac{1}{\text{x}}}\ .....(\text{i})$
Taking log on both sides,
$\log\text{y}=\log\text{x}^{\frac{1}{\text{x}}}$
$\Rightarrow\log\text{y}=\frac{1}{\text{x}}\log\text{x}\ \big[\because\log\text{a}^{\text{b}}=\text{b}\log\text{a}\big]$
Differentiating with respect to x,
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}\big(\text{x}^{-1}\big)$
[Using product rule]
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}\times\frac{1}{\text{x}}(\log\text{x})\times\Big(-\frac{1}{\text{x}^2}\Big)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}^2}-\frac{\log\text{x}}{\text{x}^2}$
$\Rightarrow \frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{(1-\log\text{x})}{\text{x}^2}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\text{x}^\frac{1}{\text{x}}\Big[\frac{1-\log\text{x}}{\text{x}}\Big]$
View full question & answer→Question 1595 Marks
Differentiate the following functions with respect to x:
$\cos^{-1}\Big\{\frac{\cos\text{x}+\sin\text{x}}{\sqrt{2}}\Big\},-\frac{\pi}{4}<\text{x}<\frac{\pi}{4}$
AnswerLet $\text{y}=\cos^{-1}\Big\{\frac{\cos\text{x}+\sin\text{x}}{\sqrt{2}}\Big\}$
$\text{y}=\cos^{-1}\Big\{\cos\text{x}\Big(\frac{1}{\sqrt{2}}\Big)+\sin\text{x}\Big(\frac{1}{\sqrt{2}}\Big)\Big\}$
$=\cos^{-1}\Big\{\cos\text{x}\cos\Big(\frac{\pi}{4}\Big)+\sin\text{x}\sin\text{x}\Big(\frac{\pi}{4}\Big)\Big\}$
$\text{y}=\cos^{-1}\Big[\cos\Big(\text{x}-\frac{\pi}{4}\Big)\Big]$
Here, $-\frac{\pi}{4}<\text{x}<\frac{\pi}{4}$
$\Rightarrow\ \Big(-\frac{\pi}{4}-\frac{\pi}{4}\Big)<\Big(\text{x}-\frac{\pi}{4}\Big)<\Big(\frac{\pi}{4}-\frac{\pi}{4}\Big)$
$\Rightarrow-\frac{\pi}{2}<\Big(\text{x}-\frac{\pi}{4}\Big)<0$
So, from equation (i),
$\text{y}=-\Big(\text{x}-\frac{\pi}{4}\Big)$
$\Big[\text{Since}, \cos^{-1}(\cos\theta)=-\theta,\text{ if }\theta\in [-\pi, 0]\Big]$
$\text{y}=-\text{x}+\frac{\pi}{4}$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=-1$
View full question & answer→Question 1605 Marks
Find $\frac{\text{dy}}{\text{dx}},$ when
$\text{x}=\text{a}(\cos\theta+\theta\sin\theta)$ and $\text{y}=\text{a}(\sin\theta-\theta\sin\theta-\theta\cos\theta)$
AnswerWe have, $\text{x}=\text{a}(\cos\theta+\theta\sin\theta)$ and $\text{y}=\text{a}(\sin\theta-\theta\sin\theta-\theta\cos\theta)$
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{a}\Big[\frac{\text{d}}{\text{d}\theta}\cos\theta+\frac{\text{d}}{\text{d}\theta}(\theta\sin\theta)\Big] $ and
$\frac{\text{dy}}{\text{d}\theta}=\text{a}\Big[\frac{\text{d}}{\text{d}\theta}\sin\theta-\frac{\text{d}}{\text{d}\theta}(\theta\cos\theta)\Big] $
$ \Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{a}\Big[-\sin\theta+\theta\frac{\text{d}}{\text{d}\theta}(\sin\theta)+\sin\theta\frac{\text{d}}{\text{d}\theta}(\theta)\Big]$ and
$\frac{\text{dy}}{\text{d}\theta}=\text{a}\Big[\cos\theta-\left\{\theta\frac{\text{d}}{\text{d}\theta}(\cos\theta)+\cos\theta\frac{\text{d}}{\text{d}\theta}(\theta)\right\}\Big]$
$ \Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{a}\big[-\sin\theta+\theta\cos\theta\big]$ and
$\frac{\text{dy}}{\text{d}\theta}=\text{a}\big[\cos\theta+\theta\sin\theta-\cos\theta\big]$
$ \Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{a}\theta\cos\theta$ and $\frac{\text{dy}}{\text{d}\theta}=\text{a}\theta\sin\theta$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}=\frac{\text{a}\theta\sin\theta}{\text{a}\theta\cos\theta}=\tan\theta$
View full question & answer→Question 1615 Marks
Find $\frac{\text{dy}}{\text{dx}}$ in the following cases:
$(x + y)^2 = 2axy$
AnswerWe Have, $(x + y)^2 = 2axy$
Differentiating with respect to x, we get,
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{x}+\text{y}\big)^2=\frac{\text{d}}{\text{dx}}\big(2\text{axy}\big)$
$\Rightarrow2(\text{x}+\text{y})\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})=2\text{a}\Big[\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\frac{\text{d}}{\text{dx}}(\text{x})\Big]$
$\Rightarrow2(\text{x}+\text{y})\Big[1+\frac{\text{dy}}{\text{dx}}\Big]=2\text{a}\Big[\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}(1)\Big]$
$\Rightarrow2(\text{x}+\text{y})+2(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}=2\text{a}\text{x}\frac{\text{dy}}{\text{dx}}+2\text{ay}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\big[2(\text{x}+\text{y})-2\text{a}\text{x}\big]=2\text{ay}-2(\text{x}+\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2[\text{ay}-\text{x}-\text{y}]}{2[\text{x}+\text{y}-\text{a}\text{x}]}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\Big(\frac{\text{ay}-\text{x}-\text{y}}{\text{x}+\text{y}-\text{a}\text{x}}\Big)$
View full question & answer→Question 1625 Marks
Find $\frac{\text{dy}}{\text{dx}}$ in the following cases:
$4\text{x}+3\text{y}=\log\big(4\text{x}-3\text{y}\big)$
AnswerWe have, $4\text{x}+3\text{y}=\log\big(4\text{x}-3\text{y}\big)$
Differentiating with respect to x, we get,
$\frac{\text{d}}{\text{dx}}\big(4\text{x}\big)+\frac{\text{d}}{\text{dx}}\big(3\text{y}\big)=\frac{\text{d}}{\text{dx}}\big\{\log\big(4\text{x}-3\text{y}\big)\big\}$
$\Rightarrow4+3\frac{\text{dy}}{\text{dx}}=\frac{1}{\big(4\text{x}-3\text{y}\big)}\frac{\text{d}}{\text{dx}}\big(4\text{x}-3\text{y}\big)$
$\Rightarrow4+3\frac{\text{dy}}{\text{dx}}=\frac{1}{\big(4\text{x}-3\text{y}\big)}\Big(4-3\frac{\text{dy}}{\text{dx}}\Big)$
$\Rightarrow3\frac{\text{dy}}{\text{dx}}+\frac{3}{\big(4\text{x}-3\text{y}\big)}\frac{\text{dy}}{\text{dx}}=\frac{4}{\big(4\text{x}-3\text{y}\big)}-4$
$\Rightarrow3\frac{\text{dy}}{\text{dx}}\Big\{1+\frac{1}{(4\text{x}-3\text{y})}\Big\}=4\Big\{\frac{1}{(4\text{x}-3\text{y})}-1\Big\}$
$\Rightarrow3\frac{\text{dy}}{\text{dx}}\Big\{\frac{4\text{x}-3\text{y}+1}{(4\text{x}+3\text{y})}\Big\}=4\Big\{\frac{1-4\text{x}-3\text{y}}{4\text{x}-3\text{y}}\Big\}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{4}{3}\Big\{\frac{1-4\text{x}+3\text{y}}{(4\text{x}-3\text{y})}\Big\}\Big(\frac{4\text{x}-3\text{y}}{4\text{x}-3\text{y}+1}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{4}{3}\Big(\frac{1-4\text{x}+3\text{y}}{4\text{x}-3\text{y}+1}\Big)$
View full question & answer→Question 1635 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=\text{x}^{\text{x}}+(\sin\text{x})^\text{x}$
AnswerWe have, $\text{y}=\text{x}^{\text{x}}+(\sin\text{x})^\text{x}$
$\Rightarrow\text{y}=\text{e}^{\log\text{x}^\text{x}}+\text{e}^{\log(\sin\text{x})^\text{x}}$
$\Rightarrow\text{y}=\text{e}^{\text{x}\log\text{x}}+\text{e}^{\text{x}\log\sin\text{x}}$
Differentiating with respect to x using chain rule and product rule,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}\log\text{x}}\big)+\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}\log\sin\text{x}}\big)$
$=\text{e}^{\text{x}\log\text{x}}\frac{\text{d}}{\text{dx}}(\text{x}\log\text{x})+\text{e}^{\text{x}\log\sin\text{x}}\frac{\text{d}}{\text{dx}}(\text{x}\log\sin\text{x})$
$=\text{e}^{\text{x}\log\text{x}}\Big[\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big] \\ +\text{e}^{\log(\sin\text{x})^\text{x}}\Big[\text{x}\frac{\text{d}}{\text{dx}}(\log\sin\text{x})+\log\sin\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big]$
$=\text{x}^{\text{x}}\Big[\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}(1)\Big] \\ +(\sin\text{x})^\text{x}\Big[\text{x}\Big(\frac{1}{\sin\text{x}}\Big)\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}\big]$
$=\text{x}^{\text{x}}\big[1+\log\text{x}\big]+(\sin\text{x})^\text{x}\Big[\text{x}\Big(\frac{1}{\sin\text{x}}\Big)(\cos\text{x})+\log\sin\text{x}\Big]$
$=\text{x}^{\text{x}}\big[1+\log\text{x}\big]+(\sin\text{x})^\text{x}\big[\text{x}\cot\text{x}+\log\sin\text{x}\big]$
View full question & answer→Question 1645 Marks
Differentiate the following functions with respect to x:
$\log(\text{cosec x}-\cot\text{x})$
AnswerConsider $\text{y}=\log(\text{cosec x}-\cot\text{x})$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\log\big(\text{cosec x}-\cot\text{x}\big)$
$=\frac{1}{(\text{cosec x}-\cot\text{x})}\times\big(-\text{cosec x}\cot\text{x}+\text{cosec}^2\text{x}\big)$
[Using chain rule]
$=\frac{1}{(\text{cosec x}-\cot\text{x})}\times\big(-\text{cosec x}\cot\text{x}+\text{cosec}^2\text{x}\big)$
$=\frac{\text{cosec x}(\text{cosec x}-\cos\text{x})}{(\text{cosec x}-\cot\text{x})}$
$=\text{cosec x}$
Hence, the solution is, $\frac{\text{d}}{\text{dx}}\big(\log(\text{cosec x}-\cot\text{x})\big)=\text{cosec x}$
View full question & answer→Question 1655 Marks
Differentiate the following functions with respect to x:
$\text{x}\sin2\text{x}+5^{\text{x}}+\text{k}^\text{k}+(\tan^2\text{x})^3$
AnswerLet $\text{y}=\text{x}\sin2\text{x}+5^{\text{x}}+\text{k}^\text{k}+(\tan^2\text{x})^3$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[\text{x}\sin2\text{x}+5^{\text{x}}+\text{k}^\text{k}+(\tan^2\text{x})^3\big]$
$=\frac{\text{d}}{\text{dx}}(\text{x}\sin2\text{x})+\frac{\text{d}}{\text{dx}}(5^\text{x})+\frac{\text{d}}{\text{dx}}(\text{k}^\text{k})+\frac{\text{d}}{\text{dx}}\big(\tan^6\text{x}\big)$
$=\Big[\text{x}\frac{\text{d}}{\text{dx}}(\sin2\text{x})+\sin2\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big] \\ +5^\text{x}\log5+0+6\tan^5\text{x}\frac{\text{d}}{\text{dx}}(\tan\text{x})$
[Using product rule and chain rule]
$=\Big[\text{x}\cos2\text{x}\frac{\text{d}}{\text{dx}}(2\text{x})+\sin2\text{x}\Big]+5^\text{x}\log5+6\tan^5\text{x}\sec^2\text{x}$
$=2\text{x}\cos2\text{x}+\sin2\text{x}+5^\text{x}\log5+6\tan^5\text{x}\sec^2\text{x}$
So,
$\frac{\text{d}}{\text{dx}}\big(\text{x}\sin2\text{x}+5^\text{x}+\text{k}^\text{k}+(\tan^2\text{x})^3\big) \\ =2\text{x}\cos2\text{x}+\sin2\text{x}+5^\text{x}\log5+6\tan^5\text{x}\sec^2\text{x}$
View full question & answer→Question 1665 Marks
If $\text{y}=1+\frac{\alpha}{\big(\frac{1}{\text{x}}-\alpha\big)}+\frac{\frac{\beta}{\text{x}}}{\big(\frac{1}{\text{x}}-\alpha\big)\big(\frac{1}{\text{x}}-\beta\big)}+\frac{\frac{\gamma}{\text{x}^2}}{\big(\frac{1}{\text{x}}-\alpha\big)\big(\frac{1}{\text{x}}-\beta\big)\big(\frac{1}{\text{x}}-\gamma\big)},$ find $\frac{\text{dy}}{\text{dx}}$
Answer$\text{y}=1+\frac{\alpha}{\big(\frac{1}{\text{x}}-\alpha\big)}+\frac{\frac{\beta}{\text{x}}}{\big(\frac{1}{\text{x}}-\alpha\big)\big(\frac{1}{\text{x}}-\beta\big)}+\frac{\frac{\gamma}{\text{x}^2}}{\big(\frac{1}{\text{x}}-\alpha\big)\big(\frac{1}{\text{x}}-\beta\big)\big(\frac{1}{\text{x}}-\gamma\big)}$
Using the theorem,
If $\text{y}=1+\frac{\text{ax}^2}{(\text{x}-\text{a})(\text{x}-\text{b})(\text{x}-\text{c})}+\frac{\text{bx}}{(\text{x}-\text{b})(\text{x}-\text{c})}+\frac{\text{c}}{(\text{x}-\text{x})}$ then,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}\Big\{\frac{\text{a}}{\text{a}-\text{x}}+\frac{\text{b}}{\text{b}-\text{x}}+\frac{\text{c}}{\text{c}-\text{x}}\Big\}$
Here, we have $\frac{1}{\text{x}}$ instead of x,
So, using above theorem we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\alpha}{\big(\frac{1}{\text{a}}-\alpha\big)}+\frac{\beta}{\big(\frac{1}{\text{x}}-\beta\big)}+\frac{\gamma}{\big(\frac{1}{\text{x}}-\gamma\big)}$
View full question & answer→Question 1675 Marks
Differentiate the following functions with respect to x:
$\text{e}^{\text{ax}}\sec\text{x}\tan2\text{x}$
AnswerLet $\text{y}=\text{e}^{\text{ax}}\sec\text{x}\tan2\text{x}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{e}^{\text{ax}}\sec\text{x}\tan2\text{x})$
$=\text{e}^{\text{ax}}\frac{\text{d}}{\text{dx}}\big\{\sec\text{x}\tan2\text{x}\big\}+\sec\text{x}\tan2\text{x}\frac{\text{d}}{\text{dx}}\big\{\text{e}^{\text{ax}}\big\}$
$=\text{e}^{\text{ax}}\big[\text{sec}\text{x}\tan\text{x}\tan2\text{x}+2\sec^2 2\text{x}\sec\text{x}\big]+\text{ae}^{\text{ax}}\sec\text{a}\tan^{2\text{x}}$
$=\text{ae}^{\text{ax}}\sec\text{x}\tan2\text{x}+\text{e}^{\text{ax }}\sec\text{x}\tan\text{x}\tan2\text{x}+2\text{e}^{\text{ax}}\sec\text{x}\sec^2 2\text{x}$
$=\text{e}^{\text{ax}}\sec\text{x}\big\{\text{a}\tan2\text{x}+\tan\text{x}\tan2\text{x}+2\sec^22\text{x}\big\}$
So,
$\frac{\text{d}}{\text{dx}}(\text{e}^{\text{ax}}\sec\text{x}\tan2\text{x})=\text{e}^{\text{ax}}\sec\text{x}\big\{\text{a}\tan2\text{x}+\tan\text{x}\tan2\text{x}+2\sec^22\text{x}\big\}$
View full question & answer→Question 1685 Marks
Differentiate the following functions with respect to x:
$\sin^{-1}\Big\{\sqrt{\frac{1-\text{x}}{2}}\Big\},0<\text{x}<1$
AnswerLet $\text{y}=\sin^{-1}\Big\{\sqrt{\frac{1-\text{x}}{2}}\Big\}$
Put $\text{x}=\cos2\theta$
$\text{y}=\sin^{-1}\Big\{\sqrt{\frac{1+\cos2\theta}{2}}\Big\}$
$=\sin^{-1}\Big\{\sqrt{\frac{2\sin^2\theta}{2}}\Big\}$
$\text{y}=\sin^{-1}(\sin\theta)\ .....(\text{i})$
Here, $0<\text{x}<1$
$\Rightarrow\ 0<\cos2\theta<1$
$\Rightarrow\ 0<2\theta<\frac{\pi}{2}$
$\Rightarrow\ 0<\theta<\frac{\pi}{4}$
So, from equation (i),
$\text{y}=\theta$
$\Big[\text{Since, } \sin^{-1}(\sin\theta)=\theta\text{ if }\theta \in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\text{y}=\frac{1}{2}\cos^{-1}\text{x}\ \big[\text{Since x}=\cos2\theta\big]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=-\frac{1}{2\sqrt{1-\text{x}^2}}$
View full question & answer→Question 1695 Marks
Differentiate the following functions with respect to x:
$\text{x}^{\sin{\text{x}}}$
AnswerLet $\text{y}=\text{x}^{\sin{\text{x}}}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=\log\text{x}^{\sin{\text{x}}}$
$\log\text{y}=\sin\text{x}\log\text{x}\ \big[\text{Since,}\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating with respect to x,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\sin\text{x}\frac{\text{d}}{\text{dx}}\log\text{x}+\log\text{x}\frac{\text{d}}{\text{dx}}\sin\text{x}$
[Using product rule]
$\frac{1}{\text{y}}\frac{\text{dt}}{\text{dx}}=\sin\text{x}\big(\frac{1}{\text{x}}\big)+\log\text{x}(\cos\text{x})$
$\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{\sin\text{x}}{\text{x}}+(\log\text{x})(\cos\text{x})\Big]$
Put the value of y,
$\frac{\text{dy}}{\text{dx}}=\text{x}^{\sin\text{x}}\Big[\frac{\sin\text{x}}{\text{x}}+(\log\text{x})(\cos\text{x})\Big]$
View full question & answer→Question 1705 Marks
Differentiate $\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)$ with respect to $\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big),$ if 0 < x < 1.
AnswerLet $\text{u}=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)$
Put $\text{x}=\tan\theta,\text{so}$
$\text{u}=\tan^{-1}\Big(\frac{2\tan\theta}{1-\tan^2\theta}\Big)$
$\text{u}=\tan^{-1}(\tan2\theta)\ .....(\text{i})$
Let $\text{v}=\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$
$=\cos^{-1}\Big(\frac{1-\tan^2\theta}{1+\tan^2\theta}\Big)$
$\text{v}=\cos^{-1}(\cos2\theta)\ .....(\text{ii})$
Hrer, 0 < x < 1
$\Rightarrow0< \tan\theta<1$
$\Rightarrow0<\theta<\frac{\pi}{4}$
So, from equation (i),
$\text{u}=2\theta\Big[\text{Since,}\tan^{-1}(\tan\theta)=\theta,\text{if }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\text{u}=2\tan^{-1}\text{x }[\text{Since,x}=\tan\theta]$
Differentiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=\frac{2}{1+\text{x}^2}\ .....(\text{iii})$
From equation (ii),
$\text{v}=\theta \big[\text{Since,} \cos^{-1}(\cos\theta)=\theta,\text{if }\theta\in[0,\pi]\big]$
$\text{v}=2\tan^{-1}\text{x }[\text{Since},\text{x}=\tan\theta]$
Differentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{2}{1+\text{x}^2}\ .....(\text{iv})$
Dividing equation (iii) by (iv)
$\frac{\frac{\text{}du}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{2}{1+\text{x}^2}\times\frac{1+\text{x}^2}{2}$
$\frac{\text{du}}{\text{dv}}=1$
View full question & answer→Question 1715 Marks
If $\text{y}=\cos^{-1}(2\text{x})+2\cos^{-1}\sqrt{1-4\text{x}^2}, -\frac{1}{2}<\text{x}<0,$ find $\frac{\text{dy}}{\text{dx}}.$
AnswerHere, $\text{y}=\cos^{-1}(2\text{x})+2\cos^{-1}\sqrt{1-4\text{x}^2}$
Put $2\text{x}=\cos\theta, \text{So},$
$\text{y}=\cos^{-1}(\cos\theta)+2\cos^{-1}\sqrt{1-\cos^2\theta}$
$=\cos^2(\cos\theta)+2\cos^{-1}(\sin\theta)$
$\text{y}=\cos^{-1}(\cos\theta)+2\cos^{-1}\Big(\cos\Big(\frac{\pi}{2}-\theta\Big)\Big)\ .....(\text{i})$
Now, $-\frac{1}{2}<\text{x}<0$
$\Rightarrow -1<2\text{x}<0$
$\Rightarrow -1<\cos\theta<0$
$\Rightarrow \frac{\pi}{2}<\theta<\pi$
And
$\Rightarrow -\frac{\pi}{2}>-\theta>-\pi$
$\Rightarrow \Big(\frac{\pi}{2}-\frac{\pi}{2}\Big)>\Big(\frac{\pi}{2}-\theta\Big)>\Big(\frac{\pi}{2}-\pi\Big)$
$\Rightarrow 0>\Big(\frac{\pi}{2}-\theta\Big)>-\frac{\pi}{2}$
So, from equation (i),
$\text{y}=\theta+2\Big[-\Big(\frac{\pi}{2}-\theta\Big)\Big]$
$\begin{bmatrix} \text{Since}, \cos^{-1}\cos(\theta)=\theta, \text{if }\theta\in[0,\pi] \\ \cos^{-1}\cos(\theta)=-\theta, \text{if }\theta\in[-\pi,0] \end{bmatrix}$
$\text{y}=\theta-2\times\frac{\pi}{2}+2\theta$
$\text{y}=-\pi+3\theta$
$\text{y}=-\pi+3\cos^{-1}(2\text{x})\ \big[\text{Since}, 2\text{x}=\cos\theta\big]$
Differentiating it with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=0+3\Big(\frac{1}{\sqrt{1-(2\text{x})^2}}\Big)\frac{\text{d}}{\text{dx}}(2\text{x})$
$=\frac{-3}{\sqrt{1-4\text{x}^2}}(2)$
$\frac{\text{dy}}{\text{dx}}=-\frac{6}{\sqrt{1-4\text{x}^2}}$
View full question & answer→Question 1725 Marks
If $\text{y}=\frac{\text{x}}{\text{x}+2},$ show that $\text{x}\frac{\text{dy}}{\text{dx}}=(1-\text{y})\text{y}$
AnswerWe have, $\text{y}=\frac{\text{x}}{\text{x}+2}$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}}{\text{x}+2}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(\text{x}+2)\frac{\text{d}}{\text{dx}}(\text{x})-\text{x}\frac{\text{d}}{\text{dx}}(\text{x}+2)}{(\text{x}+2)^2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}+2-\text{x}}{(\text{x}+2)^2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}+2}{(\text{x}+2)^2}-\frac{\text{x}}{(\text{x}+2)^2}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}+2}-\frac{\text{xy}^2}{\text{x}^2} \Big[\because\ \text{x}+2=\frac{\text{x}}{\text{y}}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}-\frac{\text{y}^2}{\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}\text{y}(1-\text{y})$
$\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}=(1-\text{y})\text{y}$
Hence, proved.
View full question & answer→Question 1735 Marks
Differentiate the following functions with respect to x:
$(1+\cos\text{x})^\text{x}$
AnswerLet $\text{y}=(1+\cos\text{x})^\text{x}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=\log(1+\cos\text{x})^\text{x}$
$\log\text{y}=\text{x}\log(1-\cos\text{x})$
Differentiating with respect to x,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}\log(1+\cos\text{x})+\log(1+\cos\text{x})\frac{\text{d}}{\text{dx}}(\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{1}{(1+\cos\text{x})}\frac{\text{d}}{\text{dx}}(1+\cos\text{x})+\log(1+\cos\text{x})(1)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{(1+\cos\text{x})}(0-\sin\text{x})+\log(1+\cos\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\log(1+\cos\text{x})-\frac{\text{x}\sin\text{x}}{(1+\cos\text{x})}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\log(1+\cos\text{x})-\frac{\text{x}\sin\text{x}}{(1+\cos\text{x})}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=(1+\cos\text{x})^\text{x}\Big[\log(1+\cos\text{x})-\frac{\text{x}\sin\text{x}}{(1+\cos\text{x})}\Big]$
[Using equation (i)]
View full question & answer→Question 1745 Marks
If $\text{y}=\log\big\{\sqrt{\text{x}-1}-\sqrt{\text{x}+1}\big\},$ show that $\frac{\text{dy}}{\text{dt}}=\frac{-1}{2\sqrt{\text{x}^2-1}}.$
AnswerHere $\text{y}=\log\big\{\sqrt{\text{x}-1}-\sqrt{\text{x}+1}\big\}$
Differentiating it with respect to x and applying the chain and product rule, we get
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}} \log\big\{\sqrt{\text{x}-1}-\sqrt{\text{x}+1}\big\}$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\big\{\sqrt{\text{x}-1}-\sqrt{\text{x}+1}\big\}}\frac{\text{d}}{\text{dx}}(\sqrt{\text{x}-1}-\sqrt{\text{x}+1})$
$=\frac{1}{\big\{\sqrt{\text{x}-1}-\sqrt{\text{x}+1}\big\}}\Big[\frac{\text{d}}{\text{dx}}\sqrt{\text{x}-1}-\frac{\text{d}}{\text{dx}}\sqrt{\text{x}+1}\Big]$
$=\frac{1}{\big\{\sqrt{\text{x}-1}-\sqrt{\text{x}+1}\big\}}\Big[\frac{1}{2}(\text{x}-1)^{\frac{1}{2}}-\frac{1}{2}(\text{x}+1)^{\frac{1}{2}}\Big]$
$=\frac{1}{2}\frac{1}{\big\{\sqrt{\text{x}-1}-\sqrt{\text{x}+1}\big\}}\Big[\frac{1}{\sqrt{\text{x}-1}}-\frac{1}{\sqrt{\text{x}+1}}\Big]$
$=\frac{1}{2}\frac{1}{\big\{\sqrt{\text{x}-1}-\sqrt{\text{x}+1}\big\}}\Bigg(\frac{-\big\{\sqrt{\text{x}-1}-\sqrt{\text{x}-1}\big\}}{\big(\sqrt{\text{x}-1}\big)\big(\sqrt{\text{x}+1}\big)}\Bigg)$
$=\frac{1}{2}\bigg(\frac{1}{\big(\sqrt{\text{x}-1}\big)\big(\sqrt{\text{x}+1}\big)}\bigg)$
$\frac{\text{dy}}{\text{dx}}=\frac{-1}{2\sqrt{\text{x}^2-1}}$
Therefore,
$\frac{\text{dy}}{\text{dx}}=\frac{-1}{2\sqrt{\text{x}^2-1}}$
View full question & answer→Question 1755 Marks
If $\text{f(x)}=\log\Big\{\frac{\text{u(x)}}{\text{v(x)}}\Big\},\text{u}(1)=\text{v}(1)$ and u'(1) = v'(1) = 2, then find the value of f(1).
AnswerWe have, $\text{f(x)}=\log\Big\{\frac{\text{u(x)}}{\text{v(x)}}\Big\}$
And,
$\text{u}(1)=\text{v}(1),\text{u}'(1)=\text{v}'(1)=2\ .....(\text{i})$
$\Rightarrow\text{f}'\text{(x)}=\frac{\text{d}}{\text{dx}}\Big[\text{f(x)}=\log\Big\{\frac{\text{u(x)}}{\text{v(x)}}\Big\}\Big]$
$\Rightarrow\text{f}'\text{(x)}=\frac{1}{\Big[\frac{\text{u(x)}}{\text{v(x)}}\Big]}\times\frac{\text{d}}{\text{dx}}\Big[\frac{\text{u(x)}}{\text{v(x)}}\Big]$
$\Rightarrow\text{f}'\text{(x)}=\frac{\text{v(x)}}{\text{u(x)}}\times\bigg[\frac{\text{v(x)}\frac{\text{d}}{\text{dx}}\{\text{u(x)}\}-\text{u(x)}\frac{\text{d}}{\text{dx}}\{\text{v(x)}\}}{\{\text{v(x)}\}^2}\bigg]$
$\Rightarrow\text{f}'\text{(x)}=\frac{\text{v(x)}}{\text{u(x)}}\times\Big[\frac{\text{u(x)}\times\text{u}'\text{(x)}-\text{u(x)}\times\text{v}'\text{(x)}}{\{\text{v(x)}\}^2}\Big]$
Putting x = 1, we get,
$\text{f}'(1)=\frac{\text{v}(1)}{\text{u}(1)}\times\Big[\frac{\text{u}(1)\times\text{u}'(1)-\text{u}1\times\text{v}'(1)}{\{\text{v}(1)\}^2}\Big]$
$\Rightarrow\text{f}'(1)=1\times\Big[\frac{\text{u}(1)\times2-\text{u}(1)\times 2}{\{\text{u}(1)\}^2}\Big]$
[Using eqn (1)]
$\Rightarrow\text{f}'(1)=\Big[\frac{0}{\{\text{u}(1)\}^2}\Big]$
$\Rightarrow\text{f}'(1)=0$
View full question & answer→Question 1765 Marks
Differentiate the following functions with respect to x:
$10^{\log\sin\text{x}}$
AnswerLet $\text{y}=10^{\log\sin\text{x}}\ .....(\text{i})$
Taking log on both sides,
$\log\text{y}=\log10^{\log\sin\text{x}}$
$\Rightarrow\log\text{y}=\log\sin\text{x}\log10$
Differentiating with respect to x,
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\log10\frac{\text{d}}{\text{dx}}\log\sin\text{x}$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\log10\frac{1}{\sin\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\log10\Big(\frac{1}{\sin\text{x}}\Big)(\cos\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}\big[\log10\times\cot\text{x}\big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=10^{\log\sin\text{x}}\times\log10\times\cot\text{x}$
[Using equation (i)]
View full question & answer→Question 1775 Marks
If $\text{y}=\text{x}\sin\text{y},$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}(1-\text{x}\cos\text{y})}$
AnswerWe have, $\text{y}=\text{x}\sin\text{y}\ .....(\text{i})$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x}\sin\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}(\sin\text{y})+\sin\text{y}\frac{\text{d}}{\text{dx}}(\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}\cos\text{y}\frac{\text{dy}}{\text{dx}}+\sin\text{y}(1)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}-\text{x}\cos\text{y}\frac{\text{dy}}{\text{dx}}=\sin\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}(1-\text{x}\cos\text{y})=\sin\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sin\text{y}}{(1-\text{x}\cos\text{y})}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}(1-\text{x}\cos\text{y})}\Big[\because\sin\text{y}=\frac{\text{y}}{\text{x}}\Big]$
View full question & answer→Question 1785 Marks
If $\text{y}=\text{e}^{\text{x}}\cos\text{x},$ Prvoe that $\frac{\text{dy}}{\text{dx}}=\sqrt{2}\text{e}^\text{x}.\cos\Big(\text{x}+\frac{\pi}{4}\Big)$
AnswerGiven, $\text{y}=\text{e}^{\text{x}}\cos\text{x}$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^\text{x}\cos\text{x}\big)$
$=\text{e}^\text{x}\frac{\text{d}}{\text{dx}}\cos\text{x}+\cos\text{x}\frac{\text{d}}{\text{dx}}\text{e}^{\text{x}}$ [Using product rule]
$=\text{e}^\text{e}(-\sin\text{x})+\text{e}^\text{x}\cos\text{x}$
$=\text{e}^\text{x}(\cos\text{x}-\sin\text{x})$
$=\sqrt{2}\text{e}^\text{x}\Big(\frac{\cos\text{x}}{\sqrt{2}}-\frac{\sin\text{x}}{\sqrt{2}}\Big)$ $\big[$Multiplying and dividing by $\sqrt{2}\big]$
$=\sqrt{2}\text{e}^\text{x}\Big(\cos\frac{\pi}{4}\cos\text{x}-\sin\frac{\pi}{4}\sin\text{x}\Big)$
$\frac{\text{dy}}{\text{dx}}=\sqrt{2}\text{e}^\text{x}\cos\Big(\text{x}+\frac{\pi}{4}\Big)$
View full question & answer→Question 1795 Marks
Differentiate the following functions with respect to x:
$3\text{e}^{-3\text{x}}\log(1+\text{x})$
AnswerConsider $\text{y}=3\text{e}^{-3\text{x}}\log(1+\text{x})$
Differentiating it with respect to x and applying the chain and product rule, we get
$\frac{\text{dy}}{\text{dx}}=3\frac{\text{d}}{\text{dx}}\big[3\text{e}^{-3\text{x}}\log(1+\text{x})\big]$
$\frac{\text{dy}}{\text{dt}}=3\Big(\text{e}^{-3\text{x}}\frac{1}{1+\text{x}}+\log(1+\text{x})\big(-3\text{e}^{-3\text{x}}\big)\Big)$
$=3\Big(\frac{\text{e}^{-3\text{x}}}{1+\text{x}}-3\log(1+\text{x})\Big)$
The solution is,
$=3\text{e}^{-3\text{e}}\Big(\frac{1}{1+\text{x}}-3\log(1-\text{x})\Big)$
View full question & answer→Question 1805 Marks
Differentiate the following functions with respect to x:
$(\log\text{x})^{\cos\text{x}}$
AnswerLet $\text{y}=(\log\text{x})^{\cos\text{x}}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=(\log\text{x})^{\cos\text{x}}$
$\Rightarrow\log\text{y}=\cos\text{x}\log(\log\text{x})$
Differentiating with respect to x,
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\cos\text{x}\frac{\text{d}}{\text{dx}}\log(\log\text{x})+\log(\log\text{x})\frac{\text{d}}{\text{dx}}(\cos\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\cos\text{x}}{\log\text{x}}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log(\log\text{x})\times(-\sin\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\cos\text{x}}{\log\text{x}}\times\big(\frac{1}{\text{x}}\big)-\sin\text{x}\log(\log\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{\cos\text{x}}{\text{x}\log\text{x}}-\sin\text{x}\log(\log\text{x})\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=(\log\text{x}^{\cos\text{x}})\Big[\frac{\cos\text{x}}{\text{x}\log\text{x}}-\sin\text{x}\log(\log\text{x})\Big]$
[Using equation (i)]
View full question & answer→Question 1815 Marks
Differentiate the following functions with respect to x:
$\text{x}^{\text{x}\cos\text{x}}+\frac{\text{x}^2+1}{\text{x}^2-1}$
AnswerLet $\text{y}=\text{x}^{\text{x}\cos\text{x}}+\frac{\text{x}^2+1}{\text{x}^2-1}$
Also, let $\text{u}=\text{x}^{\text{x}\cos\text{x}}\text{ and v}=\frac{\text{x}^2+1}{\text{x}^2-1}$
$\therefore\ \text{y}=\text{u}+\text{v}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{du}}{\text{dx}}+\frac{\text{dv}}{\text{dx}}\ .....(\text{i})$
$\text{u}=\text{x}^{\text{x}\cos\text{x}}$
$\Rightarrow\ \log\text{u}=\log(\text{x}^{\text{x}\cos\text{x}})$
$\Rightarrow\log\text{u}=\text{x}\cos\text{x}\log\text{x}$
Diffrerentiating both sides with respect to x, we obtain
$\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x}).\cos\text{x}\log\text{x}+\text{x}.\frac{\text{d}}{\text{dx}}(\cos\text{x}).\log\text{x}+\text{x}\cos\text{x}.\frac{\text{d}}{\text{dx}}(\log\text{x})$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{u}\Big[1.\cos\text{x}.\log\text{x}+\text{x}.(-\sin\text{x})\log\text{x}+\text{x}\cos\text{x}.\frac{1}{\text{x}}\Big]$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{x}^{\text{x}\cos\text{x}}(\cos\text{x}\log\text{x}-\text{x}\sin\text{x}\log\text{x}+\cos\text{x})$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{x}^{\text{x}\cos\text{x}}\big[\cos\text{x}(1+\cos\text{x})-\text{x}\sin\text{x}\log\text{x}\big]\ .....(\text{ii})$
$\text{v}=\frac{\text{x}^2+1}{\text{x}^2-1}$
$\Rightarrow\log\text{v}=\log(\text{x}^2+1)-\log(\text{x}^2-1)$
Differentiating both sides with respect to x, we obtain
$\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\frac{2\text{x}}{\text{x}^2+1}-\frac{2\text{x}}{\text{x}^2-1}$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\text{v}\Big[\frac{2\text{x}(\text{x}^2-1)-2\text{x}(\text{x}^2+1)}{(\text{x}^2+1)(\text{x}^2-1)}\Big]$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{\text{x}^2+1}{\text{x}^2-1}\times\Big[\frac{-4\text{x}}{(\text{x}^2+1)(\text{x}^2-1)}\Big]$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{-4\text{x}}{(\text{x}^2-1)^2}\ .....(\text{iii})$
From (1), (2) and (3), we obtain
$\frac{\text{dy}}{\text{dx}}=\text{x}^{\text{x}\cos\text{x}}\big[\cos\text{x}(1+\log\text{x})-\text{x}\sin\text{x}\log\text{x}\big]-\frac{4\text{x}}{(\text{x}^2-1)^2}$
View full question & answer→Question 1825 Marks
Differentiate the following functions with respect to x:
$\sin^{-1}\big\{\sqrt{1-\text{x}^2}\big\},0<\text{x}<1$
AnswerLet $\text{y}=\sin^{-1}\big\{\sqrt{1-\text{x}^2}\big\}$
Put $\text{x}=\cos2\theta$
$\text{y}=\sin^{-1}\big\{\sqrt{1-\cos^2\theta}\big\}$
$\text{y}=\sin^{-1}(\sin\theta)\ .....(\text{i})$
Here, $0<\text{x}<1$
$\Rightarrow\ 0<\cos2\theta<1$
$\Rightarrow\ 0<2\theta<\frac{\pi}{2}$
From equation (i),
$\text{y}=\theta$
$\Big[\text{Since, } \sin^{-1}(\sin\theta)=\theta\text{ if }\theta \in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\text{y}=\cos^{-1}\text{x}\ \big[\text{Since x}=\cos\theta\big]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=-\frac{1}{\sqrt{1-\text{x}^2}}$
View full question & answer→Question 1835 Marks
If $\text{y}=(\tan\text{x})^{(\tan\text{x})^{(\tan\text{x})^{....\infty}}},$ prove that $\frac{\text{dy}}{\text{dx}}=2\text{ at x}=\frac{\pi}{4}$
AnswerWe have, $\text{y}=(\tan\text{x})^{(\tan\text{x})^{(\tan\text{x})^{....\infty}}}$
$\Rightarrow\text{y}=(\tan\text{x})^{\text{y}}$
Taking log on both sides,
$\log\text{y}=\log(\tan\text{x})^\text{y}$
$\Rightarrow\log\text{y}=\text{y}\log\tan\text{x}$
Differentaiting with respect to x using chain rule,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{y}\frac{\text{d}}{\text{dx}}\big\{\log\tan\text{x}\big\}+\log\tan\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\tan\text{x}}\frac{\text{d}}{\text{dx}}(\tan\text{x})+\log\tan\frac{\text{d}}{\text{dx}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\Big(\frac{1}{\text{y}}-\log\tan\text{x}\Big)=\frac{\text{y}}{\tan\text{y}}\sec^2\text{x}$
Now, $\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=\frac{\pi}{4}}=\frac{\text{y}\sec^3\big(\frac{\pi}{4}\big)}{\tan\big(\frac{\pi}{4}\big)}\times\frac{\text{y}}{1-\text{y}\log\tan\big(\frac{\pi}{4}\big)}$
$\Rightarrow\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=\frac{\pi}{4}}=\frac{\text{y}^2\big(\sqrt{2}\big)^2}{1(1-\text{y}\log\tan1)}$
$\Rightarrow \Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=\frac{\pi}{4}}=\frac{2(1)^2}{(1-0)}\Bigg[\because(\text{y})_{\frac{\pi}{4}}=\big(\tan\frac{\pi}{4}\big)^{\big(\tan\frac{\pi}{4}\big)^{\big(\tan\frac{\pi}{4}\big)^{\ .....\infty}}}=1\Bigg]$
$\Rightarrow \Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=\frac{\pi}{4}}=2$
View full question & answer→Question 1845 Marks
Differentiate the following functions with respect to x:
$\big(\sin^{-1}\text{x}^4\big)^4$
AnswerConsider $\text{y}=\big(\sin^{-1}\text{x}^4\big)^4$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\sin^{-1}\text{x}^4\big)^4$
$=4\big(\sin^{-1}\text{x}^4\big)\frac{\text{d}}{\text{dx}}\big(\sin^{-1}\text{x}^4\big)$
[Using chain rule]
$=4\big(\sin^{-1}\text{x}^4\big)^3\frac{1}{\sqrt{1-\big(\text{x}^4\big)^2}}\frac{\text{d}}{\text{dx}}\big(\text{x}^4\big)$
$=4\big(\sin^{-1}\text{x}^4\big)^3\frac{4\text{x}^3}{\sqrt{1-\text{x}^8}}$
$=\frac{16\text{x}^3\big(\sin^{-1}\text{x}^4\big)^3}{\sqrt{1-\text{x}^8}}$
Hence, the solution is, $\frac{\text{d}}{\text{dx}}\big(\sin^{-1}\text{x}^4\big)=\frac{16\text{x}^3\big(\sin^{-1}\text{x}^4\big)^3}{\sqrt{1-\text{x}^8}}$
View full question & answer→Question 1855 Marks
Differentiate the following functions with respect to x:
$\frac{2^\text{x}\cos\text{x}}{(\text{x}^2+3)^2}$
AnswerLet $\text{y}=\frac{\text{2}^\text{x}\cos\text{x}}{(\text{x}^2+3)^3}$
Differentiate it with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big[\frac{\text{2}^\text{x}\cos\text{x}}{(\text{x}^2+3)^3}\Big]$
$=\bigg[\frac{(\text{x}^2+3)^2\frac{\text{d}}{\text{dx}}(2^\text{x}\cos\text{x})-(2^\text{x}\cos\text{x})\frac{\text{d}}{\text{dx}}(\text{x}^2+3)^2}{\big[(\text{x}^2+3)\big]^2}\bigg]$
[Using quotient rule]
$=\Bigg[\frac{(\text{x}^2+3)^2\Big\{2^\text{x}\frac{\text{d}}{\text{dx}}\cos\text{x}+\cos\text{x}\frac{\text{d}}{\text{dx}}2^\text{x}\Big\}-(2^\text{x}+3)2(\text{x}^2+3)\frac{\text{d}}{\text{dx}}(\text{x}^2+3)}{(\text{x}^2+3)^4}\Bigg]$
[Using Product rule and chain rule]
$=\bigg[\frac{(\text{x}^2+3)^2\big\{-2^\text{x}\sin\text{x}+\cos\text{x}2^\text{x}\log_\text{e}2\big\}-2(2^\text{x}\cos\text{x})(\text{x}^2+3)(2\text{x})}{(\text{x}^2+3)^4}\bigg]$
$=\bigg[\frac{2^\text{x}(\text{x}^2+3)\big\{(\text{x}^2+3)(\cos\text{x}\log_\text{e}2-\sin\text{x})-4\text{x}\cos\text{x}\big\}}{(\text{x}^2+3)^4}\bigg]$
$=\frac{2^\text{x}}{(\text{x}^2+3)^2}\bigg[\cos\text{x}\log_\text{e}2-\sin\text{x}-\frac{4\text{x}\cos\text{x}}{(\text{x}^2+3)}\bigg]$
So,
$\frac{\text{d}}{\text{dx}}\Big[\frac{2^\text{x}\cos\text{x}}{(\text{x}^2+3)^2}\Big]=\frac{2^\text{x}}{(\text{x}^2+3)^2}\bigg[\cos\text{x}\log_\text{e}2-\sin\text{x}-\frac{4\text{x}\cos\text{x}}{(\text{x}^2+3)}\bigg]$
View full question & answer→Question 1865 Marks
Differentiate the following functions with respect to x:
$\text{x}^{\sin^{-1}\text{x}}$
AnswerLet $\text{y}=\text{x}^{\sin^{-1}\text{x}}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=\log\text{x}^{\sin^{-1}\text{x}}$
$\log\text{y}=\sin^{-1}\text{x}\log\text{x}$
Differentiating it with respect to x,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\sin^{-1}\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+(\log\text{x})\frac{\text{d}}{\text{dx}}(\sin^{-1}\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\sin^{-1}\text{x}\Big(\frac{1}{\text{x}}\Big)+(\log\text{x})\Big(\frac{1}{\sqrt{1-\text{x}^2}}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{\sin^{-1}\text{x}}{\text{x}}+\frac{\log\text{x}}{\sqrt{1-\text{x}^2}}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}^{\sin^{-1}\text{x}}\Big[\frac{\sin^{-1}\text{x}}{\text{x}}+\frac{\log\text{x}}{\sqrt{1-\text{x}^2}}\Big]$
[Using equation (i)]
View full question & answer→Question 1875 Marks
If $xy^2 = 1$, prove that $2\frac{\text{dy}}{\text{dx}}+\text{y}^3=0$
AnswerWe have $xy^2 = 1 .....(i)$
Differentiating with respect to x, we get,
$\frac{\text{d}}{\text{dx}}(\text{xy}^2)=\frac{\text{d}}{\text{dx}}(1)$
$\Rightarrow\text{x}\frac{\text{d}}{\text{dx}}(\text{y}^2)+\text{y}^2\frac{\text{d}}{\text{dx}}(\text{x})=0$
$\Rightarrow\text{x}(2\text{y})\frac{\text{d}}{\text{dx}}+\text{y}^2(1)=0$
$\Rightarrow2\text{xy}\frac{\text{dy}}{\text{dx}}=-\text{y}^2$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-\text{y}^2}{2\text{xy}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-\text{y}}{2\text{x}}$
Put $\text{x}=\frac{1}{\text{y}^2}$ from equation (i)
$ \Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-\text{y}}{2\Big(\frac{1}{\text{y}^2}\Big)}$
$\Rightarrow2\frac{\text{dy}}{\text{dx}}=-\text{y}^3$
$\Rightarrow2\frac{\text{dy}}{\text{dx}}+\text{y}^3=0$
View full question & answer→Question 1885 Marks
If $\text{y}=\tan^{-1}\Big(\frac{1-\text{x}}{1+\text{x}}\Big),$, find $\frac{\text{dy}}{\text{dx}}.$
AnswerHere,
$\text{y}=\tan^{-1}\Big(\frac{1-\text{x}}{1+\text{x}}\Big)$
Differentiating it with respect to x using chain rule and quotinet rule,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{1+\big(\frac{1-\text{x}}{1+\text{x}}\big)^2}\frac{\text{d}}{\text{dx}}\Big(\frac{1-\text{x}}{1+\text{x}}\Big)$
$=\frac{(1+\text{x})^2}{(1+\text{x}^2+2\text{x}+1+\text{x}^2-2\text{x})}\bigg[\frac{(1+\text{x})\frac{\text{d}}{\text{dx}}(1-\text{x})-(1-\text{x})\frac{\text{d}}{\text{dx}}(1+\text{x})}{(1+\text{x})^2}\bigg]$
$=\frac{(1+\text{x})^2}{2\text{x}^2+2}\Big[\frac{(1+\text{x})(-1)-(1-\text{x})(1)}{(1+\text{x})^2}\Big]$
$=\frac{(1+\text{x})^2}{2(\text{x}^2+1)}\Big(\frac{-\text{x}-1-1+\text{x}}{(1+\text{x})^2}\Big)$
$=\frac{(1+\text{x})^2}{2(\text{x}^2+1)}\times\frac{-2}{(1+\text{x})^2}$
$\frac{\text{dy}}{\text{dx}}=-\frac{1}{\text{x}^2+1}$
View full question & answer→Question 1895 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\bigg[\frac{\text{x}^\frac{1}{3}+\text{a}^{\frac{1}{3}}}{1-(\text{ax})^\frac{1}{3}}\bigg]$
AnswerLet $\text{y}=\tan^{-1}\bigg[\frac{\text{x}^\frac{1}{3}+\text{a}^{\frac{1}{3}}}{1-(\text{ax})^\frac{1}{3}}\bigg]$
$\Rightarrow\text{y}=\tan^{-1}\big(\text{x}^\frac{1}{3}\big)+\tan^{-1}\big(\text{a}^\frac{1}{3}\big)$
$\Big[\text{Since},\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)\Big]$
Differentiate it with respect to x usign chain rule,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{1+\big(\text{x}^\frac{1}{3}\big)^2}\times\frac{\text{d}}{\text{dx}}\big(\text{x}^\frac{1}{2}\big)+0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\bigg(\frac{1}{2}\times\text{x}^{\frac{1}{3}-1}\bigg)}{1+\text{x}^\frac{2}{3}}$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{1}{3\text{x}^\frac{2}{3}\Big(1+\text{x}^\frac{2}{3}\Big)}$
View full question & answer→Question 1905 Marks
Differentiate the following functions with respect to x:
$\frac{\text{e}^\text{x}\log\text{x}}{\text{x}^2}$
AnswerLet $\text{y}=\frac{\text{e}^\text{x}\log\text{x}}{\text{x}^2}$
Differentiate with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}^2\frac{\text{d}}{\text{dx}}(\text{e}^\text{x}\log\text{x})-(\text{e}^\text{x}\log\text{x})\frac{\text{d}}{\text{dx}}\text{x}^2}{\big(\text{x}^2\big)^2}$
[Using quotient rule]
$=\frac{\text{x}^2\Big\{\text{e}^\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{e}^\text{x})\Big\}-\text{e}^\text{x}\log\text{x}\times2\text{x}}{\text{x}^4}$
[Using product rule]
$=\frac{\text{x}^2\Big[\frac{\text{e}^\text{x}}{\text{x}}+\text{e}^\text{x}\log\text{x}\Big]-2\text{xe}^\text{x}\log\text{x}}{\text{x}^4}$
$=\frac{\frac{\text{x}^2\text{e}^\text{z}(1+\text{x}\log\text{x})}{\text{x}}-2\text{xe}^\text{z}\log\text{x}}{\text{x}^4}$
$=\frac{\text{xe}^\text{x}[1+\text{x}\log\text{x}-2\log\text{x}]}{\text{x}^4}$
$=\frac{\text{xe}^\text{x}}{\text{x}^3}\Big[\frac{1}{\text{x}}+\frac{\text{x}\log\text{x}}{\text{x}}-\frac{2\log\text{x}}{\text{x}}\Big]$
$=\text{e}^\text{x}\text{x}^{-2}\Big[\frac{1}{\text{x}}+\log\text{x}-\frac{2}{\text{x}}\log\text{x}\Big]$
So,
$\frac{\text{d}}{\text{dx}}\Big[\frac{\text{e}^\text{x}\log\text{x}}{\text{x}^2}\Big]=\text{e}^\text{x}\text{x}^{-2}\Big[\frac{1}{\text{x}}+\log\text{x}-\frac{2}{\text{x}}\log\text{x}\Big]$
View full question & answer→Question 1915 Marks
Differentiate the following functions with respect to x:
$\sqrt{\frac{1-\text{x}^2}{1+\text{x}^2}}$
AnswerLet $\text{y}=\sqrt{\frac{1-\text{x}^2}{1+\text{x}^2}}$
$\text{y}=\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)^\frac{1}{2}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)^\frac{1}{2}$
$=\frac{1}{2}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)^{\frac{1}{2}-1}\frac{\text{d}}{\text{dx}}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$
[Using chain rule]
$=\frac{1}{2}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)^{\frac{-1}{2}}\bigg[\frac{(1+\text{x}^2)\frac{\text{d}}{\text{dx}}(1-\text{x}^2)-(1-\text{x}^2)\frac{\text{d}}{\text{dx}}(1+\text{x}^2)}{\big(1+\text{x}^2\big)^2}\bigg]$
[Using chain rule]
$=\frac{1}{2}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)^{\frac{1}{2}}\bigg[\frac{(1+\text{x}^2)(-2\text{x})-(1-\text{x}^2)(2\text{x})}{\big(1+\text{x}^2\big)^2}\bigg]$
$=\frac{1}{2}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)^{\frac{1}{2}}\bigg[\frac{-2\text{x}-2\text{x}^3-2\text{x}+2\text{x}^3}{\big(1+\text{x}^2\big)^2}\bigg]$
$=\frac{1}{2}\frac{-4\text{x}}{\sqrt{1-\text{x}^2}\big(1+\text{x}^2)^\frac{3}{2}}$
So,
$\frac{\text{d}}{\text{dx}}\bigg(\sqrt{\frac{1-\text{x}^2}{1+\text{x}^2}}\bigg)=\frac{-4\text{x}}{\sqrt{1-\text{x}^2}\big(1+\text{x}^2)^\frac{3}{2}}$
View full question & answer→Question 1925 Marks
Find $\frac{\text{dy}}{\text{dx}}$ in the following cases:
$y^3 - 3xy^2 = x^3 + 3x^2y$
AnswerHere, $y^3 - 3xy^2 = x^3 + 3x^2y$
Differentiating with respect to x,
$\Rightarrow\frac{\text{d}}{\text{dy}}(\text{y}^3)-\frac{\text{d}}{\text{dx}}(3\text{xy}^2)=\frac{\text{d}}{\text{dx}}(\text{x}^3)+\frac{\text{d}}{\text{dx}}(3\text{x}^2\text{y})$
$\Rightarrow 3\text{y}^2\frac{\text{dy}}{\text{dx}}-3\Big[\text{x}\frac{\text{d}}{\text{dx}}\text{y}^2\frac{\text{d}}{\text{dx}}(\text{x})\Big]=3\text{x}^2+3\Big[\text{x}^2\frac{\text{d}}{\text{dx}}(\text{y})+\text{y}\frac{\text{d}}{\text{dx}}(\text{x}^2)\Big]$
$\Rightarrow 3\text{y}^2\frac{\text{dy}}{\text{dx}}-3\Big[\text{x}(2\text{y})\frac{\text{dy}}{\text{dx}}+\text{y}^2\Big]=3\text{x}^2+3\Big[\text{x}^2\frac{\text{dy}}{\text{dx}}+\text{y}(2\text{x})\Big]$
$\Rightarrow 3\text{y}^2\frac{\text{dy}}{\text{dx}}-6\text{xy}\frac{\text{dy}}{\text{dx}}-3\text{y}^2+3\text{x}^2+3\text{x}^2\frac{\text{dy}}{\text{dx}}+6\text{xy}$
$\Rightarrow 3\text{y}^2\frac{\text{dy}}{\text{dx}}-6\text{xy}\frac{\text{dy}}{\text{dx}}-3\text{x}^2\frac{\text{dy}}{\text{dx}}=3\text{x}^2+6\text{xy}+3\text{y}^2$
$=3\frac{\text{dy}}{\text{dx}}(\text{y}^2-2\text{xy}-\text{x}^2)=3(\text{x}^2+2\text{xy}+\text{y}^2)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{3(\text{x}+\text{y})^2}{3(\text{y}^2-2\text{xy}-\text{x}^2)}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(\text{x}+\text{y})^2}{\text{y}^2-2\text{xy}-\text{x}^2)}$
View full question & answer→Question 1935 Marks
Differentiate the following functions with respect to x:
$\sin^{-1}\big(2\text{x}^2-1\big),0<\text{x}<1$
AnswerLet $\text{y}=\sin^{-1}\big\{2\text{x}^2-1\big\}$
Let $\text{x}=\cos\theta$
$\text{y}=\sin^{-1}\big(2\cos^2\theta-1\big)$
$=\sin^{-1}(\cos2\theta)$
$\text{y}=\sin^{-1}\Big\{\sin\Big(\frac{\pi}{2}-2\theta\Big)\Big\}\ .....(\text{i})$
Here, $0<\text{x}<1$
$\Rightarrow\ 0<\cos\theta<1$
$\Rightarrow\ 0<2\theta<\frac{\pi}{2}$
$\Rightarrow\ 0> -2\theta>-\pi$
$\Rightarrow\ \frac{\pi}{2}>\Big(\frac{\pi}{2}-2\theta\Big)>-\frac{\pi}{2}$
So, from equation (i),
$\text{y}=\frac{\pi}{2}-2\theta$
$\Big[\text{Sicne}, \sin^{-1}(\cos\theta)=\theta,\text{ if }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\text{y}=\frac{\pi}{2}-2\cos^{-1}\text{x}\ \big[\text{Since x}=\cos\theta\big]$
$\frac{\text{dy}}{\text{dx}}=0-2\frac{\text{d}}{\text{dx}}\big(\cos^{-1}\text{x}\big)$
$=-2\Big(-\frac{1}{\sqrt{1-\text{x}^2}}\Big)$
$\frac{\text{dy}}{\text{dx}}=\frac{2}{\sqrt{1-\text{x}^2}}$
View full question & answer→Question 1945 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{\text{a}+\text{bx}}{\text{b}-\text{ax}}\Big)$
AnswerLet $\text{y}=\tan^{-1}\Big(\frac{\text{a}+\text{bx}}{\text{b}-\text{ax}}\Big)$
$=\tan^{-1}\bigg(\frac{\frac{\text{a}+\text{bx}}{\text{b}}}{\frac{\text{b}-\text{ax}}{\text{b}}}\bigg)$
$=\tan^{-1}\bigg(\frac{\frac{\text{a}}{\text{b}}+\frac{\text{bx}}{\text{b}}}{\frac{\text{b}}{\text{a}}-\frac{\text{ax}}{\text{b}}}\bigg)$
$=\tan^{-1}\bigg(\frac{\frac{\text{a}}{\text{b}}+\text{x}}{1-\big(\frac{\text{a}}{\text{b}}\big)\text{x}}\bigg)$
$\text{y}=\tan^{-1}\big(\frac{\text{a}}{\text{b}}\big)+\tan^{-1}\text{x}$
$\Big[\text{Since},\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)\Big]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=0+\frac{1}{1+\text{x}^2}$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{1+\text{x}^2}$
View full question & answer→Question 1955 Marks
Differentiate the following functions with respect to x:
$\log(\text{x}+\sqrt{\text{x}^2+1})$
AnswerLet $\text{y}=\log(\text{x}+\sqrt{\text{x}^2+1})$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dt}}=\frac{\text{d}}{\text{dy}}\log\big(\text{x}+\sqrt{\text{x}^2+1}\big)$
$=\frac{1}{\text{x}+\sqrt{\text{x}^2+1}}\frac{\text{d}}{\text{dx}}\Big(\text{x}+\big(\text{x}^2+1\big)^\frac{1}{2}\Big)$
[Using chain rule]
$=\frac{1}{\text{x}+\sqrt{\text{x}^2+1}}\Big[1+\frac{1}{2}\big(\text{x}^2+1\big)^{\frac{1}{2}-1}\frac{\text{d}}{\text{dx}}\big(\text{x}^2+1\big)\Big]$
$=\frac{1}{\text{x}+\sqrt{\text{x}^2+1}}\Big[1+\frac{1}{2\sqrt{\text{x}^2+1}}\times2\text{x}\Big]$
$=\frac{1}{\text{x}+\sqrt{\text{x}^2+1}}\Big[\frac{\sqrt{\text{x}^2+1}+\text{x}}{\sqrt{\text{x}^2+1}}\Big]$
$=\frac{1}{\sqrt{\text{x}^2+1}}$
So,
$\frac{\text{d}}{\text{dx}}\Big(\log\big(\text{x}+\sqrt{\text{x}^2+1}\big)\Big)=\frac{1}{\sqrt{\text{x}^2+1}}$
View full question & answer→Question 1965 Marks
Differentiate the following functions with respect to x:
$\log_\text{x}3$
AnswerLet, $\text{y}=\log_\text{x}3$
$\Rightarrow\ \text{y}=\frac{\log3}{\log\text{x}}\ \Big[\because\ \log_\text{a}\text{b}=\frac{\log\text{b}}{\log\text{a}}\Big]$
Differentiate it with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\frac{\log3}{\log\text{x}}\Big)$
$=\log3\frac{\text{d}}{\text{dx}}(\log\text{x})^{-1}$
$=\log3\times\Big[-1(\log\text{x})^{-2}\Big]\frac{\text{d}}{\text{dx}}(\log\text{x})$
[Using chain rule]
$=-\frac{\log 3}{(\log\text{x})^2}\times\frac{1}{\text{x}}$
$=-\Big(\frac{\log 3}{\log\text{x}}\Big)^2\times\frac{1}{\text{x}}\times\frac{1}{\log3}$
$=-\frac{1}{\text{x}\log3(\log_3\text{x})^2} \Big[\because \frac{\log\text{b}}{\log\text{a}}=\log_\text{a}\text{b}\Big]$
So,
$\frac{\text{d}}{\text{dx}}(\log_\text{x}3)=-\frac{1}{\text{x}\log3(\log_3\text{x})^2}$
View full question & answer→Question 1975 Marks
If $xy = e^{x-y}$, find $\frac{\text{dy}}{\text{dx}}$
AnswerThe given function is $xy = e^{x-y}$
Taking logarithm on both the sides, we obtain
$\log(\text{xy})=\log\big(\text{e}^{\text{x}-\text{y}})$
$\Rightarrow\log\text{x}+\log\text{y}=(\text{x}-\text{y})\log\text{e}$
$\Rightarrow\log\text{x}=\log\text{y}=(\text{x}-\text{y})\times1$
$\Rightarrow\log\text{x}=\log\text{y}=\text{x}-\text{y}$
Differentiating both sides with respect to x, we obtain
$\frac{\text{d}}{\text{dx}}(\log\text{x})+\frac{\text{d}}{\text{dx}}(\log\text{y})=\frac{\text{d}}{\text{dx}}(\text{x})-\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{1}{\text{x}}+\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=1-\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\big(1+\frac{1}{\text{y}}\big)\frac{\text{dy}}{\text{dx}}=1-\frac{1}{\text{x}}$
$\Rightarrow\big(\frac{\text{y}+1}{\text{y}}\big)\frac{\text{dy}}{\text{dx}}=\frac{\text{x}-1}{\text{x}}$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\text{y}(\text{x}-1)}{\text{x}(\text{y}+1)}$
View full question & answer→Question 1985 Marks
Find $\frac{\text{dy}}{\text{dx}}$ in the following cases:
$(x^2 + y^2)^2 = xy$
AnswerGiven, $(x^2 + y^2)^2 = xy$
Differentiating with respect to x,
$\frac{\text{d}}{\text{dx}}\Big(\big(\text{x}^2+\text{y}^2\big)^2\Big)=\frac{\text{d}}{\text{dx}}(\text{xy})$
$\Rightarrow2(\text{x}^2+\text{y}^2\big)\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{y}^2\big)=\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\frac{\text{d}}{\text{dx}}\big(\text{x}\big)$
[Using chain rule]
$\Rightarrow2(\text{x}^2+\text{y}^2\big)\Big(2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}\Big)=\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}(1)$
$\Rightarrow4\text{x}(\text{x}^2+\text{y}^2\big)+4\text{y}\big(\text{x}^2+\text{y}^2\big)\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}$
$\Rightarrow4\text{y}\big(\text{x}^2+\text{y}^2\big)\frac{\text{dy}}{\text{dx}}-\text{x}\frac{\text{dy}}{\text{dx}}=\text{y}-4\text{x}(\text{x}^2+\text{y}^2\big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\big[4\text{y}\text{x}^2+4\text{y}^3-\text{x}\big]=\text{y}-4\text{x}^3-4\text{x}\text{y}^2$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\Big(\frac{\text{y}-4\text{x}^3-4\text{x}\text{y}^2}{4\text{y}\text{x}^2+4\text{y}^3-\text{x}}\Big)$
View full question & answer→Question 1995 Marks
If $\text{y}=\cos^{-1}\Big\{\frac{2\text{x}-3\sqrt{1-\text{x}^2}}{\sqrt{13}}\Big\},$ find $\frac{\text{dy}}{\text{dx}}.$
AnswerHere, $\text{y}=\cos^{-1}\Big\{\frac{2\text{x}-3\sqrt{1-\text{x}^2}}{\sqrt{13}}\Big\}$
Let $\text{x}=\cos\theta, \text{So},$
$\text{y}=\cos^{-1}\Big\{\frac{2\cos\theta-3\sqrt{1-\cos^2\theta}}{\sqrt{13}}\Big\}$
$=\cos^{-1}\Big\{\frac{2}{\sqrt{13}}\cos\theta-\frac{3}{13}\sin\theta\Big\}$
Let $\cos\phi=\frac{2}{\sqrt{13}}$
$\Rightarrow\sin\phi=\sqrt{1-\cos^2\phi}$
$=\sqrt{1-\Big(\frac{2}{\sqrt{13}}\Big)^2}$
$=\sqrt{\frac{13-4}{13}}$
$=\sqrt{\frac{9}{13}}$
$\sin\phi=\frac{3}{\sqrt{13}}$
So,
$\text{y}=\cos^{-1}\big\{\cos\phi\cos\theta-\sin\phi\sin\theta\big\}$
$=\cos^{-1}\big[\cos(\theta+\phi)\big]$
$\text{y}=\phi+\theta$
$\text{y}=\cos^{-1}\Big(\frac{2}{\sqrt{13}}\Big)+\cos^{-1}\text{x}$
$\Big[\text{Since, x}=\cos\theta,\cos\phi=\frac{2}{\sqrt{13}}\Big]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=0+\Big(-\frac{1}{\sqrt{1-\text{x}^2}}\Big)$
$\frac{\text{dy}}{\text{dx}}=-\frac{1}{\sqrt{1-\text{x}^2}}$
View full question & answer→Question 2005 Marks
Differentiate the following functions with respect to x:
$\frac{\text{x}^2(1-\text{x}^2)}{\cos2\text{x}}$
AnswerConsider $\text{y}=\frac{\text{x}^2+2}{\sqrt{\cos\text{x}}}$
Differentiate it with respect to x and applying the chain and product rule, we get
$\frac{\text{dy}}{\text{dx}}=\frac{\cos2\text{x}\frac{\text{d}}{\text{dx}}\text{x}^2(1-\text{x}^2)^3-\text{x}^2(1-\text{x}^2)^3\frac{\text{d}}{\text{dx}}\cos2\text{x}}{\cos^2 2\text{x}}$
$=\frac{\cos2\text{x}\Big[\text{x}^2\frac{\text{d}}{\text{dx}}(1-\text{x}^2)^3+(1-\text{x}^2)^3\frac{\text{d}}{\text{dx}}\text{x}^2-\text{x}^2(1-\text{x}^2)^3(-2\sin2\text{x})\Big]}{\cos^2 2\text{x}}$
$=\frac{\cos2\text{x}\Big[-6\text{x}^2(1-\text{x}^2)^2+(1-\text{x}^2)^32\text{x}+2\text{x}^2(1-\text{x}^2)^3\sin2\text{x}\Big]}{\cos^2 2\text{x}}$
$=\frac{2\text{x}(1-\text{x}^2)^2}{\cos2\text{x}}-\frac{6\text{x}^3(1-\text{x}^2)^2}{\cos2\text{x}}+\frac{2\text{x}^2(1-\text{x}^2)^3\sin2\text{x}}{\cos^2 2\text{x}}$
$=2\text{x}(1-\text{x}^2)\sec2\text{x}\big\{1-4\text{x}^2+\text{x}(1-\text{x}^2)\tan2\text{x}\big\}$
Therefore,
$\frac{\text{dy}}{\text{dx}}=2\text{x}(1-\text{x}^2)\sec2\text{x}\big\{1-4\text{x}^2+\text{x}(1-\text{x}^2)\tan2\text{x}\big\}$
View full question & answer→