RD Sharma Solutions Class 12 Mathematics Chapter 10 MCQ

# RD Sharma Solutions Class 12 Mathematics Chapter 10 MCQ

Edited By Satyajeet Kumar | Updated on Jan 20, 2022 06:06 PM IST

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Also Read - RD Sharma Solution for Class 9 to 12 Maths

The RD Sharma class 12 chapter 10 exercise MCQ is one of the best NCERT solutions trusted and recommended by hundreds of students in their exams. It can be used for both school and board exams. The 10th chapter of the maths book deals with Differentiations in trigonometric equations. It will teach students about Logarithmic differentiation, differentiation of inverse trigonometric functions, Differentiation of a function. Exercise MCQ includes 33 questions on all the concepts that students have learned in this chapter.

## Differentiation Excercise: MCQ

Differentiation exercise Multiple choice question 1

$\frac{1}{2e}$
Hint:
Use quotient rule to differentiate the function
Given:
$f(x)=\log _{x^{2}}\left(\log _{e}\right) x$
Solution:
$f(x)=\log _{x^{2}}\left(\log _{e}\right) x$
$=\frac{\log (\log x)}{(\log (x))^{2}}$
$f(x)=\frac{\log (\log x)}{2 \log x}$
Use quotient rule
$f^{\prime}(x)=\frac{\frac{1}{2}\left[\log \left(\frac{1}{\log x}\right)\left(\frac{1}{x}\right)-\log (\log (x))\left(\frac{1}{x}\right)\right]}{(\log x)^{2}}$
$=\frac{1}{2} \frac{\left[\frac{(1-\log (\log x))}{x}\right]}{(\log x)^{2}}$
Put x=e
$f^{\prime}(e)=\left(\frac{1}{2}\right) \frac{\left[\frac{(1-\log (\log e))}{e}\right]}{(\log e)^{2}}$
\begin{aligned} &=\left(\frac{1}{2}\right) \frac{(1-0)}{e} \quad[\text { Since } \log e=1] \\\\ &f^{\prime}(e)=\frac{1}{2 e} \end{aligned}

Differentiation exercise Multiple choice question 2

$(x \log )^{-1}$
Hint:
Use differential formula
Given:
$f(x)=\log x$
Solution:
We have
\begin{aligned} &f(x)=\log x \\\\ &f(\log x)=\log (\log x) \\\\ &f^{\prime}(\log x)=\frac{1}{\log x} \frac{d}{d x}(\log x) \end{aligned}
$=\frac{1}{x \log x}$
$f^{\prime}(\log x)=(x \log x)^{-1}$

Differentiation exercise Multiple choice question 3

$\left(\frac{2}{3}\right)^{\frac{1}{2}}$
Hint:
Differentiate the given function and replace x by $\frac{\pi }{6}$ and solve
Given:
$\cot ^{-1}\left\{(\cos 2 x)^{\frac{1}{2}}\right\} \text { at } x=\frac{\pi}{6}$
Solution:
\begin{aligned} &y=\cot ^{-1}(\sqrt{\cos 2 x}) \\\\ &\frac{d y}{d x}=\frac{-1}{1+\cos 2 x} \frac{d}{d x} \sqrt{\cos 2 x} \end{aligned}
$\frac{d y}{d x}=\left[\frac{-1}{1+\cos 2 x}\right]\left[\frac{1}{2}(\sqrt{\cos 2 x})\right](-2 \sin 2 x)$
$=\frac{\sin 2 x}{(1+\cos 2 x) \sqrt{\cos 2 x}}$
$\text { At } x=\frac{\pi}{6}$
$\frac{d y}{d x}=\frac{\sin \frac{\pi}{3}}{\left(1+\cos \frac{\pi}{3}\right) \sqrt{\cos \frac{\pi}{3}}}$
$=\frac{\left(\frac{\sqrt{3}}{2}\right)}{\left(1+\frac{1}{2}\right) \sqrt{\frac{1}{2}}}$
\begin{aligned} &=\sqrt{\frac{2}{3}} \\\\ &=\left(\frac{2}{3}\right)^{\frac{1}{2}} \end{aligned}

Differentiation exercise Multiple choice question 4

$\frac{x}{\sqrt{1+x^{2}}}$
Hint:
Differentiate the function w.r.t x
Given:
$\sec \left(\tan ^{-1} x\right)$
Solution:
\begin{aligned} &y=\sec \left(\tan ^{-1} x\right) \\\\ &\frac{d y}{d x}=\sec \left(\tan ^{-1} x\right) \tan \left(\tan ^{-1} x\right) \times \frac{d}{d x}\left(\tan ^{-1} x\right) \end{aligned}
\begin{aligned} &=\sec \left(\tan ^{-1} x\right) \tan \left(\tan ^{-1} x\right) \times \frac{1}{\sqrt{1+x^{2}}} \\\\ &=y \tan \left(\tan ^{-1} x\right) \times \frac{1}{\sqrt{1+x^{2}}} \end{aligned}
$=y\left(\frac{x}{\sqrt{1+x^{2}}}\right) \; \; \; \; \; \; \quad\left[\tan \left(\tan ^{-1} x\right)=x\right]$
This is the equation of differential equation which have co-efficient= $\left(\frac{x}{\sqrt{1+x^{2}}}\right)$

Differentiation exercise Multiple choice question 5

$\frac{1}{2}$
Hint:
Differentiate the function and replace x by $\frac{\pi }{6}$ solve
Given:
$f(x)=\tan ^{-1} \sqrt{\frac{1+\sin x}{1-\sin x}}, 0 \leq x \leq \frac{\pi}{2}, f^{\prime}\left(\frac{\pi}{6}\right)$
Solution:
$y=\tan ^{-1}\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right)$
$=\tan ^{-1}\left(\sqrt{\frac{1-\cos \left(\frac{\pi}{2}+x\right)}{1+\cos \left(\frac{\pi}{2}+x\right)}}\right)$
$=\tan ^{-1}\left(\sqrt{\frac{2 \sin ^{2}\left(\frac{\pi}{4}+\frac{x}{2}\right)}{2 \cos ^{2}\left(\frac{\pi}{4}+\frac{x}{2}\right)}}\right)$
$y=\tan ^{-1}\left(\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right)=\frac{\pi}{4}+\frac{x}{2} \; \; \; \; \; \; \; \quad\left[\tan ^{-1}(\tan x)=x\right]$
$\frac{d y}{d x}=\frac{1}{2}$

Differentiation exercise Multiple choice question 6

$\left(1+\frac{1}{x}\right)^{x}\left(1+\frac{1}{x}\right)-\frac{1}{x+1}$
Hint:
Differentiate the function w.r.t x
Given:
$y=\left(1+\frac{1}{x}\right)^{x}$
Solution:
Let
$y=\left(1+\frac{1}{x}\right)^{x}$
Taking log on both sides
\begin{aligned} &\log y=x \log \left(1+\frac{1}{x}\right) \\\\ &\frac{1}{y} \frac{d y}{d x}=x \frac{d}{d x} \log \left(1+\frac{1}{x}\right)+\log \left(1+\frac{1}{x}\right) \frac{d}{d x}(x) \end{aligned}
$=x\left(\frac{1}{1+\frac{1}{x}}\right) \frac{d}{d x}\left(1+\frac{1}{x}\right)+\log \left(1+\frac{1}{x}\right)$
\begin{aligned} &=x \times \frac{x}{x+1}\left(\frac{-1}{x^{2}}\right)+\log \left(1+\frac{1}{x}\right) \\\\ &=\frac{x^{2}}{x+1} \times \frac{-1}{x^{2}}+\log \left(1+\frac{1}{x}\right) \end{aligned}
\begin{aligned} &\frac{1}{y} \frac{d y}{d x}=\frac{-1}{x+1}+\log \left(1+\frac{1}{x}\right) \\\\ &\frac{d y}{d x}=y\left[\frac{-1}{x+1}+\log \left(1+\frac{1}{x}\right)\right] \end{aligned}
$=\left(1+\frac{1}{x}\right)^{x}\left[\log \left(1+\frac{1}{x}\right)-\frac{1}{x+1}\right]$

Differentiation exercise Multiple choice question 7

$\frac{\log x}{(1+\log x)^{2}}$
Hint:
Differentiate the function w.r.t x
Given:
$x^{y}=e^{x-y}$
Solution:
$x^{y}=e^{x-y}$
Taking log on both sides
$\log x^{y}=\log \left(e^{x-y}\right)$
$y \log x=(x-y) \log e \quad\left[\because \log m^{n}=n \log m\right]$
\begin{aligned} &y \log x=(x-y) \cdot 1\quad\quad\quad[\because \log e=1] \\\\ &y \log x=x-y \end{aligned}
\begin{aligned} &x=y+y \log x \\\\ &x=y(1+\log x) \\\\ &y=\frac{x}{1+\log x} \end{aligned}
Differentiating y w.r.t x then
$\frac{d y}{d x}=\frac{d}{d x}\left(\frac{x}{1+\log x}\right)$
$=\frac{(1+\log x) \frac{d(x)}{d x}-x \frac{d}{d x}(1+\log x)}{(1+\log x)^{2}} \quad\left[\because \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}\right]$
$=\frac{(1+\log x) \cdot 1-x\left\{\frac{d(1)}{d x}+\frac{d(\log x)}{d x}\right\}}{(1+\log x)^{2}}\left[\because \frac{d}{d x}\{f(x)+g(x)\}=\frac{d}{d x} f(x)+\frac{d}{d x} g(x), \frac{d(x)}{d x}=1\right]$
$=\frac{1+\log x-x\left(0+\frac{1}{x}\right)}{(1+\log x)^{2}} \quad\left[\because \frac{d}{d x} \log x=\frac{1}{x}, \frac{d}{d x}(\operatorname{con} s \tan t)=0\right]$
$=\frac{1+\log x-x \cdot \frac{1}{x}}{(1+\log x)^{2}}=\frac{1+\log x-1}{(1+\log x)^{2}}$
$=\frac{\log x}{(1+\log x)^{2}}$

Differentiation exercise Multiple choice question 9

$|\sec \theta|$
Hint:
Differentiate x and y w.r.t $\theta$, then divide and solve
Given:
$x=a \cos ^{3} \theta, y=a \sin ^{3} \theta, \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=$
Solution:
\begin{aligned} &x=a \cos ^{3} \theta \\\\ &\frac{d x}{d \theta}=a \frac{d}{d \theta}\left(\cos ^{3} \theta\right) \\\\ &\frac{d x}{d \theta}=a(3) \cos ^{2} \frac{d}{d \theta}(\cos \theta) \end{aligned}
$\frac{d x}{d \theta}=-3 a \cos ^{2} \theta \sin \theta$ ....................(1)
\begin{aligned} &y=a \sin \theta \\\\ &\frac{d y}{d \theta}=a \frac{d}{d \theta}\left(\sin ^{3} \theta\right) \end{aligned}
$=3 a \sin ^{2} \theta \frac{d}{d \theta}(\sin \theta)$
$\frac{d y}{d \theta}=3 a \sin ^{2} \theta \cos \theta$ ......................(2)
Dividing (2) by (1) we get
$\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{3 a \sin ^{2} \theta \cos \theta}{-3 a \cos ^{2} \theta \sin \theta}$
\begin{aligned} &\frac{d y}{d x}=\frac{\sin \theta}{-\cos \theta}=-\tan \theta \\\\ &\sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=\sqrt{1+\tan ^{2} \theta}=\sqrt{\sec ^{2} \theta}=|\sec \theta| \end{aligned}

Differentiation exercise Multiple choice question 10

$\frac{-2}{1+x^{2}}$
Hint:
Differentiate y function w.r.t x
Given:
$y=\sin ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$
Solution:
Let
$y=\sin ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$
Differentiate y w.r.t x then
$\frac{d y}{d x}=\frac{d}{d x}\left(\sin ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\right)$
$=\frac{1}{\sqrt{1-\left(\frac{1-x^{2}}{1+x^{2}}\right)^{2}}}\left\{\frac{\left(1+x^{2}\right) \frac{d}{d x}\left(1-x^{2}\right)-\left(1-x^{2}\right) \frac{d}{d x}\left(1+x^{2}\right)}{\left(1+x^{2}\right)^{2}}\right\}$ $\left[\because \frac{d}{d x}\left(\sin ^{-1} x\right)=\frac{1}{\sqrt{1-x^{2}}}, \frac{d}{d x}\left(\frac{u}{v}\right)=v \frac{d u}{d x}-u \frac{d v}{d x}\right]$
$=\frac{1}{\sqrt{\frac{\left(1+x^{2}\right)^{2}-\left(1-x^{2}\right)^{2}}{\left(1+x^{2}\right)^{2}}}}\left[\frac{\left(1+x^{2}\right)\left\{\frac{d}{d x}(1)-\frac{d}{d x}\left(x^{2}\right)\right\}-\left(1-x^{2}\right)\left\{\frac{d}{d x}(1)+\frac{d}{d x}\left(x^{2}\right)\right\}}{\left(1+x^{2}\right)^{2}}\right]$ $\left[\because \frac{d}{d x}\{f(x)+g(x)\}=\frac{d}{d x} f(x)+\frac{d}{d x} g(x)\right]$
$=\frac{\sqrt{\left(1+x^{2}\right)^{2}}}{\sqrt{\left(1+x^{4}+2 x^{2}\right)-\left(1+x^{4}-2 x^{2}\right)}}\left[\frac{\left(1+x^{2}\right)(0-2 x)-\left(1-x^{2}\right)(0+2 x)}{\left(1+x^{2}\right)^{2}}\right]$
$\left[\frac{d}{d x}\left(x^{n}\right)=n x^{n-1},\left(a^{2}+b^{2}+2 a b\right)=(a+b)^{2} \&(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$
$=\frac{\left(1+x^{2}\right)}{\sqrt{1+x^{4}+2 x^{2}-1-x^{4}+2 x^{2}}}\left[\frac{-2 x-2 x^{3}-\left(2 x-2 x^{3}\right)}{\left(1+x^{2}\right)^{2}}\right]$
$=\frac{1+x^{2}}{\sqrt{4 x^{2}}}\left[\frac{-2 x-2 x^{3}-2 x+2 x^{3}}{\left(1+x^{2}\right)^{2}}\right]$
\begin{aligned} &\left.=\frac{1+x^{2}}{\sqrt{(2 x)^{2}}\left[\frac{-4 x}{\left(1+x^{2}\right)^{2}}\right.}\right]=\frac{\left(1+x^{2}\right)}{2 x} \times \frac{-4 x}{\left(1+x^{2}\right)^{2}}=\frac{-2}{\left(1+x^{2}\right)} \\\\ &\therefore \frac{d y}{d x}=\frac{-2}{1+x^{2}} \end{aligned}

Differentiation exercise Multiple choice question 11

Does not exist
Hint:
$\sec ^{-1} \alpha$is not defined
Given:
$\sec ^{-1}\left(\frac{1}{2 x^{2}+1}\right) \text { W.r.t } \sqrt{1+3 x} \text { at } x=-\frac{1}{3}$
Solution:
We know that $\sec ^{-1} \alpha$ is not defined for $\alpha \in(-1,1)$
Here for $x=-\frac{1}{3}, \frac{1}{2 x^{2}+1}=\frac{9}{11} \in(-1,1)$
$\therefore \sec ^{-1}\left(\frac{1}{2 x^{2}+1}\right)$ Is not defined at $x=-\frac{1}{3}$
Derivative of $\sec ^{-1}\left(\frac{1}{2 x^{2}+1}\right)$ does not exist at $x=-\frac{1}{3}$

Differentiation exercise Multiple choice question 12

$-1$
Hint:
Differentiate the function w.r.t x
Given:
$\sqrt{x}+\sqrt{y}=1, \frac{d y}{d x} a t\left(\frac{1}{4}, \frac{1}{4}\right)$
Solution:
$\sqrt{x}+\sqrt{y}=1$
Differentiating w.r.t x
\begin{aligned} &\frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{y}} \frac{d y}{d x}=0 \\\\ &\frac{1}{2 \sqrt{y}} \frac{d y}{d x}=\frac{-1}{2 \sqrt{x}} \end{aligned}
$\frac{d y}{d x}=-\frac{1}{2 \sqrt{x}} \times \frac{2 \sqrt{y}}{1}=\frac{-\sqrt{y}}{\sqrt{x}}$
Now,
$\left[\frac{d y}{d x}\right]_{\left(\frac{1}{4} \cdot \frac{1}{4}\right)}=\frac{-\sqrt{\frac{1}{4}}}{\sqrt{\frac{1}{4}}}=-1$

Differentiation exercise Multiple choice question 13

$-1$
Hint:
Differentiate the function w.r.t x
Given:
$\sin (x+y)=\log (x+y)$
Solution:
\begin{aligned} &\sin (x+y)=\log (x+y) \\\\ &\cos (x+y)\left(1+\frac{d y}{d x}\right)=\frac{1}{(x+y)}\left(1+\frac{d y}{d x}\right) \end{aligned}
\begin{aligned} &\cos (x+y)+\cos (x+y) \frac{d y}{d x}=\frac{1}{(x+y)}+\frac{1}{(x+y)} \frac{d y}{d x} \\\\ &\cos (x+y) \frac{d y}{d x}-\frac{1}{(x+y)} \frac{d y}{d x}=\frac{1}{(x+y)}-\cos (x+y) \end{aligned}
\begin{aligned} &{\left[\cos (x+y)-\frac{1}{(x+y)}\right] \frac{d y}{d x}=\frac{1}{(x+y)}-\cos (x+y)} \\\\ &-\left[\frac{1}{(x+y)}-\cos (x+y)\right] \frac{d y}{d x}=\frac{1}{(x+y)}-\cos (x+y) \\\\ &\frac{d y}{d x}=-1 \end{aligned}

Differentiation exercise Multiple choice question 14

1
Hint:
Differentiate the function w.r.t x
Given:
$u=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right), v=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$
Solution:
$u=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right), v=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$
Differentiating u w.r.t x then
$\frac{d y}{d x}=\frac{d}{d x}\left(\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\right)$
$=\frac{1}{\sqrt{1-\left(\frac{2 x}{1+x^{2}}\right)^{2}}}\left[\frac{\left(1+x^{2}\right) \frac{d}{d x}(2 x)-2 x \frac{d}{d x}\left(1+x^{2}\right)}{\left(1+x^{2}\right)^{2}}\right]$ $\left[\begin{array}{l} \frac{d}{d x}\left(\sin ^{-1} x\right)=\frac{1}{\sqrt{1-x^{2}}} \\\\ \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}} \end{array}\right]$
$=\frac{1}{\sqrt{\frac{\left(1+x^{2}\right)^{2}-(2 x)^{2}}{\left(1+x^{2}\right)^{2}}}}\left[\frac{\left(1+x^{2}\right) 2 \cdot \frac{d(x)}{d x}-2 x\left\{\frac{d}{d x}(1)+\frac{d}{d x}\left(x^{2}\right)\right\}}{\left(1+x^{2}\right)^{2}}\right]$ $\left[\begin{array}{l} \because \frac{d}{d x}\{f(x)+g(x)\}=\frac{d}{d x} f(x)+\frac{d}{d x} g(x) \\\\ \frac{d}{d x}\left(a x^{n}\right)=a \frac{d}{d x}\left(x^{n}\right) \end{array}\right]$
$=\frac{\sqrt{\left(1+x^{2}\right)^{2}}}{\sqrt{1+x^{4}+2 x^{2}-4 x^{2}}} \times\left[\frac{\left(1+x^{2}\right) 2 \cdot 1-2 x(0+2 x)}{\left(1+x^{2}\right)^{2}}\right]$ $\left[\because \frac{d}{d x}(\operatorname{cons} \tan t)=0, \frac{d}{d x}\left(x^{n}\right)=n x^{n-1}\right]$
$=\frac{\left(1+x^{2}\right)}{\sqrt{1+x^{2}-2 x^{2}}}\left[\frac{\left(1+x^{2}\right) 2-2 x(2 x)}{\left(1+x^{2}\right)^{2}}\right]$
$=\frac{\left(1+x^{2}\right)}{\sqrt{\left(1-x^{2}\right)^{2}}}\left[\frac{2+2 x^{2}-4 x^{2}}{\left(1+x^{2}\right)^{2}}\right] \quad\quad\quad\left[\because a^{2}+b^{2}-2 a b=(a-b)^{2}\right]$
\begin{aligned} &=\frac{\left(1+x^{2}\right)}{\left(1-x^{2}\right)} \times \frac{2-2 x^{2}}{\left(1+x^{2}\right)^{2}}=\frac{2\left(1-x^{2}\right)}{\left(1-x^{2}\right)\left(1+x^{2}\right)}=\frac{2}{1+x^{2}}\\\\ &\therefore \frac{d u}{d x}=\frac{2}{1+x^{2}} &\text { } \end{aligned}................(1)
Differentiating v w.r.t x then,
$\frac{d v}{d x}=\frac{d}{d x}\left(\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)\right)$
$=\frac{1}{1+\left(\frac{2 x}{1-x^{2}}\right)^{2}}\left[\frac{\left(1-x^{2}\right) \frac{d}{d x}(2 x)-\frac{d}{d x}\left(1-x^{2}\right) \cdot 2 x}{\left(1-x^{2}\right)^{2}}\right]$ $\left[\because \frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^{2}}, \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}\right]$
$=\frac{1}{\frac{\left(1-x^{2}\right)^{2}+(2 x)^{2}}{\left(1-x^{2}\right)^{2}}}\left[\frac{\left(1-x^{2}\right) 2 \frac{d}{d x}(x)-2 x\left\{\frac{d}{d x}(1)-\frac{d}{d x}\left(x^{2}\right)\right\}}{\left(1-x^{2}\right)^{2}}\right]$ $\left[\begin{array}{l} \frac{d}{d x}\{f(x)+g(x)\}=\frac{d}{d x} f(x)+\frac{d}{d x} g(x) \\\\ \frac{d}{d x}\left(a x^{n}\right)=a \frac{d}{d x}\left(x^{n}\right) \end{array}\right]$
$=\frac{\left(1-x^{2}\right)^{2}}{1+x^{4}-2 x^{2}+4 x^{2}}\left[\frac{\left(1-x^{2}\right) \cdot 2 \cdot 1-2 x(0-2 x)}{\left(1-x^{2}\right)^{2}}\right]$ $\left[\begin{array}{l} \because \frac{d}{d x}\left(x^{n}\right)=n x^{n-1}, \frac{d}{d x}(\operatorname{cons} \tan t)=0, \frac{d(x)}{d x}=1 \\\\ (a-b)^{2}=a^{2}+b^{2}-2 a b \end{array}\right]$
$=\frac{1}{1+x^{4}+2 x^{2}}\left[\frac{2-2 x^{2}+4 x^{2}}{1}\right]$
$=\frac{2+2 x^{2}}{\left(1+x^{2}\right)^{2}}=\frac{2\left(1+x^{2}\right)}{\left(1+x^{2}\right)^{2}} \quad\quad\quad\left[\because a^{2}+b^{2}+2 a b=(a+b)^{2}\right]$
$\therefore \frac{d v}{d x}=\frac{2}{1+x^{2}}$ ..............................(2)
Thus
$\frac{d y}{d x}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{d u}{d x} \times \frac{d x}{d v}$
$=\frac{2}{1+x^{2}} \times \frac{1+x^{2}}{2}$ [Using (1) and (2)]
$\therefore \frac{d y}{d x}=1$

Differentiation exercise Multiple choice question 15

$-\frac{1}{2}$
Hint:
Differentiate the function w.r.t x
Given:
$\frac{d}{d x}\left[\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)\right]$
Solution:
$u=\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)$
$=\tan ^{-1}\left(\frac{\cos ^{2}\left(\frac{x}{2}\right)-\sin ^{2}\left(\frac{x}{2}\right)}{\cos ^{2}\left(\frac{x}{2}\right)-\sin ^{2}\left(\frac{x}{2}\right)+2 \sin \frac{x}{2} \cos \frac{x}{2}}\right)$
$=\tan ^{-1} \frac{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^{2}}$
$=\tan ^{-1}\left[\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}\right]$
$u=\tan ^{-1}\left[\frac{\tan \frac{\pi}{4}-\tan \frac{x}{2}}{1+\tan \frac{\pi}{4} \times \tan \frac{x}{2}}\right]$
$=\tan ^{-1}\left[\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right]$
\begin{aligned} &u=\frac{\pi}{4}-\frac{x}{2} \\\\ &\frac{d u}{d x}=0-\frac{1}{2} \\\\ &\frac{d u}{d x}=-\frac{1}{2} \end{aligned}

Differentiation exercise Multiple choice question 16

$\frac{x^{2}-1}{x^{2}-4}$
Hint:
Differentiate the function w.r.t x
Given:
$\frac{d}{d x}\left[\log \left\{e^{x}\left(\frac{x-2}{x+2}\right)^{\frac{3}{4}}\right\}\right]$
Solution:
$y=\frac{d}{d x}\left[\log \left(e^{x}\left(\frac{x-2}{x+2}\right)^{2}\right)\right]$
$=\frac{d}{d x}\left[x \log e+\frac{3}{4} \log \left(\frac{x-2}{x+2}\right)\right]$
$y=\frac{d}{d x}\left[x+\frac{3}{4} \log \left(\frac{x-2}{x+2}\right)\right]$
$\frac{d y}{d x}=1+\frac{3}{4\left(\frac{x-2}{x+2}\right)} \times \frac{(x+2)-(x-2)}{(x+2)^{2}}$
\begin{aligned} &=1+\frac{3}{4} \frac{(x+2)}{(x-2)} \times \frac{x+2-x+2}{(x+2)^{2}} \\\\ &=1+\frac{3}{4} \frac{(x+2)}{(x-2)} \times \frac{4}{(x+2)} \end{aligned}
\begin{aligned} &=1+\frac{3}{\left(x^{2}-4\right)} \\\\ &\frac{d y}{d x}=\frac{x^{2}-4+3}{x^{2}-4}=\frac{x^{2}-1}{x^{2}-4} \end{aligned}

Differentiation exercise Multiple choice question 17

$\frac{\cos 2 x}{2 y-1}$
Hint:
Differentiate the function w.r.t x
Given:
$y=\sqrt{\sin x+y}$
Solution:
$y=\sqrt{\sin x+y}$
Squaring both side
\begin{aligned} &y^{2}=\sin x+y \\\\ &y^{2}-y=\sin x \\\\ &2 y \frac{d y}{d x}-\frac{d y}{d x}=\cos x \end{aligned}

Differentiation exercise Multiple choice question 18

$-\frac{y}{x}$
Hint:
Differentiate the function w.r.t x
Given:
$3 \sin (x y)+4 \cos (x y)=5$
Solution:
$3 \sin (x y)+4 \cos (x y)=5$
\begin{aligned} &3 \cos (x y)\left[x \frac{d y}{d x}+y\right]-4 \sin (x y)\left[x \frac{d y}{d x}+y\right]=0 \\\\ &{\left[x \frac{d y}{d x}+y\right][3 \cos (x y)-4 \sin (x y)]=0} \end{aligned}
\begin{aligned} &x \frac{d y}{d x}+y=0 \\\\ &x \frac{d y}{d x}=-y \\\\ &\frac{d y}{d x}=-\frac{y}{x} \end{aligned}

Differentiation exercise Multiple choice question 19

$\frac{\sin ^{2}(a+y)}{\sin a}$
Hint:
Differentiate the function w.r.t x
Given:
$\sin y=x \sin (a+y), \text { then } \frac{d y}{\alpha x}$
Solution:
\begin{aligned} &\sin y=x \sin (a+y) \\\\ &\frac{d}{d x}(\sin y)=\frac{d}{d x}[(a+y) \sin x] \end{aligned}
$\cos y \frac{d y}{d x}=\sin (a+y) \frac{d}{d x}(x)+x \frac{d}{d x}[\sin (a+y)]$
\begin{aligned} &=\sin (a+y) \times 1+x \cos (a+y) \frac{d y}{d x} \\ &=\sin (a+y)+x \cos (a+y) \frac{d y}{d x} \end{aligned}
$\cos y \frac{d y}{d x}=-x \cos (a+y) \frac{d y}{d x}=\sin (a+y)$
$\left(\cos y-\frac{\sin y}{\sin (a+y)} \times \cos (a+y)\right) \frac{d y}{d x}=\sin (a+y)$ $\left[\begin{array}{l} \sin y=2 \sin x \cos x \\\\ x=\frac{\sin y}{\sin (a+y)} \end{array}\right]$
$\left(\frac{\sin (a+y) \cos y-\sin y \cos (a+y)}{\sin (a+y)}\right) \frac{d y}{d x}=\sin (a+y)$
\begin{aligned} &\frac{\sin (a+y-y)}{\sin (a+y)} \times \frac{d y}{d x}=\sin (a+y) \\\\ &\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin a} \end{aligned}

Differentiation exercise Multiple choice question 20

2
Hint:
Differentiate the function w.r.t $\cos ^{-1}x$
Given:
$\cos ^{-1}(2x^{2}-1)$ With respect to $\cos ^{-1}x$
Solution:
\begin{aligned} &u=\cos ^{-1}\left(2 x^{2}-1\right) \\\\ &x=\cos \theta \\\\ &\theta=\cos ^{-1} x \\\\ &\frac{d \theta}{d x}=\frac{-1}{\sqrt{1-x^{2}}} \end{aligned}
\begin{aligned} &u=\cos ^{-1}(\cos 2 \theta) \\\\ &u=2 \theta \\\\ &\frac{d u}{d x}=2 \frac{d \theta}{d x} \\\\ &\frac{d u}{d x}=\frac{-2}{\sqrt{1-x^{2}}} \end{aligned} ...............(1)
\begin{aligned} &v=\cos ^{-1} x \\\\ &v=\cos ^{-1}(\cos \theta), v=\theta \\\\ &\frac{d v}{d x}=\frac{d \theta}{d x} \\\\ &\frac{d v}{d x}=\frac{-1}{\sqrt{1-x^{2}}} \end{aligned} ..................(2)
Divide (1) by (2)
\begin{aligned} &\frac{d u}{d x}=\frac{-2}{\frac{d v}{d x}}{\sqrt{1-x^{2}}} \times \frac{\sqrt{1-x^{2}}}{-1} \\\\ &\frac{d u}{d v}=2 \end{aligned}

Differentiation exercise Multiple choice question 21

$-1 \text { for } x<-3$
Hint:
Differentiate the function w.r.t x
Given:
$f(x)=\sqrt{x^{2}+6 x+9}, f^{\prime}(x)$
Solution:
\begin{aligned} &f(x)=\sqrt{x^{2}+6 x+9} \\\\ &x^{2}+6 x+9=(x+3)(x+3)=(x+3)^{2} \end{aligned}
\begin{aligned} &f(x)=\sqrt{(x+3)^{2}}=|x+3| \\\\ &f(x)=\left\{\begin{array}{cc} x+3 & x \geq-3 \\\\ -x-3 & x<-3 \end{array}\right. \end{aligned}
\begin{aligned} &f^{\prime}(x)=\left\{\begin{array}{cl} 1 & x \geq-3 \\\\ -1 & x<-3 \end{array}\right. \\\\ &f^{\prime}(x)=-1 \text { For } x<-3 \end{aligned}

Differentiation exercise Multiple choice question 22

$-2 x+9, \text { If } 4
Hint:
Differentiate the function w.r.t x
Given:
$f(x)=\left|x^{2}-9 x+20\right| \text { Then } f^{\prime}(x)=$
Solution:
$f(x)=\left|x^{2}-9 x+20\right|$
$f(x)=\left\{\begin{array}{cc} x^{2}-9 x+20 & -\infty
$f^{\prime}(x)= \begin{cases}2 x-9 & \infty
$f^{\prime}(x)=-2 x+9 \text { for } 4

Differentiation exercise Multiple choice question 23

None of these
Hint:
In case of function of one variable it’s a function that doesn’t have finite derivation
Given:
$f(x)=\sqrt{x^{2}-10 x+25}, f^{\prime}(x)=$
Solution:
$f(x)=\sqrt{x^{2}-10 x+25}$
\begin{aligned} &=\sqrt{(x-5)^{2}} \\\\ &=|x-5| \end{aligned}
$f(x)=\left\{\begin{array}{cll} x-5 & \text { for } & x>5 \\\\ -(x-5) & \text { for } & x<5 \end{array}\right.$
$L H D=\lim _{x \rightarrow 5^{-}} \frac{f(x)-f(a)}{x-a}$
$=\lim _{x \rightarrow 5^{-}} \frac{\sqrt{x^{2}-10 x+25}-\sqrt{5^{2}-10(5)+25}}{x-5}$
$=\lim _{x \rightarrow 5^{-}} \frac{|x-5|}{(x-5)}=\lim _{x \rightarrow 5^{-}} \frac{-(x-5)}{x-5}=-1$
$R H D=\lim _{x \rightarrow 5^{+}} \frac{f(x)-f(a)}{x-a}$
$=\lim _{x \rightarrow 5^{+}} \frac{\sqrt{x^{2}-10 x+25}-\sqrt{5^{2}-10(5)+25}}{x-5}$
\begin{aligned} &=\lim _{x \rightarrow 5^{+}} \frac{|x-5|}{x-5} \\\\ &=\lim _{x \rightarrow 5^{+}} \frac{x-5}{x-5} \\\\ &=1 \end{aligned}
$L H D \neq R H D,$ so function is not differentiable .

Differentiation exercise Multiple choice question 25

0
Hint:
Differentiate the function w.r.t x
Given:
$f(x)=\left(\frac{x^{l}}{x^{m}}\right)^{l+m}\left(\frac{x^{m}}{x^{n}}\right)^{m+n}\left(\frac{x^{n}}{x^{3}}\right)^{n+l}$
Solution:
$f(x)=\left(\frac{x^{l}}{x^{m}}\right)^{l+m}\left(\frac{x^{m}}{x^{n}}\right)^{m+n}\left(\frac{x^{n}}{x^{l}}\right)^{n+l}$
\begin{aligned} &=x^{(l-m)(l+m)} \times x^{(m-n)(m+n)} \times x^{(n-l)(n+l)} \\\\ &=x^{l^{2}-m^{2}} \times x^{m^{2}-n^{2}} \times x^{n^{2}-l^{2}} \end{aligned}
\begin{aligned} &f(x)=x^{\left(l^{2}-m^{2}+m^{2}-n^{2}+n^{2}-l^{2}\right)}=x^{0} \\\\ &f(x)=1 \\\\ &f^{\prime}(x)=0 \end{aligned}

Differentiation exercise Multiple choice question 26

0
Hint:
Differentiate the function w.r.t x
Given:
$y=\frac{1}{1+x^{a-b}+x^{c-b}}+\frac{1}{1+x^{b-c}+x^{a-c}}+\frac{1}{1+x^{b-a}+x^{c-a}}$
Solution:
$y=\frac{1}{1+x^{a-b}+x^{c-b}}+\frac{1}{1+x^{b-c}+x^{a-c}}+\frac{1}{1+x^{b-a}+x^{c-a}}$
$=\frac{1}{1+\frac{x^{a}}{x^{b}}+\frac{x^{c}}{x^{b}}}+\frac{1}{1+\frac{x^{b}}{x^{c}}+\frac{x^{a}}{x^{c}}}+\frac{1}{1+\frac{x^{b}}{x^{a}}+\frac{x^{c}}{x^{a}}}$
$=\frac{x^{b}}{x^{a}+x^{b}+x^{c}}+\frac{x^{c}}{x^{a}+x^{b}+x^{c}}+\frac{x^{a}}{x^{a}+x^{b}+x^{c}}$
\begin{aligned} &=\frac{x^{b}+x^{c}+x^{a}}{x^{a}+x^{b}+x^{c}} \\\\ &y=1 \\\\ &\frac{d y}{d x}=\frac{d(1)}{d x}=0 \end{aligned}

Differentiation exercise Multiple choice question 27

$\frac{x^{2}}{y^{2}} \sqrt{\frac{1-y^{6}}{1-x^{6}}}$
Hint:
Differentiate the function w.r.t x
Given:
$\sqrt{1-x^{6}}+\sqrt{1-y^{6}}=a^{3}\left(x^{3}-y^{3}\right)$
Solution:
We have, $\sqrt{1-x^{6}}+\sqrt{1-y^{6}}=a^{3}\left(x^{3}-y^{3}\right)$
Putting $x^{3}=\sin A, y^{3}=\sin B$
\begin{aligned} &\sqrt{1-\sin ^{2} A}+\sqrt{1-\sin ^{2} B}=a(\sin A-\sin B) \\\\ &\cos A+\cos B=a(\sin A-\sin B) \end{aligned}
\begin{aligned} &2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)=2 a \sin \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right) \\\\ &\cot \left(\frac{A-B}{2}\right)=a^{3} \end{aligned}
\begin{aligned} &\frac{A-B}{2}=\cot ^{-1}\left(a^{3}\right) \\\\ &A-B=2 \cot ^{-1}\left(a^{3}\right) \\\\ &\sin ^{-1} x^{3}-\sin ^{-1} y^{3}=2 \cot ^{-1}\left(a^{3}\right) \end{aligned}
\begin{aligned} &\frac{1}{\sqrt{1-x^{6}}} \times \frac{d}{d x}\left(x^{3}\right)-\frac{1}{\sqrt{1-y^{6}}} \times \frac{d}{d x}\left(y^{3}\right)=0 \\\\ &\frac{1}{\sqrt{1-x^{6}}} \times 3 x^{2}-\frac{1}{\sqrt{1-y^{6}}} \times 3 y^{2} \times \frac{d y}{d x}=0 \end{aligned}
$\frac{d y}{d x}=\frac{x^{2}}{y^{2}} \sqrt{\frac{1-y^{6}}{1-x^{6}}}$

Differentiation exercise Multiple choice question 28

1
Hint:
Differentiate the function w.r.t x
Given:
$y=\log \sqrt{\tan x}$
Solution:
\begin{aligned} &y=\log \sqrt{\tan x} \\\\ &\frac{d y}{d x}=\frac{1}{\sqrt{\tan x}} \times \frac{d}{d x}(\sqrt{\tan x}) \end{aligned}
$=\frac{1}{\sqrt{\tan x}} \times \frac{1}{2 \sqrt{\tan x}} \times \frac{d}{d x}(\tan x)$
$\frac{d y}{d x}=\frac{\sec ^{2} x}{2 \tan x}$
Now
$\left(\frac{d y}{d x}\right)_{x=\frac{\pi}{4}}=\frac{\left(\sec \frac{\pi}{4}\right)^{2}}{2 \tan \left(\frac{\pi}{4}\right)}=\frac{2}{2 \times 1}=1$

Differentiation exercise Multiple choice question 29

$\frac{y}{x}$
Hint:
Differentiate the function w.r.t x
Given:
$\sin ^{-1}\left(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\right)=\log a$
Solution:
$\sin ^{-1}\left(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\right)=\log a$
$\frac{x^{2}-y^{2}}{x^{2}+y^{2}}=\sin \log a$
$\frac{\left(x^{2}+y^{2}\right)\left(2 x-2 y \frac{d y}{d x}\right)-\left(x^{2}-y^{2}\right)\left(2 x+2 y \frac{d y}{d x}\right)}{\left(x^{2}+y^{2}\right)^{2}}=0$
$\frac{2 x^{3}-2 x^{2} y d y+2 x y^{2}-2 y^{3} \frac{d y}{d x}-2 x^{3}-2 x^{2} y \frac{d y}{d x}+2 x y^{2}+2 y^{3} \frac{d y}{d x}}{\left(x^{2}+y^{2}\right)^{2}}=0$
\begin{aligned} &-4 x^{2} y \frac{d y}{d x}+4 x y^{2}=0 \\\\ &-4 x^{2} y \frac{d y}{d x}=-4 x y^{2} \\\\ &\frac{d y}{d x}=\frac{4 x y^{2}}{4 x^{2} y}=\frac{y}{x} \end{aligned}

Differentiation exercise Multiple choice question 30

$\frac{\cos ^{2}(a+y)}{\cos a}$
Hint:
Differentiate the function w.r.t x
Given:
$\sin y=x \cos (a+y)$
Solution:
\begin{aligned} &\sin y=x \cos (a+y) \\\\ &\frac{d}{d x}(\sin y)=\frac{d}{d x}[x \cos (a+y)] \end{aligned}
\begin{aligned} &\cos y \frac{d y}{d x}=1 \times \cos (a+y)-x \sin (a+y) \frac{d}{d x}(a+y) \\\\ &\cos y \frac{d y}{d x}=\cos (a+y)-x \sin (a+y) \frac{d y}{d x} \end{aligned}
\begin{aligned} &\cos y \frac{d y}{d x}+x \sin (a+y) \frac{d y}{d x}=\cos (a+y) \\\\ &{[\cos y+x \sin (a+y)] \frac{d y}{d x}=\cos (a+y)} \end{aligned}
$\left[\cos y+\frac{\sin y}{\cos (a+y)} \times \sin (a+y)\right] \frac{d y}{d x}=\cos (a+y)$ $\left[\begin{array}{l} \sin y=x \cos (a+y) \\\\ x=\frac{\sin y}{\cos (a+y)} \end{array}\right]$
$\left[\frac{\cos (a+y) \cos y+\sin y \sin (a+y)}{\cos (a+y)}\right] \frac{d y}{d x}=\cos (a+y)$
\begin{aligned} &\frac{\cos (a+y-y)}{\cos (a+y)} \times \frac{d y}{d x}=\cos (a+y) \\\\ &\frac{d y}{d x}=\frac{\cos ^{2}(a+y)}{\cos a} \end{aligned}

Differentiation exercise Multiple choice question 31

$-\frac{4 x}{1-x^{4}}$
Hint:
Differentiate the function w.r.t x
Given:
$y=\log \left(\frac{1-x^{2}}{1+x^{2}}\right)$
Solution:
$y=\log \left(\frac{1-x^{2}}{1+x^{2}}\right)$
$\frac{d y}{d x}=\frac{1}{\frac{1-x^{2}}{1+x^{2}}} \frac{d}{d x}\left(\frac{1-x^{2}}{1+x^{2}}\right)$
$=\frac{1+x^{2}}{1-x^{2}}\left[\frac{\left(1+x^{2}\right)(-2 x)-\left(1-x^{2}\right)(2 x)}{\left(1+x^{2}\right)^{2}}\right]$
\begin{aligned} &=\frac{1}{1-x^{2}}\left[\frac{-2 x-2 x^{3}-2 x+2 x^{3}}{1+x^{4}}\right] \\\\ &\frac{d y}{d x}=\frac{-4 x}{1-x^{4}} \end{aligned}

Differentiation exercise Multiple choice question 32

1
Hint:
Differentiate the function w.r.t x
Given:
$y=\tan ^{-1}\left(\frac{\sin x+\cos x}{\cos x-\sin x}\right)$
Solution:
$y=\tan ^{-1}\left(\frac{\sin x+\cos x}{\cos x-\sin x}\right)$
$\frac{d y}{d x}=\frac{1}{1+\left(\frac{\sin x+\cos x}{\cos x-\sin x}\right)^{2}} \frac{d}{d x}\left(\frac{\sin x+\cos x}{\cos x-\sin x}\right)$
$\frac{d y}{d x}=\frac{(\cos x-\sin x)^{2}}{(\cos x-\sin x)^{2}+(\sin x+\cos x)^{2}} \times$ $\left[\frac{(\cos x-\sin x) \frac{d}{d x}(\sin x+\cos x)-(\sin x+\cos x) \frac{d}{d x}(\cos x-\sin x)}{(\cos x-\sin x)^{2}}\right]$
$=\frac{(\cos x-\sin x)^{2}}{(\cos x-\sin x)^{2}+(\sin x+\cos x)^{2}}\left[\frac{(\cos x-\sin x)(\cos x-\sin x)-(\sin x+\cos x)(-\sin x-\cos x)}{(\cos x-\sin x)^{2}}\right]$
$=\frac{(\cos x-\sin x)^{2}}{(\cos x-\sin x)^{2}+(\sin x+\cos x)^{2}}\left[\frac{(\cos x-\sin x)(\cos x-\sin x)+(\sin x+\cos x)(\sin x+\cos x)}{(\cos x-\sin x)^{2}}\right]$
\begin{aligned} &=\frac{(\cos x-\sin x)^{2}}{(\cos x-\sin x)^{2}(\sin x+\cos x)^{2}} \times \frac{(\cos x-\sin x)^{2}+(\sin x+\cos x)^{2}}{(\cos x-\sin x)^{2}} \\ &\frac{d y}{d x}=1 \end{aligned}

Differentiation exercise Multiple choice question 33

$\text { (c) } \frac{d y}{d x}=\frac{y-1}{x+1}$
Hint:
Differentiate the function w.r.t x
Given:
$\sec ^{-1}\left(\frac{1+x}{1-y}\right)$
Solution:
$\sec ^{-1}\left(\frac{1+x}{1-y}\right)=a$
\begin{aligned} &\sec a=\frac{1+x}{1-y} \\\\ &(1-y) \sec a=1+x \end{aligned}
Differentiate w.r.t x
$-\sec a \frac{d y}{d x}=1 \quad\quad\quad\quad\left[\frac{d(1)}{d x}=0, \frac{d(x)}{d x}=1\right]$
$\frac{d y}{d x}=\frac{-1}{\sec a}$
\begin{aligned} &=\frac{-(1-y)}{(1+x)} \\\\ &=\frac{y-1}{1+x} \end{aligned}
Hence, option (c) is correct.

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In RD Sharma class 12 solutions chapter 10 ex MCQ, there are 33 questions that include concepts from the entire Differentiations chapter.

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