RD Sharma Class 12 Exercise 10.2 Differentiation Solutions Maths-Download PDF Free Online

RD Sharma Class 12 Exercise 10.2 Differentiation Solutions Maths - Download PDF Free Online

Edited By Satyajeet Kumar | Updated on Jan 20, 2022 05:16 PM IST

RD Sharma Class 12th Exercise 10.2 is very important because RD Sharma has always been the best book for every student. RD Sharma Mathematics has set a standard where all the essential concepts and theorems are mentioned to become the teacher's favorite. Moreover, the questions in board exams and competitive exams have come from RD Sharma's book for many years.

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RD Sharma Class 12 Solutions Chapter 10 Differentiation - Other Exercise
Differentiation Excercise: 10.2
RD Sharma Chapter-wise Solutions

Rd Sharma Class 12th Exercise 10.2 has solved every problem of a student regarding differentiation. Differentiation Ex. 10.2 Solutions has 76 questions including subparts, that are formatted in a way that students will enjoy doing them. The exercise includes differentiating the functions w.r.t. to x, recapitulation of the product rule, quotient rule differentiation of the constant, differentiation of inverse trigonometric functions, logarithmic differentiation, etc.

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

RD Sharma Class 12 Solutions Chapter 10 Differentiation - Other Exercise

Differentiation Excercise: 10.2

Differentiation exercise 10.2 question 1

Answer: $3 \cos (3 x+5)$

Hint: You must know the values of solving derivative problems.
Given: $\sin (3 x+5)$
Solution: $\sin (3 x+5)$
Let : $y=\sin (3 x+5)$

Differentiating with respect to x,
$\frac{d y}{d x}=\frac{d}{d x}[\sin (3 x+5)]$
$\frac{d y}{d x}=\cos (3 x+5) \frac{d}{d x}(3 x+5)$ [using chain rule]
$\begin{aligned} &\frac{d y}{d x}=\cos (3 x+5) \times(3) \\ &\frac{d y}{d x}=3 \cos (3 x+5) \end{aligned}$

So,
$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}[\sin (3 x+5)] \\ &\frac{d y}{d x}=3 \cos (3 x+5) \end{aligned}$

Differentiation exercise 10.2 question 2

Answer: $2 \tan x \sec ^{2} x$
Hint: You must know the rules of solving derivative of trigonometric functions.
Given: $\tan ^{2} x$
Solution:
Let $y=\tan ^{2} x$
Differentiating with respect to x,
$\frac{d y}{d x}=2 \tan \mathrm{x} \frac{d}{d x}(\tan x)$
$\frac{d y}{d x}=2 \tan x \times \sec ^{2} x$
So, $\frac{d}{d x}\left(\tan ^{2} x\right)=2 \tan x \sec ^{2} x$ [ using chain rule]

Differentiation exercise 10.2 question 3

Answer: $\frac{\pi}{180} \sec ^{2}\left(x^{\circ}+45^{\circ}\right)$
Hint: You must know the rules of solving derivative of trigonometric function.
Given: $\tan \left(x^{\circ}+45^{\circ}\right)$
Solution:
$y=\tan \left(x^{\circ}+45^{\circ}\right)$

$y=\left[\tan (x+45) \cdot \frac{\pi}{180}\right]$ ...To convert degree into radian multiply by $\frac{\pi }{180}$
Differentiating with respect to x,
$\frac{d y}{d x}=\frac{d}{d x}\left[\tan (x+45) \cdot \frac{\pi}{180}\right]$
$\frac{d y}{d x}=\frac{\pi}{180} \cdot \sec ^{2}[\mathrm{x}+45] \times \frac{d}{d x}(x+45) \frac{\pi}{180}$ [ using chain rule]
$\frac{d y}{d x}=\frac{\pi}{180} \sec ^{2}\left(x^{\circ}+45^{\circ}\right)$
$\text { So, } \frac{d}{d x}\left[\tan \left(x^{\circ}+45^{\circ}\right)\right]=\frac{\pi}{180} \sec ^{2}\left(x^{\circ}+45^{\circ}\right)$ .

Differentiation exercise 10.2 question 4

Answer: $\frac{1}{x} \cos (\log x)$
Hint: You must know the rules of solving derivative of logarithm function.
Given: $\sin (\log x)$
Solution:
$y=\sin (\log x)$
Differentiating with respect to x,
$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x} \sin (\log x) \\ &\frac{d y}{d x}=\cos (\log x) \frac{d}{d x}(\log x) \end{aligned}$ [ using chain rule]
$\frac{d y}{d x}=\frac{1}{x} \cos (\log x)$

Differentiation exercise 10.2 question 5

Answer: $: \frac{\cos \sqrt{x} e^{\sin \sqrt{x}}}{2 \sqrt{x}}$
Hint: You must know the rules of solving derivation of exponential function and trigonometric function.
Given: $e^{\sin \sqrt{x}}$
Solution:
Let $y=e^{\sin \sqrt{x}}$
Differentiating with respect to x,
$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(e^{\sin \sqrt{x}}\right) \\ &\frac{d y}{d x}=e^{\sin \sqrt{x}} \frac{d}{d x}(\sin \sqrt{x}) \end{aligned}$ [using chain rule]
$\frac{d y}{d x}=e^{\sin \sqrt{x}} \times \cos \sqrt{x} \frac{d}{d x} \sqrt{x}$ [again using chain rule]
$\begin{aligned} &\frac{d y}{d x}=e^{\sin \sqrt{x}} \times \cos \sqrt{x} \times \frac{1}{2 \sqrt{x}} \\ &\frac{d y}{d x}=\frac{\cos \sqrt{x} e^{\sin \sqrt{x}}}{2 \sqrt{x}} \end{aligned}$

Differentiation exercise 10.2 question 6

Answer: $e^{\tan x} \times \sec ^{2} x$
Hint:You must know the rules of solving derivation of exponential function and trigonometric function.
Given: $e^{\tan x}$
Solution:
$y=e^{\tan x}$
Differentiating with respect to x,
$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x} e^{\tan x} \\ &\frac{d y}{d x}=e^{\tan x} \frac{d}{d x}(\tan x) \end{aligned}$ [ using chain rule] and $\frac{d}{d x}(\tan x)=\sec ^{2} x$
$\frac{d y}{d x}=e^{\tan x} \times \sec ^{2} x$

Differentiation exercise 10.2 question 7

Answer: $2 \sin (4 x+2)$
Hint: You must know the rules of solving derivation of trigonometric function.
Given: $\sin ^{2}(2 x+1)$
Solution:
Let $y=\sin ^{2}(2 x+1)$
Differentiating with respect to x,
$\frac{d y}{d x}=\frac{d}{d x}\left[\sin ^{2}(2 x+1)\right]$
$\frac{d y}{d x}=2 \sin (2 x+1) \frac{d}{d x} \sin (2 x+1)$ [ using chain rule ]
$\frac{d y}{d x}=2 \sin (2 x+1) \cos (2 x+1) \frac{d}{d x}(2 x+1)$ $\left[\frac{d}{d x} \sin x=\cos x\right]$
$\begin{aligned} &\frac{d y}{d x}=4 \sin (2 x+1) \cos (2 x+1) \\ &\frac{d y}{d x}=2 \sin (4 x+2) \end{aligned}$ $[\therefore \sin 2 A=2 \sin A \cos A]$
$\frac{d y}{d x}=2 \sin (4 x+2)$ .

Differentiation exercise 10.2 question 8

Answer: $\frac{2}{(2 x-3) \log 7}$
Hint: You must know the rules of solving derivative of logarithm function.
Given: $\log _{7}(2 x-3)$
Solution:
Let $y=\log _{7}(2 x-3)$
Differentiating with respect to x,
$\frac{d y}{d x}=\frac{d}{d x}\left[\log _{7}(2 x-3)\right]$
$\frac{d y}{d x}=\frac{d}{d x}\left[\frac{\log (2 x-3)}{\log 7}\right]$ $\left[\therefore \log _{a} b=\frac{\log b}{\log a}\right]$
$\begin{aligned} &\frac{d y}{d x}=\frac{1}{\log 7} \frac{d}{d x}[\log (2 x-3)] \\ &\frac{d y}{d x}=\frac{1}{\log 7} \times \frac{1}{(2 x-3)} \frac{d}{d x}(2 x-3) \end{aligned}$ [ using chain rule ]
$\frac{d y}{d x}=\frac{2}{(2 x-3) \log 7}$

Differentiation exercise 10.2 question 9

Answer: $\frac{5 \pi}{180} \sec ^{2}\left(5 x^{\circ}\right)$
Hint: You must know the rules of solving derivative of logarithm function.
Given: $\tan 5 x^{\circ}$
Solution:
Let $y=\tan 5 x^{\circ}$ or $y=\left(\tan 5 x \times \frac{\pi}{180}\right)$
Differentiating with respect to x
$\frac{d y}{d x}=\frac{d y}{d x} \tan \left(5 x \times \frac{\pi}{180}\right)$
$\frac{d y}{d x}=\sec ^{2}\left(5 x \times \frac{\pi}{180}\right) \frac{d}{d x}\left(5 x \times \frac{\pi}{180}\right)$ [ using chain rule ]
$\begin{aligned} &\frac{d y}{d x}=\left(\frac{5 \pi}{180}\right) \sec ^{2}\left(5 x \times \frac{\pi}{180}\right) \\ &\frac{d y}{d x}=\frac{5 \pi}{180} \sec ^{2}\left(5 x^{\circ}\right) \end{aligned}$

Differentiation exercise 10.2 question 10

Answer: $3 x^{2} \times 2^{x^{3}} \times \log _{e} 2$
Hint: You must know the rules of solving derivative of polynomial function.
Given: $2^{x^{8}}$
Solution:
Let $y=2^{x^{3}}$
Differentiating with respect to x
$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(2^{x^{3}}\right) \\ &\frac{d y}{d x}=2^{x^{3}} \times \log _{e} 2 \frac{d}{d x}\left(x^{3}\right) \end{aligned}$ $\frac{d}{d x} a^{x}=a^{x} \log a$ [using chain rule]
$\frac{d y}{d x}=3 x^{2} \times 2^{x^{3}} \times \log _{e} 2$

Differentiation exercise 10.2 question 11

Answer: $e^{x} \times 3^{e^{x}} \log (3)$
Hint: You must know the rules of solving derivative of exponential function.
Given: $3^{e^{x}}$
Solution:
Let $y=3^{e^{x}}$
Differentiating with respect to x
$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(3^{x^{x}}\right) \\ &\frac{d y}{d x}=3^{e^{x}} \log (3) \frac{d}{d x}\left(e^{x}\right) \end{aligned}$ $\frac{d}{d x} a^{x}=a^{x} \log a$ [ using chain rule]
$\frac{d y}{d x}=e^{x} \times 3^{e^{x}} \log (3)$

Differentiation exercise 10.2 question 12

Answer: $\frac{-1}{x \log 3\left(\log _{8} x\right)^{2}}$
Hint: You must know the rules of solving derivative of logarithm function
Given: $\log _{x} 3$
Solution:
Let $y=\log _{x} 3$
$y=\frac{\log 3}{\log x}$ $\left[\therefore \log _{a} b=\frac{\log b}{\log a}\right]$
Differentiating with respect to x
$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(\frac{\log 3}{\log x}\right) \\ &\frac{d y}{d x}=\log 3 \frac{d}{d x}(\log x)^{-1} \end{aligned}$ [ using chain rule]
$\begin{aligned} &\frac{d y}{d x}=\log 3 \times\left[-1(\log x)^{-2}\right] \frac{d}{d x}(\log x) \\ &\frac{d y}{d x}=\frac{-\log 3}{(\log x)^{2}} \times \frac{1}{x} \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}=-\left(\frac{\log 3}{\log x}\right)^{2} \times \frac{1}{x} \times \frac{1}{\log 3} \\ &\frac{d y}{d x}=\frac{-1}{x \log 3\left(\log _{3} x\right)^{2}} \end{aligned}$ $\left[\therefore \frac{\log b}{\log a}=\log _{a} b\right]$

Differentiation exercise 10.2 question 13

Answer: $(2 x+2) 3^{x^{2}+2 x} \log _{e} 3$

Hint: You must know the rules of solving derivative of polynomial function
Given: $3^{x^{2}+2 x}$
Solution:
Let $y=3^{x^{2}+2 x}$
Differentiating with respect to x
$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(3^{x^{2}+2 x}\right) \\ &\frac{d y}{d x}=3^{x^{2}+2 x} \times \log _{e} 3 \frac{d}{d x}\left(x^{2}+2 x\right) \end{aligned}$ $\frac{d}{d x} a^{x}=a^{x} \log a$ [using chain rule]
$\frac{d y}{d x}=(2 x+2) 3^{x^{2}+2 x} \log _{e} 3$

Differentiation exercise 10.2 question 14

Answer: $\frac{-2 x a^{2}}{\sqrt{a^{2}-x^{2}}\left(a^{2}+x^{2}\right)^{\frac{3}{2}}}$
Hint: You must know the rules of solving derivative of polynomial function
Given: $\sqrt{\frac{a^{2}-x^{2}}{a^{2}+x^{2}}}$
Solution:
Let $y=\sqrt{\frac{a^{2}-x^{2}}{a^{2}+x^{2}}}$
$y=\left(\frac{a^{2}-x^{2}}{a^{2}+x^{2}}\right)^{\frac{1}{2}}$
Differentiating with respect to x
$\frac{d y}{d x}=\frac{d}{d x}\left(\frac{a^{2}-x^{2}}{a^{2}+x^{2}}\right)^{\frac{1}{2}}$
$\frac{d y}{d x}=\frac{1}{2}\left(\frac{a^{2}-x^{2}}{a^{2}+x^{2}}\right)^{\frac{1}{2}-1} \times \frac{d}{d x}\left(\frac{a^{2}-x^{2}}{a^{2}+x^{2}}\right)$

$\frac{d y}{d x}=\frac{1}{2}\left(\frac{a^{2}-x^{2}}{a^{2}+x^{2}}\right)^{-\frac{1}{2}} \times\left\{\frac{\left(a^{2}+x^{2}\right) \frac{d}{d x}\left(a^{2}-x^{2}\right)-\left(a^{2}-x^{2}\right) \frac{d}{d x}\left(a^{2}+x^{2}\right)}{\left(a^{2}+x^{2}\right)^{2}}\right\}$ $..........\frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}$
$\frac{d y}{d x}=\frac{1}{2}\left(\frac{a^{2}+x^{2}}{a^{2}-x^{2}}\right)^{\frac{1}{2}}\left\{\frac{-2 x\left(a^{2}+x^{2}\right)-2 x\left(a^{2}-x^{2}\right)}{\left(a^{2}+x^{2}\right)^{2}}\right\}$
$\frac{d y}{d x}=\frac{1}{2}\left(\frac{a^{2}+x^{2}}{a^{2}-x^{2}}\right)^{\frac{1}{2}}\left\{\frac{-2 x a^{2}-2 x^{3}-2 x a^{2}+2 x^{3}}{\left(a^{2}+x^{2}\right)^{2}}\right\}$
$\frac{d y}{d x}=\frac{1}{2}\left(\frac{a^{2}+x^{2}}{a^{2}-x^{2}}\right)^{\frac{1}{2}}\left\{\frac{-4 x a^{2}}{\left(a^{2}+x^{2}\right)^{2}}\right\}$
$\frac{d y}{d x}=\frac{-2 x a^{2}}{\sqrt{a^{2}-x^{2}}\left(a^{2}+x^{2}\right)^{\frac{3}{2}}}$

Differentiation exercise 10.2 question 15

Answer: $3^{x \log ^{x}}(1+\log x) \times \log _{e} 3$
Hint: You must know the rules of solving derivative of logarithm and polynomial function.
Given: $3^{x \log x}$
Solution:
Let $y=3^{x \log x}$
Differentiating with respect to x
$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(3^{x \log x}\right) \\ &\frac{d y}{d x}=3^{x \log x} \times \log _{e} 3 \frac{d}{d x}(x \log x) \end{aligned}$
$\frac{d y}{d x}=3^{x \log x} \times \log _{e} 3\left[x \frac{d}{d x}(\log x)+\log x \frac{d}{d x}(x)\right]$
$\frac{d y}{d x}=3^{x \log x} \times \log _{e} 3\left[x \times \frac{1}{x}+\log x(1)\right]$
$\begin{aligned} &\frac{d y}{d x}=3^{x \log x} \times \log _{e} 3[1+\log x] \\ &\frac{d y}{d x}=3^{x \log x}(1+\log x) \times \log _{e} 3 \end{aligned}$

Differentiation exercise 10.2 question 16

Answer: $\sec x(\sec x+\tan x)$
Hint: You must know the rules of solving derivative of trigonometric function.
Given: $\sqrt{\frac{1+\sin x}{1-\sin x}}$
Solution:
Let $y=\sqrt{\frac{1+\sin x}{1-\sin x}}$
Differentiating with respect to x
$\frac{d y}{d x}=\frac{d}{d x}\left(\frac{1+\sin x}{1-\sin x}\right)^{\frac{1}{2}}$
$\frac{d y}{d x}=\frac{1}{2}\left(\frac{1+\sin x}{1-\sin x}\right)^{\frac{1}{2}-1} \times \frac{d}{d x}\left(\frac{1+\sin x}{1-\sin x}\right)$
$\frac{d y}{d x}=\frac{1}{2}\left(\frac{1-\sin x}{1+\sin x}\right)^{\frac{1}{2}} \times\left\{\frac{(1-\sin x)(\cos x)-(1+\sin x)(-\cos x)}{(1-\sin x)^{2}}\right\}$ $\ldots \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}$
$\frac{d y}{d x}=\frac{1}{2}\left(\frac{1-\sin x}{1+\sin x}\right)^{\frac{1}{2}}\left\{\frac{(\cos x)(1-\sin x)-(1+\sin x)(-\cos x)}{(1-\sin x)^{2}}\right\}$
$\frac{d y}{d x}=\frac{1}{2}\left(\frac{1-\sin x}{1+\sin x}\right)^{\frac{1}{2}}\left\{\frac{2 \cos x}{(1-\sin x)^{2}}\right\}$
$\frac{d y}{d x}=\frac{\cos x}{\sqrt{1+\sin x}(1-\sin x)^\frac{3}{2}}$
$\frac{d y}{d x}=\frac{\cos x}{\sqrt{1+\sin x} \sqrt{1-\sin x}(1-\sin x)}$
$\frac{d y}{d x}=\frac{\cos x}{\sqrt{1-\sin ^{2} x} \times(1-\sin x)}$
$\frac{d y}{d x}=\frac{\cos x}{\cos x(1-\sin x)}$
$\begin{aligned} &\frac{d y}{d x}=\frac{1}{(1-\sin x)} \times\left(\frac{1+\sin x}{1-\sin x}\right) \\ &\frac{d y}{d x}=\frac{(1+\sin x)}{\left(1-\sin ^{2} x\right)} \end{aligned}$
$\frac{d y}{d x}=\frac{(1+\sin x)}{\left(\cos ^{2} x\right)}$
$\frac{d y}{d x}=\frac{1}{(\cos x)}\left(\frac{1}{\cos x}+\frac{\sin x}{\cos x}\right)$
$\frac{d y}{d x}=\sec x(\sec x+\tan x)$

Differentiation exercise 10.2 question 17

Answer: $\frac{-2 x}{\sqrt{1-x^{2}}\left(1+x^{2}\right)^{\frac{3}{2}}}$
Hint: You must know the rules of solving derivative of polynomial function
Given: $\sqrt{\frac{1-x^{2}}{1+x^{2}}}$
Solution:
Let $y=\sqrt{\frac{1-x^{2}}{1+x^{2}}}$
$y=\left(\frac{1-x^{2}}{1+x^{2}}\right)^{\frac{1}{2}}$
Differentiating with respect to x
$\frac{d y}{d x}=\frac{d}{d x}\left(\frac{1-x^{2}}{1+x^{2}}\right)^{\frac{1}{2}}$
$\frac{d y}{d x}=\frac{1}{2}\left(\frac{1-x^{2}}{1+x^{2}}\right)^{\frac{1}{2}-1} \times \frac{d}{d x}\left(\frac{1-x^{2}}{1+x^{2}}\right)$ [ using chain rule]
$\frac{d y}{d x}=\frac{1}{2}\left(\frac{1-x^{2}}{1+x^{2}}\right)^{-\frac{1}{2}} \times\left\{\frac{\left(1+x^{2}\right) \frac{d}{d x}\left(1-x^{2}\right)-\left(1-x^{2}\right) \frac{d}{d x}\left(1+x^{2}\right)}{\left(1+x^{2}\right)^{2}}\right\}$ $\cdot \cdot \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}$
$\frac{d y}{d x}=\frac{1}{2}\left(\frac{1+x^{2}}{1-x^{2}}\right)^{\frac{1}{2}} \times\left\{\frac{-2 x\left(1+x^{2}\right)-2 x\left(1-x^{2}\right)}{\left(1+x^{2}\right)^{2}}\right\}$
$\frac{d y}{d x}=\frac{1}{2}\left(\frac{1+x^{2}}{1-x^{2}}\right)^{\frac{1}{2}} \times\left\{\frac{-2 x-2 x^{3}-2 x+2 x^{3}}{\left(1+x^{2}\right)^{2}}\right\}$
$\frac{d y}{d x}=\frac{1}{2}\left(\frac{1+x^{2}}{1-x^{2}}\right)^{\frac{1}{2}} \times\left\{\frac{-4 x}{\left(1+x^{2}\right)^{2}}\right\}$
$\frac{d y}{d x}=\frac{-2 x}{\sqrt{1-x^{2}}\left(1+x^{2}\right)^{\frac{3}{2}}}$

Differentiation exercise 10.2 question 18

Answer: $2(\log \sin x) \cot x$
Hint: You must know the value of solving logarithm and trigonometric function.
Given: $(\log \sin x)^{2}$
Solution:
Let $y=(\log \sin x)^{2}$
Differentiating with respect to x
$\frac{d y}{d x}=\frac{d}{d x}(\log \sin x)^{2}$
$\frac{d y}{d x}=2(\log \sin x) \frac{d}{d x}(\log \sin x)$
$\frac{d y}{d x}=2(\log \sin x) \times \frac{1}{\sin x} \frac{d}{d x}(\sin x)$
$\begin{aligned} &\frac{d y}{d x}=2(\log \sin x) \times \frac{1}{\sin x} \times \cos x \\ &\frac{d y}{d x}=2(\log \sin x) \times \cot x \end{aligned}$

Differentiation exercise 10.2 question 19

Answer: $\frac{1}{\sqrt{1+x}(1-x)^{\frac{3}{2}}}$
Hint: You must know the rule of solving derivative of polynomial function.
Given: $\sqrt{\frac{1+x}{1-x}}$
Solution:
Let $y=\sqrt{\frac{1+x}{1-x}}$
$y=\left(\frac{1+x}{1-x}\right)^{\frac{1}{2}}$
Differentiating with respect to x
$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(\frac{1+x}{1-x}\right)^{\frac{1}{2}} \\ &\frac{d y}{d x}=\frac{1}{2}\left(\frac{1+x}{1-x}\right)^{\frac{1}{2}-1} \times \frac{d}{d x}\left(\frac{1+x}{1-x}\right) \end{aligned}$ [ using chain rule]
$\frac{d y}{d x}=\frac{1}{2}\left(\frac{1+x}{1-x}\right)^{-\frac{1}{2}} \times\left\{\frac{(1-x) \frac{d}{d x}(1+x)-(1+x) \frac{d}{d x}(1-x)}{(1-x)^{2}}\right\}$ $...\frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}$
$\frac{d y}{d x}=\frac{1}{2}\left(\frac{1-x}{1+x}\right)^{\frac{1}{2}} \times\left\{\frac{(1-x)(1)-(1+x)(-1)}{(1-x)^{2}}\right\}$
$\frac{d y}{d x}=\frac{1}{2}\left(\frac{1-x}{1+x}\right)^{\frac{1}{2}} \times\left\{\frac{1-x+1+x}{(1-x)^{2}}\right\}$
$\begin{aligned} &\frac{d y}{d x}=\frac{1}{2}\left(\frac{1-x}{1+x}\right)^{\frac{1}{2}} \times\left\{\frac{2}{(1-x)^{2}}\right\} \\ &\frac{d y}{d x}=\frac{1}{\sqrt{1+x}(1-x)^{\frac{3}{2}}} \end{aligned}$

Differentiation exercise 10.2 question 20

Answer: $\frac{4 x}{\left(1-x^{2}\right)^{2}} \cos \left(\frac{1+x^{2}}{1-x^{2}}\right)$
Hint: You must know the rules of solving derivative of trigonometric functions.
Given: $\sin \left(\frac{1+x^{2}}{1-x^{2}}\right)$
Solution:
Let $y=\sin \left(\frac{1+x^{2}}{1-x^{2}}\right)$
Differentiating with respect to x
$\frac{d y}{d x}=\frac{d}{d x}\left[\sin \left(\frac{1+x^{2}}{1-x^{2}}\right)\right]$
$\frac{d y}{d x}=\cos \left(\frac{1+x^{2}}{1-x^{2}}\right) \frac{d}{d x}\left(\frac{1+x^{2}}{1-x^{2}}\right)$ [ using chain rule]
$\frac{d y}{d x}=\cos \left(\frac{1+x^{2}}{1-x^{2}}\right)\left[\frac{\left(1-x^{2}\right) \frac{d}{d x}\left(1+x^{2}\right)-\left(1+x^{2}\right) \frac{d}{d x}\left(1-x^{2}\right)}{\left(1-x^{2}\right)^{2}}\right]$ $...\frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}$
$\frac{d y}{d x}=\cos \left(\frac{1+x^{2}}{1-x^{2}}\right)\left[\frac{\left(1-x^{2}\right)(2 x)-\left(1+x^{2}\right)(-2 x)}{\left(1-x^{2}\right)^{2}}\right]$
$\frac{d y}{d x}=\cos \left(\frac{1+x^{2}}{1-x^{2}}\right)\left[\frac{2 x-2 x^{3}+2 x+2 x^{3}}{\left(1-x^{2}\right)^{2}}\right]$
$\frac{d y}{d x}=\frac{4 x}{\left(1-x^{2}\right)^{2}} \cos \left(\frac{1+x^{2}}{1-x^{2}}\right)$

Differentiation exercise 10.2 question 21

Answer: $e^{3 x} \cos 2 x$
Hint: You must know the rules of solving exponential and trigonometric functions.
Given: $e^{3 x} \cos 2 x$
Solution:
Let $y=e^{3 x} \cos 2 x$
Differentiating with respect to x
$\frac{d y}{d x}=\frac{d}{d x} e^{3 x} \cos 2 x$
$\frac{d y}{d x}=e^{3 x} \times \frac{d}{d x}(\cos 2 x)+\cos 2 x \frac{d}{d x}\left(e^{3 x}\right)$
$\frac{d y}{d x}=e^{3 x} \times(-\sin 2 x) \frac{d}{d x}(2 x)+\cos 2 x e^{3 x} \frac{d}{d x}(3 x)$
$\begin{aligned} &\frac{d y}{d x}=-2 e^{3 x} \sin 2 x+3 e^{3 x} \cos 2 x \\ &\frac{d y}{d x}=e^{3 x}[3 \cos 2 x-2 \sin 2 x] \end{aligned}$

Differentiation exercise 10.2 question 22

Answer: $\cos (\log \sin x) \cdot \cot x$
Hint: You must know the rules of solving derivative of trigonometric and logarithm function.
Given: $\sin (\log \sin x)$
Solution:
Let $y=\sin (\log \sin x)$
Differentiating with respect to x
$\frac{d y}{d x}=\frac{d}{d x} \sin (\log \sin x)$
Using chain rule
$\begin{aligned} &\frac{d y}{d x}=\cos (\log \sin x) \frac{d}{d x}(\log \sin x) \\ &\frac{d y}{d x}=\cos (\log \sin x) \cdot \frac{1}{\sin x} \frac{d}{d x}(\sin x) \end{aligned}$
$\frac{d y}{d x}=\cos (\log \sin x) \cdot \frac{1}{\sin x} \cdot \cos x$
$\frac{d y}{d x}=\cos (\log \sin x) \cdot \cot x$

Differentiation exercise 10.2 question 23

Answer: $3 e^{\tan 3 x} \sec ^{2} 3 x$
Hint: You must know the rules of solving derivative of exponential and trigonometric
Given: $e^{\tan 3 x}$
Solution:
Let $y=e^{\tan 3 x}$
Differentiating with respect to x
$\frac{d y}{d x}=\frac{d}{d x}\left(e^{\tan 3 x}\right)$
$\frac{d y}{d x}=e^{\tan 3 x} \frac{d}{d x}(\tan 3 x)$ $\left[\therefore \frac{d}{d x} e^{x}=e^{x}\right] e^{a x}=e^{x} \frac{d}{d x}(a)$
$\frac{d y}{d x}=e^{\tan 3 x} \sec ^{2} 3 x \times \frac{d}{d x}(3 x)$ $\left[\therefore \frac{d}{d x} \tan a x=s \sec ^{2} a x\right]$
$\begin{aligned} &\frac{d y}{d x}=e^{\tan 3 x} \cdot \sec ^{2} 3 x \times 3 \\ &\frac{d y}{d x}=3 e^{\tan 3 x} \sec ^{2} 3 x \end{aligned}$ $\left[\therefore \frac{d}{d x} \tan a x=a \sec ^{2} a x\right]$

Differentiation exercise 10.2 question 24

Answer: $\frac{e^{\sqrt{\cot x}} \times(\operatorname{cosec})^{2} x}{2 \sqrt{\cot x}}$
Hint: You must know the rules of solving derivative of trigonometric and exponential function.
Given: $e^{\sqrt{\cot x}}$
Solution:
Let $y=e^{\sqrt{\cot x}}$
$y=e^{(\cot x)^{\frac{1}{2}}}$
Differentiate both sides
$\frac{d y}{d x}=\frac{d}{d x} e^{(\cot x)^{\frac{1}{2}}}$
Using Chain Rule,
$\frac{d y}{d x}=e^{(\cot x)^{\frac{1}{2}}} \times \frac{1}{2}(\cot x)^{\frac{1}{2}-1} \frac{d}{d x}(\cot x)$
$\frac{d y}{d x}=\frac{e^{\sqrt{\cot x}} \times(\operatorname{cosec})^{2} x}{2 \sqrt{\cot x}}$

Differentiation exercise 10.2 question 25

Answer: $\operatorname{cosec} x$

Hint: You must know the rules of solving derivative of logarithm trigonometric function.
Given: $\log \left(\frac{\sin x}{1+\cos x}\right)$
Solution:
Let $y=\log \left(\frac{\sin x}{1+\cos x}\right)$
Differentiating with respect to x
$\frac{d y}{d x}=\frac{d}{d x} \log \left(\frac{\sin x}{1+\cos x}\right)$
$\frac{d y}{d x}=\frac{1}{\left(\frac{\sin x}{1+\cos x}\right)} \cdot\left[\frac{(1+\cos x) \frac{d}{d x} \sin x-\sin x \frac{d}{d x}(1+\cos x)}{(1+\cos x)^{2}}\right]$ $\cdot \cdot \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}$
$\frac{d y}{d x}=\frac{1+\cos x}{\sin x} \cdot\left[\frac{(1+\cos x)(\cos x)-\sin x(-\sin x)}{(1+\cos x)^{2}}\right]$
$\frac{d y}{d x}=\left(\frac{1+\cos x}{\sin x}\right) \cdot\left[\frac{\cos x+\cos ^{2} x+\sin ^{2} x}{(1+\cos x)^{2}}\right]$
$\frac{d y}{d x}=\left(\frac{1+\cos x}{\sin x}\right)\left(\frac{\cos x+1}{(1+\cos x)^{2}}\right)$
$\begin{aligned} &\frac{d y}{d x}=\frac{(1+\cos x)^{2}}{\sin x(1+\cos x)^{2}} \\ &\frac{d y}{d x}=\frac{1}{\sin x} \end{aligned}$
$\frac{d y}{d x}=\operatorname{cosec} x$

Differentiation exercise 10.2 question 26

Answer: $\operatorname{cosec} x$
Hint: You must know the rules of solving derivative of logarithm and trigonometric function.
Given: $\log \sqrt{\frac{1-\cos x}{1+\cos x}}$
Solution:
Let $y=\log \sqrt{\frac{1-\cos x}{1+\cos x}}$
$y=\frac{1}{2} \log \left(\frac{1-\cos x}{1+\cos x}\right)$ using $\log a^{b}=b \log a$

Differentiate with respect to x
$\frac{d y}{d x}=\frac{d}{d x}\left\{\frac{1}{2} \log \left(\frac{1-\cos x}{1+\cos x}\right)\right\}$
$\frac{d y}{d x}=\frac{1}{2} \times \frac{1}{\left(\frac{1-\cos x}{1+\cos x}\right)} \times \frac{d}{d x}\left(\frac{1-\cos x}{1+\cos x}\right)$
$\frac{d y}{d x}=\frac{1}{2} \times\left(\frac{1+\cos x}{1-\cos x}\right)\left[\frac{(1+\cos x) \frac{d}{d x}(1-\cos x)-(1-\cos x) \frac{d}{d x}(1+\cos x)}{(1+\cos x)^{2}}\right]$ $\cdot \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}$

Using quotient rule
$\frac{d y}{d x}=\frac{1}{2}\left(\frac{1+\cos x}{1-\cos x}\right)\left[\frac{(1+\cos x)(\sin x)-(1-\cos x)(-\sin x)}{(1+\cos x)^{2}}\right]$
$\frac{d y}{d x}=\frac{1}{2}\left(\frac{1+\cos x}{1-\cos x}\right)\left[\frac{2 \sin x}{(1+\cos x)^{2}}\right]$
$\frac{d y}{d x}=\frac{\sin x}{(1-\cos x)(1+\cos x)}$
$\begin{aligned} &\frac{d y}{d x}=\frac{\sin x}{\left(1-\cos ^{2} x\right)} \\ &\frac{d y}{d x}=\frac{\sin x}{\sin ^{2} x} \end{aligned}$ $\begin{aligned} &\frac{d y}{d x}=\frac{1}{\sin x} \\ &\frac{d y}{d x}=\operatorname{cosec} x \end{aligned}$

Differentiation exercise 10.2 question 27

Answer: $\cos x \sec ^{2}\left(e^{\sin x}\right) e^{\sin x}$
Hint: You must know the rules of solving derivative of exponential and trigonometric function.
Given: $\tan \left(e^{\sin x}\right)$
Solution:
Let $y=\tan \left(e^{\sin x}\right)$
Differentiate with respect to x
$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left[\tan \left(e^{\sin x}\right)\right] \\ &\frac{d y}{d x}=\sec ^{2}\left(e^{\sin x}\right) \frac{d}{d x}\left(e^{\sin x}\right) \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}=\sec ^{2}\left(e^{\sin x}\right) \times e^{\sin x} \frac{d}{d x}(\sin x) \\ &\frac{d y}{d x}=\cos x \cdot \sec ^{2}\left(e^{\sin x}\right) \cdot e^{\sin x} \end{aligned}$

Differentiation exercise 10.2 question 28

Answer: $\frac{1}{\sqrt{x^{2}+1}}$
Hint: You must know the rules of solving derivative of logarithm function.
Given: $\log \left(x+\sqrt{x^{2}+1}\right)$
Solution:
Differentiate with respect to x
$\frac{d y}{d x}=\frac{d}{d x}\left[\log \left(x+\sqrt{x^{2}+1}\right)\right]$
$\frac{d y}{d x}=\frac{1}{x+\sqrt{x^{2}+1}} \frac{d}{d x}\left(x+\left(x^{2}+1\right)^{\frac{1}{2}}\right)$
$\frac{d y}{d x}=\frac{1}{x+\sqrt{x^{2}+1}}\left[1+\frac{1}{2}\left(x^{2}+1\right)^{\frac{1}{2}-1} \frac{d}{d x}\left(x^{2}+1\right)\right]$
$\frac{d y}{d x}=\frac{1}{x+\sqrt{x^{2}+1}}\left[1+\frac{1}{2 \sqrt{x^{2}+1}} \times 2 x\right]$
$\begin{aligned} &\frac{d y}{d x}=\frac{1}{x+\sqrt{x^{2}+1}}\left[\frac{\sqrt{x^{2}+1}+x}{\sqrt{x^{2}+1}}\right] \\ &\frac{d y}{d x}=\frac{1}{\sqrt{x^{2}+1}} \end{aligned}$

Differentiation exercise 10.2 question 29

Answer: $e^{x} x^{-2}\left[\frac{1}{x}+\log x-\frac{2}{x} \log x\right]$
Hint: You must know about the rules of solving derivative of exponential and logarithm functions.
Given: $\frac{e^{x} \log x}{x^{2}}$
Solution:
Let $y=\frac{e^{x} \log x}{x^{2}}$
Differentiate with respect to x,
$\frac{d y}{d x}=\frac{x^{2} \frac{d}{d x}\left(e^{x} \log x\right)-\left(e^{x} \log x\right) \frac{d}{d x} x^{2}}{\left(x^{2}\right)^{2}} \cdot \cdot \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}$ [using quotient rule]
$\frac{d y}{d x}=\frac{x^{2}\left\{e^{x} \frac{d}{d x}(\log x)+(\log x) \frac{d}{d x}\left(e^{x}\right)\right\}-e^{x} \log x \times 2 x}{x^{4}}$
$\frac{d y}{d x}=\frac{x^{2}\left\{\frac{e^{x}}{x}+e^{x}(\log x)\right\}-2 x e^{x} \log x}{x^{4}}$
$\frac{d y}{d x}=\frac{x^{2} e^{x}\left\{\frac{(1+x \log x)}{x}-2 x e^{x} \log x\right\}}{x^{4}}$
$\frac{d y}{d x}=\frac{x e^{x}\{1+x \log x-2 \log x\}}{x^{4}}$
$\frac{d y}{d x}=\frac{x e^{x}}{x^{3}}\left[\frac{1}{x}+\frac{x \log x}{x}-\frac{2 \log x}{x}\right]$
$\frac{d y}{d x}=e^{x} x^{-2}\left[\frac{1}{x}+\log x-\frac{2}{x} \log x\right]$

Differentiation exercise 10.2 question 30

Answer: $\operatorname{cosec} x$
Hint: You must know the rules of solving derivation of logarithm and trigonometric function.
Given: $\log (\operatorname{cosec} x-\cot x)$
Solution:
Let $y=\log (\operatorname{cosec} x-\cot x)$

Differentiate both sides,

$\frac{d y}{d x}=\frac{d}{d x} \log (\operatorname{cosec} x-\cot x)$

$\frac{d y}{d x}=\frac{1}{(\operatorname{cosec} x-\cot x)} \frac{d}{d x}(\operatorname{cosec} x-\cot x)$

$\frac{d y}{d x}=\frac{1}{(\operatorname{cosec} x-\cot x)} \times\left(-\operatorname{cosec} x \cot x+\operatorname{cosec}^{2} x\right)$

$\begin{aligned} &\frac{d y}{d x}=\frac{\operatorname{cosec} x(\operatorname{cosec} x-\cot x)}{(\operatorname{cosec} x-\cot x)} \\ &\frac{d y}{d x}=\operatorname{cosec} x \end{aligned}$

Differentiation exercise 10.2 question 31

Answer: $\frac{-8}{\left(e^{2 x}-e^{-2 x}\right)^{2}}$
Hint: You must know the rules of solving derivative of exponential function.
Given: $\frac{e^{2 x}+e^{-2 x}}{e^{2 x}-e^{-2 x}}$
Solution:
Let $y=\frac{e^{2 x}+e^{-2 x}}{e^{2 x}-e^{-2 x}}$

Differentiate with respect to x,

$\frac{d y}{d x}=\frac{d}{d x}\left[\frac{e^{2 x}+e^{-2 x}}{e^{2 x}-e^{-2 x}}\right]$

$\frac{d y}{d x}=\left\{\frac{\left(e^{2 x}-e^{-2 x}\right) \frac{d}{d x}\left(e^{2 x}+e^{-2 x}\right)-\left(e^{2 x}+e^{-2 x}\right) \frac{d}{d x}\left(e^{2 x}-e^{-2 x}\right)}{\left(e^{2 x}+e^{-2 x}\right)^{2}}\right\}$ $\cdot \cdot \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}$

Using quotient rule,
$\frac{d y}{d x}=\frac{\left(e^{2 x}-e^{-2 x}\right)\left(2 e^{2 x}-2 e^{-2 x}\right)-\left(e^{2 x}+e^{-2 x}\right)\left(2 e^{2 x}+2 e^{-2 x}\right)}{\left(e^{2 x}+e^{-2 x}\right)^{2}}$
$\frac{d y}{d x}=\frac{2\left(e^{4 x}+e^{-4 x}-2 e^{2 x} e^{-2 x}-e^{4 x}-e^{-4 x}-2 e^{2 x} e^{-2 x}\right)}{\left(e^{2 x}+e^{-2 x}\right)^{2}}$
$\frac{d y}{d x}=\frac{-8}{\left(e^{2 x}+e^{-2 x}\right)^{2}}$

Differentiation exercise 10.2 question 32

Answer: $\frac{-2\left(x^{2}-1\right)}{x^{4}+x^{2}+1}$
Hint: You must know about the rules of solving derivative of logarithm functions.
Given: $\log \left(\frac{x^{2}+x+1}{x^{2}-x+1}\right)$
Solution:
Let $y=\log \left(\frac{x^{2}+x+1}{x^{2}-x+1}\right)$

Differentiate with respect to x,
$\frac{d y}{d x}=\frac{d}{d x}\left[\log \left(\frac{x^{2}+x+1}{x^{2}-x+1}\right)\right]$
$\frac{d y}{d x}=\frac{1}{\left(\frac{x^{2}+x+1}{x^{2}-x+1}\right)} \frac{d}{d x}\left(\frac{x^{2}+x+1}{x^{2}-x+1}\right)$

Now apply quotient rule, $\frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}$
$\frac{d y}{d x}=\left(\frac{x^{2}-x+1}{x^{2}+x+1}\right)\left[\frac{\left(x^{2}-x+1\right) \frac{d}{d x}\left(x^{2}+x+1\right)-\left(x^{2}+x+1\right) \frac{d}{d x}\left(x^{2}-x+1\right)}{\left(x^{2}-x+1\right)^{2}}\right]$
$\frac{d y}{d x}=\left(\frac{x^{2}-x+1}{x^{2}+x+1}\right)\left[\frac{\left(x^{2}-x+1\right)(2 x+1)-\left(x^{2}+x+1\right)(2 x-1)}{\left(x^{2}-x+1\right)^{2}}\right]$
$\frac{d y}{d x}=\left(\frac{x^{2}-x+1}{x^{2}+x+1}\right)\left[\frac{\left(2 x^{3}-2 x^{2}+2 x+x^{2}-x+1-2 x^{3}-2 x^{2}-2 x+x^{2}+x+1\right)}{\left(x^{2}-x+1\right)^{2}}\right]$
$\frac{d y}{d x}=\frac{-4 x^{2}+2 x^{2}+2}{\left(x^{2}+x+1\right)\left(x^{2}-x+1\right)}$
$\frac{d y}{d x}=\frac{-4 x^{2}+2 x^{2}+2}{\left(x^{2}+1\right)^{2}-(x)^{2}}$
$\begin{aligned} &\frac{d y}{d x}=\frac{-2\left(x^{2}-1\right)}{x^{4}+1+2 x^{2}-x^{2}} \\ &\frac{d y}{d x}=\frac{-2\left(x^{2}-1\right)}{x^{4}+x^{2}+1} \end{aligned}$

Differentiation exercise 10.2 question 33

Answer: $\frac{e^{x}}{1+e^{2 x}}$
Hint: You must know about the rules of solving derivative of Inverse trigonometric function and exponential
Given: $\tan ^{-1}\left(e^{x}\right)$
Solution:
Let $y=\tan ^{-1}\left(e^{x}\right)$
Differentiate with respect to x,
$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left[\tan ^{-1}\left(e^{x}\right)\right] \\ \\&\frac{d y}{d x}=\frac{1}{1+\left(e^{x}\right)^{2}} \frac{d}{d x}\left(e^{x}\right) \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}=\frac{1}{1+e^{2 x}} \times e^{x} \\\\ &\frac{d y}{d x}=\frac{e^{x}}{1+e^{2 x}} \end{aligned}$

Differentiation exercise 10.2 question 34

Answer: $\frac{2 e^{\sin ^{-1} 2 x}}{\sqrt{1-4 x^{2}}}$
Hint: You must know about the rules of solving derivative of Inverse trigonometric function.
Given: $e^{\sin ^{-1} 2 x}$
Solution:
Let $y=e^{\sin ^{-1} 2 x}$
Differentiate with respect to x,
$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left[e^{\sin ^{-1} 2 x}\right] \\\\ &\frac{d y}{d x}=e^{\sin ^{-1} 2 x} \times \frac{d}{d x}\left(\sin ^{-1} 2 x\right) \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}=e^{\sin ^{-1} 2 x} \times \frac{1}{\sqrt{1-(2 x)^{2}}} \frac{d}{d x}(2 x) \\\\ &\frac{d y}{d x}=\frac{2 e^{\sin ^{-1} 2 x}}{\sqrt{1-4 x^{2}}} \end{aligned}$

Differentiation exercise 10.2 question 35

Answer: $\frac{2 \cos \left(2 \sin ^{-1} x\right)}{\sqrt{1-x^{2}}}$
Hint: You must know about the rules of solving derivative of Trigonometry and Inverse trigonometric function
Given: $\sin \left(2 \sin ^{-1} x\right)$
Solution:
Let $y=\sin \left(2 \sin ^{-1} x\right)$
Differentiate with respect to x,
$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left[\sin \left(2 \sin ^{-1} x\right)\right] \\\\ &\frac{d y}{d x}=\cos \left(2 \sin ^{-1} x\right) \frac{d}{d x}\left(2 \sin ^{-1} x\right) \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}=\cos \left(2 \sin ^{-1} x\right) \times 2 \times \frac{1}{\sqrt{1-x^{2}}} \\\\ &\frac{d y}{d x}=\frac{2 \cos \left(2 \sin ^{-1} x\right)}{\sqrt{1-x^{2}}} \end{aligned}$

Differentiation exercise 10.2 question 36

Answer: $\frac{e^{\tan ^{-1 \sqrt{x}}}}{2 \sqrt{x}(1+x)}$
Hint: You must know about the rules of solving derivative of Exponential and Inverse trigonometric function.
Given: $e^{\tan ^{-1} \sqrt{x}}$
Solution:
Let $y=e^{\tan ^{-1} \sqrt{x}}$
Differentiate with respect to x,
$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left[e^{\tan ^{-1} \sqrt{x}}\right] \\\\ &\frac{d y}{d x}=e^{\tan ^{-1} \sqrt{x}} \frac{d}{d x}\left(\tan ^{-1} \sqrt{x}\right) \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}=\frac{e^{\tan ^{-1} \sqrt{x}}}{1+x} \times \frac{1}{2 \sqrt{x}}\\\\ &\frac{d y}{d x}=\frac{e^{\tan ^{-1 \sqrt{x}}}}{2 \sqrt{x}(1+x)} \end{aligned}$

Differentiation exercise 10.2 question 37

Answer: $\frac{1}{\left(4+x^{2}\right) \sqrt{\tan ^{-1}\left(\frac{x}{2}\right)}}$
Hint: You must know about the rules of solving derivative of Inverse trigonometric function.
Given: $\sqrt{\tan ^{-1}\left(\frac{x}{2}\right)}$
Solution:
Let $y=\sqrt{\tan ^{-1}\left(\frac{x}{2}\right)}$
$y=\left\{\tan ^{-1}\left(\frac{x}{2}\right)\right\}^{\frac{1}{2}}$
Differentiate with respect to x,
$\frac{d y}{d x}=\frac{d}{d x}\left\{\tan ^{-1}\left(\frac{x}{2}\right)\right\}^{\frac{1}{2}}$
$\frac{d y}{d x}=\frac{1}{2}\left\{\tan ^{-1}\left(\frac{x}{2}\right)\right\}^{\frac{1}{2}-1} \frac{d}{d x} \tan ^{-1}\left(\frac{x}{2}\right)$
$\frac{d y}{d x}=\frac{1}{2}\left\{\tan ^{-1}\left(\frac{x}{2}\right)\right\}^{-\frac{1}{2}} \times \frac{1}{1+\left(\frac{x}{2}\right)^{2}} \times \frac{d}{d x}\left(\frac{x}{2}\right)$
$\begin{aligned} &\frac{d y}{d x}=\frac{4}{4\left(4+x^{2}\right) \sqrt{\tan ^{-1}\left(\frac{x}{2}\right)}} \\\\ &\frac{d y}{d x}=\frac{1}{\left(4+x^{2}\right) \sqrt{\tan ^{-1}\left(\frac{x}{2}\right)}} \end{aligned}$

Differentiation exercise 10.2 question 38

Answer: $\frac{1}{\left(1+x^{2}\right) \tan ^{-1}(x)}$
Hint: You must know about the rules of solving derivative of logarithm and Inverse trigonometric function.
Given: $\log \left(\tan ^{-1} x\right)$
Solution:
Let $y=\log \left(\tan ^{-1} x\right)$
Differentiate with respect to x,
$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x} \log \left(\tan ^{-1} x\right) \\\\ &\frac{d y}{d x}=\frac{1}{\left(\tan ^{-1} x\right)} \times \frac{d}{d x}\left(\tan ^{-1} x\right) \\\\ &\frac{d y}{d x}=\frac{1}{\left(1+x^{2}\right) \tan ^{-1}(x)} \end{aligned}$

Differentiation exercise 10.2 question 39

Answer: $\frac{2^{x}}{\left(x^{2}+3\right)^{2}}\left[\cos x \log _{e} 2-\sin x-\frac{4 x \cos x}{\left(x^{2}+3\right)}\right]$
Hint: You must know about the rules of solving derivative of trigonometric function.
Given: $\frac{2^{x} \cos x}{\left(x^{2}+3\right)^{2}}$
Solution:
Let $y=\frac{2^{x} \cos x}{\left(x^{2}+3\right)^{2}}$
Differentiate with respect to x,
$\frac{d y}{d x}=\frac{d}{d x}\left[\frac{2^{x} \cos x}{\left(x^{2}+3\right)^{2}}\right]$
$\frac{d y}{d x}=\left[\frac{\left(x^{2}+3\right)^{2} \frac{d}{d x}\left(2^{x} \cos x\right)-\left(2^{x} \cos x\right) \frac{d}{d x}\left(x^{2}+3\right)^{2}}{\left[\left(x^{2}+3\right)^{2}\right]^{2}}\right]$ $... \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}$
$\frac{d y}{d x}=\left[\frac{\left(x^{2}+3\right)^{2}\left\{2^{x} \frac{d}{d x} \cos x+\cos x \frac{d}{d x} 2^{x}\right\}-\left(2^{x} \cos x\right) 2\left(x^{2}+3\right) \frac{d}{d x}\left(x^{2}+3\right)}{\left[x^{2}+3\right]^{4}}\right]$
$\frac{d y}{d x}=\left[\frac{\left(x^{2}+3\right)^{2}\left\{-2^{x} \sin x+\cos x 2^{x} \log _{e} 2\right\}-2\left(2^{x} \cos x\right)\left(x^{2}+3\right)(2 x)}{\left[x^{2}+3\right]^{4}}\right]$
$\frac{d y}{d x}=\left[\frac{2^{x}\left(x^{2}+3\right)^{2}\left\{\left(\cos x \log _{e} 2-\sin x\right\}-\frac{4 x \cos x}{\left(x^{2}+3\right)}\right.}{\left[x^{2}+3\right]^{4}}\right]$
$\frac{d y}{d x}=\frac{2^{x}}{\left(x^{2}+3\right)^{2}}\left[\cos x \log _{e} 2-\sin x-\frac{4 x \cos x}{\left(x^{2}+3\right)}\right]$

Differentiation exercise 10.2 question 40

Answer: $2 x \cos 2 x+\sin 2 x+5^{x} \log _{e} 5+6 \tan ^{5} x \sec ^{2} x$
Hint: You must know about the rules of solving derivative of trigonometric function.
Given: $x \sin 2 x+5^{x}+k^{k}+\left(\tan ^{6} x\right)$
Solution:
Let $y=x \sin 2 x+5^{x}+k^{k}+\left(\tan ^{6} x\right)$
Differentiate with respect to x,
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left[\mathrm{x} \sin 2 \mathrm{x}+5^{\mathrm{x}}+\mathrm{k}^{\mathrm{k}}+\left(\tan ^{6} \mathrm{x}\right)\right]$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{xsin} 2 \mathrm{x})+\frac{\mathrm{d}}{\mathrm{dx}}\left(5^{\mathrm{x}}\right)+\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{k}^{\mathrm{k}}\right)+\frac{\mathrm{d}}{\mathrm{dx}}\left(\tan ^{6} \mathrm{x}\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\left[\mathrm{x}\left\{\cos 2 \mathrm{x} \frac{\mathrm{d}}{\mathrm{dx}}(2 \mathrm{x})\right\}+\sin 2 \mathrm{x}\right]+5^{\mathrm{x}} \log _{\mathrm{e}} 5+6 \tan ^{5} \mathrm{x}+\frac{\mathrm{d}}{\mathrm{dx}}(\tan \mathrm{x})$
$\frac{d y}{d x}=\left[x\left\{\cos 2 x \frac{d}{d x}(2 x)\right\}+\sin 2 x\right]+5^{x} \log _{e} 5+6 \tan ^{5} x \sec ^{2} x$
$\frac{d y}{d x}=2 x \cos 2 x+\sin 2 x+5^{x} \log _{e} 5+6 \tan ^{5} x \sec ^{2} x$

Differentiation exercise 10.2 question 41

Answer: $\frac{3}{3 x+2}-\frac{2 x^{2}}{(2 x-1)}-2 x \log (2 x-1)$
Hint: You must know about the rules of solving derivative of logarithm function.
Given: $\log (3 x+2)-x^{2} \log (2 x-1)$
Solution:
Let $y=\log (3 x+2)-x^{2} \log (2 x-1)$
Differentiate with respect to x,
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left[\log (3 x+2)-\mathrm{x}^{2} \log (2 x-1)\right]$
$\frac{d y}{d x}=\frac{d}{d x} \log (3 x+2)-\frac{d}{d x}\left\{x^{2} \log (2 x-1)\right\}$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{(3 x+2)} \frac{\mathrm{d}}{\mathrm{dx}}(3 x+2)-\left\{\mathrm{x}^{2} \frac{\mathrm{d}}{\mathrm{dx}} \log (2 x-1)+\log (2 x-1) \frac{\mathrm{d}}{\mathrm{dx}} \mathrm{x}^{2}\right\}$
$\frac{d y}{d x}=: \frac{3}{3 x+2}-\frac{2 x^{2}}{(2 x-1)}-2 x \log (2 x-1)$

Differentiation exercise 10.2 question 42

Answer: $\left[\frac{6 x \sin x+3 x^{2} \cos x}{\sqrt{\left(7-x^{2}\right)}}+\frac{3 x^{3} \sin x}{\left(7-x^{2}\right)^{\frac{3}{2}}}\right]$
Hint: You must know about the rules of solving derivative of trigonometric function.
Given: $y=\frac{3 x^{2} \sin x}{\sqrt{\left(7-x^{2}\right)}}$
Solution:
Let $y=\frac{3 x^{2} \sin x}{\sqrt{\left(7-x^{2}\right)}}$
Differentiate with respect to x,
$\frac{d y}{d x}=\frac{d}{d x}\left[\frac{3 x^{2} \sin x}{\left(7-x^{2}\right)^{\frac{1}{2}}}\right]$
$\frac{dy}{dx}=\frac{\left(7-x^{2}\right)^{\frac{1}{2}} \times \frac{\mathrm{d}}{\mathrm{dx}}\left(3 x^{2} \sin x\right)-\left(3 x^{2} \sin x\right) \frac{\mathrm{d}}{\mathrm{dx}}\left(7-x^{2}\right)^{\frac{1}{2}}}{\left[\left(7-x^{2}\right)^{\frac{1}{2}}\right]^{2}} \ldots \frac{d}{d x}\left(\frac{u}{v}\right)=$ $\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\left[\frac{\left(7-x^{2}\right)^{\frac{1}{2}}\left(3 x^{2} \cos x+6 x \sin x\right)-3 x^{2} \sin x \times \frac{1}{2}\left(7-x^{2}\right)^{\frac{1}{2}-1}(-2 x)}{7-x^{2}}\right]$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\left[\frac{\left(7-x^{2}\right)^{\frac{1}{2}}\left(3 x^{2} \cos x+6 x \sin x\right)-3 x^{2} \sin x \times \frac{1}{2}\left(7-x^{2}\right)^{-\frac{1}{2}}(-2 x)}{7-x^{2}}\right]$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\left[\frac{6 x \sin x+3 x^{2} \cos x}{\sqrt{\left(7-x^{2}\right)}}+\frac{3 x^{3} \sin x}{\left(7-x^{2}\right)^{\frac{3}{2}}}\right]$

Differentiation exercise 10.2 question 43

Answer: $\sin \{2 \log (2 x+3)\}\left(\frac{2}{(2 x+3)}\right)$
Hint: you must know the rules of solving derivative of trigonometric and logarithm function,
Given: $\sin ^{2}\{\log (2 x+3)\}$
Solution:
Let $y=\sin ^{2}\{\log (2 x+3)\}$
Differentiate with respect to x,
$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left[\sin ^{2} \log (2 x+3)\right] \\ \\&=2 \sin \{\log (2 x+3)\} \frac{d}{d x} \sin \{\log (2 x+3)\} \end{aligned}$
$\begin{aligned} &=2 \sin \{\log (2 x+3)\} \cos \{\log (2 x+3)\} \frac{\mathrm{d}}{\mathrm{dx}} \log (2 x+3) \\\\ &\frac{\mathrm{dy}}{\mathrm{dx}}=\sin \{2 \log (2 x+3)\}\left(\frac{2}{(2 x+3)}\right) \end{aligned}$

Differentiation exercise 10.2 question 44

Answer: $2 e^{x} \cot 2 x+e^{x} \log \sin 2 x$
Hint: you must know about the rules of solving derivative of exponential logarithm and trigon function
Given: $e^{x} \log \sin 2 x$
Solution:
Let $y=e^{x} \log \sin 2 x$
Differentiate with respect to x
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left[e^{x} \log \sin 2 x\right]$
$\begin{aligned} &=e^{x} \frac{\mathrm{d}}{\mathrm{dx}}(\log \sin 2 x)+(\log \sin 2 x) \frac{\mathrm{d}}{\mathrm{dx}}\left(e^{x}\right) \\\\ &=e^{x} \frac{1}{\sin 2 x} \frac{\mathrm{d}}{\mathrm{dx}} \sin 2 x+\log \sin 2 x\left(e^{x}\right) \end{aligned}$
$\begin{aligned} &=\frac{e^{x}}{\sin 2 x} \cos 2 x \frac{\mathrm{d}}{\mathrm{dx}}(2 x)+e^{x} \log \sin 2 x \\\\ &=\frac{2 \cos 2 x e^{x}}{\sin 2 x}+e^{x} \log \sin 2 x \\\\ &\Rightarrow 2 e^{x} \cot 2 x+e^{x} \log \sin 2 x \end{aligned}$

Differentiation exercise 10.2 question 45

Answer: $2 x+\frac{2 e^{x}}{\sqrt{x^{4}-1}}$
Hint: you must know the rules of solving derivative of polynomials
Given: $\frac{\sqrt{x^{2}+1}+\sqrt{x^{2}-1}}{\sqrt{x^{2}+1}-\sqrt{x^{2}-1}}$
Solution:
Let $y=\frac{\sqrt{x^{2}+1}+\sqrt{x^{2}-1}}{\sqrt{x^{2}+1}-\sqrt{x^{2}-1}}$
By rationalizing,
$\frac{\sqrt{x^{2}+1}+\sqrt{x^{2}-1}}{\sqrt{x^{2}+1}-\sqrt{x^{2}-1}} \times \frac{\sqrt{x^{2}+1}+\sqrt{x^{2}-1}}{\sqrt{x^{2}+1}+\sqrt{x^{2}-1}}$
$\Rightarrow \frac{\left(\sqrt{x^{2}+1}+\sqrt{\left.x^{2}-1\right)}\right)^{2}}{\left(\left(\sqrt{\left.x^{2}+1\right)^{2}}-\left(\sqrt{\left.\left.x^{2}-1\right)^{2}\right)}\right.\right.\right.}$
$\Rightarrow \frac{\left(\sqrt{x^{2}+1}\right)^{2}+\left(\sqrt{x^{2}-1}\right)^{2}+2\left(\sqrt { x ^ { 2 } + 1 ) } \left(\sqrt{\left.x^{2}-1\right)}\right.\right.}{x^{2}+1-x^{2}+1}$
$=\frac{x^{2}+1+x^{2}-1+2 \sqrt{x^{4}-1}}{2}$
$=x^{2}+\sqrt{x^{4}-1}$
Now, let $y=x^{2}+\sqrt{x^{4}-1}$
Now, differentiate
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(x^{2}+\sqrt{x^{4}-1}\right)$
$\begin{aligned} &\Rightarrow 2 x+\frac{1}{2 \sqrt{x^{4}-1}} \times\left(4 x^{3}\right) \\\\ &\Rightarrow 2 x+\frac{2 x^{3}}{\sqrt{x^{4}-1}} \end{aligned}$

Differentiation exercise 10.2 question 46

Answer: $\frac{1}{\sqrt{x^{2}+4 x+1}}$
Hint: you must know the rules of solving derivative of logarithm function
Given: $\log \left[x+2+\sqrt{x^{2}+4 x+1}\right]$
Solution:
Let $y=\log \left[x+2+\sqrt{x^{2}+4 x+1}\right]$
Differentiate both side with respect to x
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}} \log \left[x+2+\sqrt{x^{2}+4 x+1}\right]$
$=\frac{1}{\left[x+2+\sqrt{x^{4}+4 x+1}\right]} \times\left[1+0+\frac{1}{2}\left(x^{2}+4 x+1\right)^{\frac{-1}{2}} \frac{\mathrm{d}}{\mathrm{dx}}\left(x^{2}+4 x+1\right)\right]$
$\Rightarrow \frac{1+\frac{2 x+4}{2\left(\sqrt{x^{2}+4 x+1}\right)}}{\left[x+2+\sqrt{x^{2}+4 x+1}\right]}$
$=\frac{\sqrt{x^{2}+4 x+1}+x+2}{\left[x+2+\sqrt{x^{2}+4 x+1}\right] \times \sqrt{x^{2}+4 x+1}}$
$\Rightarrow \frac{1}{\sqrt{x^{2}+4 x+1}}$

Differentiation exercise 10.2 question 47

Answer: $\frac{16 x^{3}\left(\sin ^{-1} x^{4}\right)^{3}}{\sqrt{1-x^{8}}}$
Hint: you must know the rules of solving derivative of inverse trigonometric function
Given: $\left(\sin ^{-1} x^{4}\right)^{4}$
Solution:
Let $y=\left(\sin ^{-1} x^{4}\right)^{4}$
Differentiate with respect to x
$\frac{d y}{d x}=\frac{d}{d x}\left(\sin ^{-1} x^{4}\right)^{4}$
$\begin{aligned} &=4\left(\sin ^{-1} x^{4}\right)^{3} \frac{\mathrm{d}}{\mathrm{dx}}\left(\sin ^{-1} x^{4}\right) \\\\ &=4\left(\sin ^{-1} x^{4}\right)^{3} \frac{1}{\sqrt{1-\left(x^{4}\right)^{2}}} \frac{\mathrm{d}}{\mathrm{dx}}\left(x^{4}\right) \end{aligned}$
$\begin{aligned} &=4\left(\sin ^{-1} x^{4}\right)^{3} \frac{4 x^{3}}{\sqrt{1-x^{8}}} \\\\ &\frac{d y}{d x} \Rightarrow \frac{16 x^{3}\left(\sin ^{-1} x^{4}\right)^{3}}{\sqrt{1-x^{8}}} \end{aligned}$

Differentiation exercise 10.2 question 48

Answer: $\frac{a}{\left(x^{2}+a^{2}\right)}$
Hint: you must know the rules of solving derivative of inverse trigonometric function
Given: $\sin ^{-1}\left(\frac{x}{\sqrt{x^{2}+a^{2}}}\right)$
Solution:
Let $y=\sin ^{-1}\left(\frac{x}{\sqrt{x^{2}+a^{2}}}\right)$
Differentiate with respect to x
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left\{\sin ^{-1}\left(\frac{x}{\sqrt{x^{2}+a^{2}}}\right)\right\}$
$=\frac{1}{\sqrt{1-\left(\frac{x}{\sqrt{x^{2}+a^{2}}}\right)^{2}}} \times\left[\frac{\left(x^{2}+a^{2}\right)^{\frac{1}{2} \frac{\mathrm{d}}{\mathrm{dx}}(x)-x \frac{\mathrm{d}}{\mathrm{dx}}\left(x^{2}+a^{2}\right)^{\frac{1}{2}}}}{\left[\left(x^{2}+a^{2}\right)^{\frac{1}{2}}\right]^{2}}\right] \cdot \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}$
$=\frac{\sqrt{x^{2}+a^{2}}}{a}\left[\frac{\sqrt{x^{2}+a^{2}}-\frac{x}{2 \sqrt{x^{2}+a^{2}}} \frac{\mathrm{d}}{\mathrm{dx}}\left(x^{2}+a^{2}\right)}{\left(x^{2}+a^{2}\right)}\right]$
$=\frac{\sqrt{x^{2}+a^{2}}}{a\left(x^{2}+a^{2}\right)} \quad\left[\sqrt{x^{2}+a^{2}}-\frac{x}{2 \sqrt{x^{2}+a^{2}}} \times 2 x\right]$
$\Rightarrow \frac{\sqrt{x^{2}+a^{2}}}{a\left(x^{2}+a^{2}\right)} \quad\left[\frac{2\left(x^{2}+a^{2}-x^{2}\right)}{2 \sqrt{x^{2}+a^{2}}}\right]$
$\begin{aligned} &\Rightarrow \frac{a^{2}}{a\left(x^{2}+a^{2}\right)} \\\\ &\Rightarrow \frac{a}{\left(x^{2}+a^{2}\right)} \end{aligned}$

Differentiation exercise 10.2 question 49

Answer: $\frac{e^{x} \sin x+e^{x} \cos x}{\left(x^{2}+2\right)^{3}}-\frac{6 x e^{x} \sin x}{\left(x^{2}+2\right)^{4}}$
Hint: you must know the rules of solving exponential derivative
Given: $\frac{e^{x} \sin x}{\left(x^{2}+2\right)^{3}}$
Solution:
Let $y=\frac{e^{x} \sin x}{\left(x^{2}+2\right)^{3}}$
Differentiate with respect to x
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\left(x^{2}+2\right)^{3} \frac{\mathrm{d}}{\mathrm{dx}}\left(e^{x} \sin x\right)-e^{x} \sin x \frac{\mathrm{d}}{\mathrm{dx}}\left(x^{2}+2\right)^{3}}{\left[\left(x^{2}+2\right)^{3}\right]^{2}} \cdot \cdot \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}$
$=\frac{\left(x^{2}+2\right)^{3}\left[e^{x} \cos x+\sin x e^{x}\right]-e^{x} \sin x 3\left(x^{2}+2\right)^{2}(2 x)}{\left(x^{2}+2\right)^{6}}$
$=\frac{\left(x^{2}+2\right)^{2}\left[\left(x^{2}+2\right)\left[e^{x} \cos x+e^{x} \sin x\right]-6 x e^{x} \sin x\right]}{\left(x^{2}+2\right)^{6}}$
$\Rightarrow \frac{\left(x^{2}+2\right)\left(e^{x} \cos x+e^{x} \sin x\right)-6 x e^{x} \sin x}{\left(x^{2}+2\right)^{4}}$
$\Rightarrow \frac{e^{x} \sin x+e^{x} \cos x}{\left(x^{2}+2\right)^{3}}-\frac{6 x e^{x} \sin x}{\left(x^{2}+2\right)^{4}}$

Differentiation exercise 10.2 question 50

Answer: $3 e^{-3 x}\left\{\frac{1}{1+x}-3 \log (1+x)\right\}$
Hint: you must know the rule of solving exponential and logarithm functions
Given: $3 e^{-3 x} \log (1+x)$
Solution:
Let $y=3 e^{-3 x} \log (1+x)$
Differentiate with respect to x
$\frac{\mathrm{dy}}{\mathrm{dx}}=3 \frac{\mathrm{d}}{\mathrm{dx}}\left[e^{-3 x} \log (1+x)\right]$
$\begin{aligned} &=3\left\{e^{-3 x} \frac{1}{(1+x)}+\log (1+x)\left(-3 e^{-3 x}\right)\right\} \\\\ &\Rightarrow 3\left\{\frac{e^{-3 x}}{1+x}-3 e^{-3 x} \log (1+x)\right\} \\\\ &\Rightarrow 3 e^{-3 x}\left\{\frac{1}{1+x}-3 \log (1+x)\right\} \end{aligned}$

Differentiation exercise 10.2 question 51

Answer: $\frac{1}{\sqrt{\cos x}}\left\{2 x+\frac{x^{2}}{2 \cos x}+\tan x\right\}$
Hint: you must know the rule of solving derivative of trigonometric function
Given: $\frac{x^{2}+2}{\sqrt{\cos x}}$
Solution:
Let $y= \frac{x^{2}+2}{\sqrt{\cos x}}$
Differentiate with respect to x
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\sqrt{\cos x} \frac{\mathrm{d}}{\mathrm{dx}}\left(x^{2}+2\right)-\left(x^{2}+2\right) \frac{\mathrm{d}}{\mathrm{dx}}(\sqrt{\cos x})}{(\sqrt{\cos x})^{2}} \cdot \cdot \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}$
$\begin{aligned} &=\frac{2 x \sqrt{\cos x}-\left(x^{2}+2\right)\left(\frac{1}{2} \frac{-\sin x}{\sqrt{\cos x}}\right)}{\cos x} \\\\ &\Rightarrow \frac{2 x \sqrt{\cos x}+\frac{\left(x^{2}+2\right) \sin x}{2 \sqrt{\cos x}}}{\cos x} \end{aligned}$
$\begin{aligned} &\Rightarrow \frac{4 x \cos x+\left(x^{2}+2\right) \sin x}{2 \cos x^{\frac{3}{2}}} \\\\ &\Rightarrow \frac{2 x}{\sqrt{\cos x}}+\frac{1}{2} \frac{\left(x^{2}+2\right) \sin x}{(\cos x)^{\frac{3}{2}}} \end{aligned}$
$\begin{aligned} &\Rightarrow \frac{1}{\sqrt{\cos x}}\left\{2 x+\frac{1}{2} \frac{\left(x^{2}+2\right) \sin x}{\cos x}\right\} \\\\ &\Rightarrow \frac{1}{\sqrt{\cos x}}\left\{2 x+\frac{x^{2}}{2 \cos x}+\tan x\right\} \end{aligned}$

Differentiation exercise 10.2 question 52

Answer: $2 x\left(1-x^{2}\right)^{2} \sec 2 x\left\{\left(1-x^{2}\right)-3 x^{2}+x\left(1-x^{2}\right) \tan 2 x\right\}$
Hint: you must know the rule of solving derivative of trigonometric functions
Given: $\frac{x^{2}\left(1-x^{2}\right)^{3}}{\cos 2 x}$
Solution:
Let $y=\frac{x^{2}\left(1-x^{2}\right)^{3}}{\cos 2 x}$
Differentiate with respect to x
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\cos 2 x \frac{\mathrm{d}}{\mathrm{dx}}\left\{x^{2}\left(1-x^{2}\right)^{3}-x^{2}\left(1-x^{2}\right)^{3} \frac{\mathrm{d}}{\mathrm{dx}} \cos 2 x\right\}}{\cos ^{2} 2 x} \cdot \cdot \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}$
$\Rightarrow \frac{\cos 2 x\left\{x^{2} \frac{\mathrm{d}}{\mathrm{dx}}\left(1-x^{2}\right)^{3}+\left(1-x^{2}\right)^{3} \frac{\mathrm{d}}{\mathrm{dx}} x^{2}\right\}-x^{2}\left(1-x^{2}\right)^{3}(-2 \sin 2 x)}{\cos ^{2} 2 x}$
$\Rightarrow \frac{\cos 2 x\left\{-6 x^{3}\left(1-x^{2}\right)^{2}+\left(1-x^{2}\right)^{3} 2 x\right\}+2 x^{2}\left(1-x^{2}\right)^{3} \sin 2 x}{\cos ^{2} 2 x}$
$\Rightarrow-\frac{6 x^{3}\left(1-x^{2}\right)^{2}}{\cos 2 x}+\frac{2 x\left(1-x^{2}\right)^{3}}{\cos 2 x}+\frac{2 x^{2}\left(1-x^{2}\right)^{3} \sin 2 x}{\cos ^{2} 2 x}$
$\Rightarrow 2 x\left(1-x^{2}\right)^{2} \sec 2 x\left\{\left(1-x^{2}\right)-3 x^{2}+x\left(1-x^{2}\right) \tan 2 x\right\}$

Differentiation exercise 10.2 question 53

Answer: $-\sec x$
Hint: you must know the rule of solving derivative of logarithm and trigonometric functions
Given: $\log \left\{\cot \left(\frac{\pi}{4}+\frac{x}{2}\right)\right\}$
Solution:
Let $y=\log \left\{\cot \left(\frac{\pi}{4}+\frac{x}{2}\right)\right\}$
Differentiate with respect to x
$\frac{d y}{d x}=\frac{1}{\cot \left(\frac{\pi}{4}+\frac{x}{2}\right)} \cdot \frac{d}{d x} \cot \left(\frac{x}{2}+\frac{\pi}{4}\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\cot \left(\frac{\pi}{4}+\frac{x}{2}\right)} \cdot-\operatorname{cosec}^{2}\left(\frac{\pi}{4}+\frac{x}{2}\right) \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\pi}{4}+\frac{x}{2}\right)$
$\begin{aligned} &\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\operatorname{cosec}^{2}\left(\frac{\pi}{4}+\frac{x}{2}\right)}{\cot \left(\frac{\pi}{4}+\frac{x}{2}\right)} \times \frac{1}{2} \\\\ &\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\operatorname{cosec}^{2}\left(\frac{\pi}{4}+\frac{x}{2}\right)}{2 \cot \left(\frac{\pi}{4}+\frac{x}{2}\right)} \end{aligned}$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-1}{\sin ^{2}\left(\frac{\pi}{4}+\frac{x}{2}\right)} \times \frac{\sin \left(\frac{\pi}{4}+\frac{x}{2}\right)}{2 \cos \left(\frac{\pi}{4}+\frac{x}{2}\right)}$
$\frac{d y}{d x}=\frac{-1}{2 \cos \left(\frac{\pi}{4}+\frac{x}{2}\right) \sin \left(\frac{\pi}{4}+\frac{x}{2}\right)}$ $[\therefore 2 \sin x \cos x=\sin 2 x]$
$\frac{d y}{d x}=\frac{-1}{\sin \left(\frac{\pi}{2}+x\right)}$
$\begin{aligned} &\frac{d y}{d x}=-\frac{1}{\cos x} \\\\ &\frac{d y}{d x}=-\sec x \end{aligned}$

Differentiation exercise 10.2 question 54

Answer: $e^{a x} \sec x\left\{\mathrm{a} \tan 2 x+\tan x \tan 2 x+2 \sec ^{2} 2 x\right\}$
Hint: you must know the rule of solving derivative of exponential and trigonometric functions
Given: $e^{a x} \sec x \tan 2 x$
Solution:
Let $y=e^{a x} \sec x \tan 2 x$
Differentiate with respect to x
$\begin{aligned} &\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}} e^{a x} \sec x \tan 2 x \\\\ &\frac{\mathrm{dy}}{\mathrm{dx}}=e^{a x} \frac{d}{d x}\{\sec x \tan 2 x\}+\sec x \tan 2 x \frac{d}{d x}\left\{e^{a x}\right\} \end{aligned}$
$\begin{aligned} &\frac{\mathrm{dy}}{\mathrm{dx}}=e^{a x}\left[\sec x \tan x \tan 2 x+2 \sec ^{2} 2 x \sec x\right]+a e^{a x} \sec x \tan 2 x \\\\ &\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{a} e^{a x} \sec x \tan 2 x+e^{a x} \sec x \tan x \tan 2 x+2 \sec ^{2} 2 x \sec x e^{a x} \end{aligned}$
$\frac{\mathrm{dy}}{\mathrm{dx}}=e^{a x} \sec x\left\{\mathrm{a} \tan 2 x+\tan x \tan 2 x+2 \sec ^{2} 2 x\right\}$

Differentiation exercise 10.2 question 55

Answer: $-2 x \tan x^{2}$
Hint: you must know the rule of solving derivative of logarithm and trigonometric functions
Given: $\log \left(\cos x^{2}\right)$
Solution:
Let $y=\log \left(\cos x^{2}\right)$
Differentiate with respect to x
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{d}{d x}\left\{\log \left(\cos x^{2}\right)\right\}$ $\left[\therefore \frac{d}{d x} \log x=\frac{1}{x}\right]$
$\frac{d y}{d x}=\frac{-2 x \sin x^{2}}{\cos x^{2}}$ [ Using chain rule ]
$\frac{d y}{d x}=-2 x \tan x^{2}$

Differentiation exercise 10.2 question 56

Answer: $\frac{-2 \log x \sin (\log x)^{2}}{x}$
Hint: you must know the rule of solving derivative of logarithm and trigonometric functions
Given: $\cos (\log x)^{2}$
Solution:
Let $y=\cos (\log x)^{2}$
Differentiate with respect to x
$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left[\cos (\log x)^{2}\right] \\\\ &\frac{d y}{d x}=-\sin (\log x)^{2} \cdot \frac{d}{d x}\left[(\log x)^{2}\right] \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}=-\sin (\log x)^{2} \cdot 2 \log x \frac{d}{d x} \log x \\\\ &\frac{d y}{d x}=-\sin (\log x)^{2} \cdot \frac{2 \log x}{x} \\\\ &\frac{d y}{d x}=\frac{-2 \log x \sin (\log x)^{2}}{x} \end{aligned}$

Differentiation exercise 10.2 question 57

Answer: $\frac{1}{x^{2}-1}$
Hint: you must know the rule of solving derivative of logarithm functions
Given: $\log \sqrt{\frac{x-1}{x+1}}$
Solution:
Let $y=\log \left(\frac{x-1}{x+1}\right)^{\frac{1}{2}}$
$\begin{aligned} &y=\frac{1}{2} \log \left(\frac{x-1}{x+1}\right) \\\\ &y=\frac{1}{2}\{\log (x-1)-\log (x+1)\} \end{aligned}$
Differentiate with respect to x
$\frac{d y}{d x}=\frac{1}{2}\left\{\frac{d}{d x}[\log (x-1)]-\frac{d}{d x}[\log (x+1)]\right\}$
$\begin{aligned} &\frac{d y}{d x}=\frac{1}{2}\left[\frac{1}{(x-1)}-\frac{1}{(x+1)}\right] \\\\ &\frac{d y}{d x}=\frac{1}{2}\left[\frac{x+1-(x-1)}{\left(x^{2}-1\right)}\right] \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}=\frac{1}{2}\left[\frac{x+1-x+1}{\left(x^{2}-1\right)}\right] \\\\ &\frac{d y}{d x}=\frac{1}{2}\left[\frac{2}{\left(x^{2}-1\right)}\right] \\\\ &\frac{d y}{d x}=\frac{1}{\left(x^{2}-1\right)} \end{aligned}$

Differentiation exercise 10.2 question 58

Answer: Proved
Hint: you must know the rule of solving derivative of logarithm functions
Given: $\log (\sqrt{x-1}-\sqrt{x+1})$
Solution:
Let $y=\log (\sqrt{x-1}-\sqrt{x+1})$
Differentiate with respect to x
$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x} \log (\sqrt{x-1}-\sqrt{x+1}) \\\\ &\frac{d y}{d x}=\frac{1}{(\sqrt{x-1}-\sqrt{x+1})} \cdot \frac{d}{d x}(\sqrt{x-1}-\sqrt{x+1}) \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}=\frac{1}{(\sqrt{x-1}-\sqrt{x+1})} \cdot\left[\frac{d}{d x}(\sqrt{x-1})-\frac{d}{d x} \sqrt{x+1}\right] \\\\ &\frac{d y}{d x}=\frac{1}{(\sqrt{x-1}-\sqrt{x+1})} \cdot\left[\frac{1}{2}(x-1)^{\frac{-1}{2}}-\frac{1}{2}(x+1)^{\frac{-1}{2}}\right] \end{aligned}$
$\frac{d y}{d x}=\frac{1}{2} \frac{1}{(\sqrt{x-1}-\sqrt{x+1})}\left(\frac{1}{\sqrt{x-1}}-\frac{1}{\sqrt{x+1}}\right)$
$\begin{aligned} &\frac{d y}{d x}=\frac{1}{2(\sqrt{x-1}-\sqrt{x+1})}\left(\frac{\sqrt{x+1}-\sqrt{x-1}}{(\sqrt{x-1})(\sqrt{x+1})}\right) \\\\ &\frac{d y}{d x}=\frac{-1}{2 \sqrt{x^{2}-1}} \end{aligned}$
∴ Proved

Differentiation exercise 10.2 question 59

Answer: Proved
Hint: you must know the rule of solving derivation of functions
Given: $\mathrm{y}=\sqrt{x+1}+\sqrt{x-1}$
Prove : $\sqrt{x^{2}-1} \frac{d y}{d x}=\frac{1}{2} y$
Solution:
Let $\mathrm{y}=\sqrt{x+1}+\sqrt{x-1}$
Differentiate with respect to x
$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}(\sqrt{x+1})+\frac{d}{d x}(\sqrt{x-1}) \\\\ &\frac{d y}{d x}=\frac{1}{2}(x+1)^{\frac{-1}{2}}+\frac{1}{2}(x-1)^{\frac{-1}{2}} \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}=\frac{1}{2}\left(\frac{1}{\sqrt{x+1}}+\frac{1}{\sqrt{x-1}}\right) \\\\ &\frac{d y}{d x}=\frac{1}{2}\left(\frac{\sqrt{x-1}+\sqrt{x+1}}{\sqrt{x+1} \sqrt{x-1}}\right) \end{aligned}$ $[\mathrm{y}=\sqrt{x+1}+\sqrt{x-1}]$
$\begin{aligned} &\frac{d y}{d x}=\frac{1}{2}\left(\frac{\mathrm{y}}{\sqrt{x+1} \sqrt{x-1}}\right) \\\\ &\frac{d y}{d x}=\frac{1}{2}\left(\frac{\mathrm{y}}{\sqrt{\mathrm{x}^{2}-1}}\right) \\\\ &\sqrt{\mathrm{x}^{2}-1} \frac{d y}{d x}=\frac{1}{2} \mathrm{y} \end{aligned}$
∴ Proved

Differentiation exercise 10.2 question 60

Answer: Proved
Hint: you must know the rule of solving derivation of functions.
Given: $y=\frac{x}{x+2}$
Prove : $x \frac{d y}{d x}=(1-y) y$
Solution:
Let $y=\frac{x}{x+2}$
Differentiate with respect to x and apply quotient rule
$\frac{d y}{d x}=\frac{d}{d x}\left(\frac{x}{x+2}\right)$
$\frac{d y}{d x}=\frac{(x+2) \frac{d}{d x}(x)-x \frac{d}{d x}(x+2)}{(x+2)^{2}} \ldots \frac{d}{d x} \text { u. } v=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}$
$\begin{aligned} &\frac{d y}{d x}=\frac{x+2-x}{(x+2)^{2}} \\\\ &\frac{d y}{d x}=\frac{x+2}{(x+2)^{2}}-\frac{x}{(x+2)^{2}} \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}=\frac{1}{x+2}-\frac{x y^{2}}{x^{2}} \\\\ &\frac{d y}{d x}=\frac{y}{x}-\frac{y^{2}}{x} \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}=\frac{y-y^{2}}{x} \\\\ &x \frac{d y}{d x}=y(1-y) \end{aligned}$
∴ Proved

Differentiation exercise 10.2 question 61

Answer: Proved
Hint: you must know the rule of solving logarithm functions.
Given: $\mathrm{y}=\log \left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)$
Prove $\frac{d y}{d x}=\frac{x-1}{2 x(x+1)}$
Solution:
Let $\mathrm{y}=\log \left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)$
Differentiate with respect to x
$\begin{aligned} &\frac{d y}{d x}=\frac{1}{\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)} \cdot \frac{d}{d x}\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right) \\\\ &\frac{d y}{d x}=\frac{\sqrt{x}}{x+1}\left(\frac{1}{2 \sqrt{x}}-\frac{1}{2 x \sqrt{x}}\right) \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}=\frac{1}{2} \frac{\sqrt{x}}{(x+1)}\left(\frac{x-1}{x \sqrt{x}}\right) \\\\ &\frac{d y}{d x}=\frac{x-1}{2 x(x+1)} \end{aligned}$
∴ Proved

Differentiation exercise 10.2 question 62

Answer: Proved
Hint:: you must know the rules of solving derivation of logarithm functions.
Given: $y=\log \left(\sqrt{\frac{1+\tan x}{1-\tan x}}\right)$
Prove $\frac{d y}{d x}=\sec 2 x$
Solution:
Let $y=\log \left(\sqrt{\frac{1+\tan x}{1-\tan x}}\right)$
$\mathrm{y}=\log \left(\frac{1+\tan x}{1-\tan x}\right)^{\frac{1}{2}}$ $\left[\therefore \log a^{x}=x \log a\right]$
$\mathrm{y}=\frac{1}{2}[\log (1+\tan \mathrm{x})-\log (1-\tan \mathrm{x})]$ $\left[\therefore \log \frac{m}{n}=\log m-\log n\right]$
Differentiate with respect to x,
$\begin{aligned} &\frac{d y}{d x}=\frac{1}{2}\left\{\frac{d}{d x}\left[\log (1+\tan x)-\frac{d}{d x} \log (1-\tan x)\right]\right\} \\\\ &\frac{d y}{d x}=\frac{1}{2}\left\{\frac{1}{1+\tan x} \frac{d}{d x}(1+\tan x)-\frac{1}{1-\tan x} \frac{d}{d x}(1-\tan x)\right\} \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}=\frac{1}{2}\left\{\frac{1}{1+\tan x}\left(\sec ^{2} x\right)-\frac{1}{1-\tan x}\left(-\sec ^{2} x\right)\right\} \\\\ &\frac{d y}{d x}=\frac{1}{2}\left\{\frac{\left(\sec ^{2} x\right)}{1+\tan x}+\frac{\sec ^{2} x}{1-\tan x}\right\} \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}=\frac{\sec ^{2} x}{2}\left[\frac{1-\tan x+1+\tan x}{1-\tan ^{2} x}\right] \\\\ &\frac{d y}{d x}=\frac{1}{2} \sec ^{2} x\left[\frac{2}{1-\tan ^{2} x}\right] \end{aligned}$
$\frac{d y}{d x}=\frac{\sec ^{2} x}{1-\tan ^{2} x}$ $\left[\therefore \sec ^{2} x=1+\tan ^{2} x\right]$
$\begin{aligned} &\frac{d y}{d x}=\frac{1+\tan ^{2} x}{1-\tan ^{2} x} \\\\ &\frac{d y}{d x}=\frac{1}{\left(\frac{1-\tan ^{2} x}{1+\tan ^{2} x}\right)} \end{aligned}$ $\left[\therefore \text { Tigonometric Property } \frac{1-\tan ^{2} x}{1+\tan ^{2} \mathrm{x}}=\cos 2 x\right]$
$\begin{aligned} &\frac{d y}{d x}=\frac{1}{\cos 2 \mathrm{x}} \\\\ &\frac{d y}{d x}=\sec 2 x \end{aligned}$
∴ Proved

Differentiation exercise 10.2 question 63

Answer: Proved
Hint: you must know the rule of solving derivation of functions.
Given: $y=\sqrt{x}+\frac{1}{\sqrt{x}}$
Prove $2 x \frac{d y}{d x}=\sqrt{x}-\frac{1}{\sqrt{x}}$
Solution:
Let $y=\sqrt{x}+\frac{1}{\sqrt{x}}$
Differentiate with respect to x,
$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left\{\sqrt{x}+\frac{1}{\sqrt{x}}\right\} \\\\ &\frac{d y}{d x}=\frac{d}{d x}(\sqrt{x})+\frac{d}{d x}\left(\frac{1}{\sqrt{x}}\right) \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}=\frac{1}{2 \sqrt{x}}+\left(\frac{-1}{2 x \sqrt{x}}\right) \\\\ &\frac{d y}{d x}=\frac{1}{2 \sqrt{x}}-\frac{1}{2 x \sqrt{x}} \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}=\frac{x-1}{2 x \sqrt{x}} \\\\ &2 x \frac{d y}{d x}=\frac{x-1}{\sqrt{x}} \end{aligned}$
$\begin{aligned} &2 x \frac{d y}{d x}=\frac{x}{\sqrt{x}}-\frac{1}{\sqrt{x}} \\\\ &2 x \frac{d y}{d x}=\sqrt{x}-\frac{1}{\sqrt{x}} \end{aligned}$
$\therefore$ Proved

Differentiation exercise 10.2 question 64

Answer: Proved
Hint: you must know the rules of derivative of inverse trigonometric functions.
Given: $y=\frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}}$
Prove $\left(1-x^{2}\right) \frac{d y}{d x}=x+\frac{y}{x}$
Solution:
Let $y=\frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}}$
Differentiate with respect to x,
$\frac{d y}{d x}=\frac{d}{d x}\left(\frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}}\right)$ [ quotient rule ]
$\frac{d y}{d x}=\frac{d}{d x}\left[\frac{\sqrt{1-x^{2}} \frac{d}{d x}\left(x \sin ^{-1} x\right)-\left(x \sin ^{-1} x\right) \frac{d}{d x}\left(\sqrt{1-x^{2}}\right)}{\left(\sqrt{1-x^{2}}\right)^{2}}\right] . \cdot \frac{d}{d x} u \cdot v=$ $\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}$
$\frac{d y}{d x}=\left[\frac{\sqrt{1-x^{2}}\left\{x \frac{d}{d x}\left(\sin ^{-1} x\right)+\left(\sin ^{-1} x\right) \frac{d}{d x}(x)\right\}-\left(x \sin ^{-1} x\right)\left(\frac{1}{2 \sqrt{1-x^{2}}}\right) \frac{d\left(1-x^{2}\right)}{d x}}{\left(\sqrt{1-x^{2}}\right)^{2}}\right]$
$\frac{d y}{d x}=\left[\frac{\sqrt{1-x^{2}}\left\{\frac{x}{\sqrt{1-x^{2}}}+\sin ^{-1} x\right\}-\frac{x \sin ^{-1} x(-2 x)}{2\left(\sqrt{1-x^{2}}\right)}}{\left(\sqrt{1-x^{2}}\right)^{2}}\right]$
$\frac{d y}{d x}=\left[\frac{x+\sqrt{1-x^{2}}\left(\sin ^{-1} x\right)+\frac{x^{2} \sin ^{-1} x}{\sqrt{1-x^{2}}}}{\left(1-x^{2}\right)}\right]$
$\left(1-x^{2}\right) \frac{d y}{d x}=x+\frac{\sqrt{1-x^{2}}\left(\sin ^{-1} x\right)}{1}+\frac{x^{2} \sin ^{-1} x}{\sqrt{1-x^{2}}}$
$\left(1-x^{2}\right) \frac{d y}{d x}=x+\left(\frac{\sin ^{-1} x-x^{2}\left(\sin ^{-1} x\right)+x^{2}\left(\sin ^{-1} x\right)}{\sqrt{1-x^{2}}}\right)$
$\left(1-x^{2}\right) \frac{d y}{d x}=x+\left(\frac{\sin ^{-1} x}{\sqrt{1-x^{2}}}\right)$
Where $y=\frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}}$
$\left(1-x^{2}\right) \frac{d y}{d x}=x+\frac{y}{x}$
∴ Proved

Differentiation exercise 10.2 question 65

Answer: Proved

Hint: you must know the rules of derivative of exponential functions.
Given: $y=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}$
Prove $\frac{d y}{d x}=1-y^{2}$
Solution:
Let $y=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}$
Differentiate with respect to x,use quotient rule
$\frac{d y}{d x}=\frac{d}{d x}\left(\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\right)$

$\frac{d y}{d x}=\left[\frac{\left(e^{x}+e^{-x}\right) \frac{d}{d x}\left(e^{x}-e^{-x}\right)-\left(e^{x}-e^{-x}\right) \frac{d}{d x}\left(e^{x}+e^{-x}\right)}{\left(e^{x}+e^{-x}\right)^{2}}\right] \cdot \cdot \frac{d}{d x} u \cdot v=$ $\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}$
$\frac{d y}{d x}=\left\{\frac{\left(e^{x}+e^{-x}\right)\left[\left(e^{x}-e^{-x}(-1)\right)\right]-\left(e^{x}-e^{-x}\right)\left[\left(e^{x}+e^{-x}(-1)\right)\right]}{\left(e^{x}+e^{-x}\right)^{2}}\right\}$
$\frac{d y}{d x}=\left[\frac{\left(e^{x}+e^{-x}\right)\left(e^{x}+e^{-x}\right)-\left(e^{x}-e^{-x}\right)\left(e^{x}-e^{-x}\right)}{\left(e^{x}+e^{-x}\right)^{2}}\right]$
$\frac{d y}{d x}=\left[\frac{\left(e^{x}+e^{-x}\right)^{2}-\left(e^{x}-e^{-x}\right)^{2}}{\left(e^{x}+e^{-x}\right)^{2}}\right]$
$\frac{d y}{d x}=1-\frac{\left(e^{x}-e^{-x}\right)^{2}}{\left(e^{x}+e^{-x}\right)^{2}}$ $\left[y=\frac{\left(e^{x}-e^{-x}\right)}{e^{x}+e^{-x}}\right]$
$\frac{d y}{d x}=1-y^{2}$
∴ Proved

Differentiation exercise 10.2 question 66

Answer: Proved
Hint: you must know the rules of derivative of logarithm functions.
Given: $\mathrm{y}=(\mathrm{x}-1) \log (x-1)-(x+1) \log (x+1)$
Prove: $\frac{d y}{d x}=\log \left(\frac{x-1}{1+x}\right)$
Solution:
Let $\mathrm{y}=(\mathrm{x}-1) \log (x-1)-(x+1) \log (x+1)$
Differentiate with respect to x, use product rule
$\frac{d y}{d x}=\frac{d}{d x}[(\mathrm{x}-1) \log (x-1)-(x+1) \log (x+1)]$
$\frac{d y}{d x}=\left[(x-1) \times \frac{1}{(x-1)}+\log (x-1)\right]-\left[(x+1) \times \frac{1}{(x+1)}+\log (x+1)\right]$ [ use product rule]
$\frac{d y}{d x}=[1+\log (x-1)]-[1+\log (x+1)]$
$\frac{d y}{d x}=\log \left(\frac{x-1}{1+x}\right)$
∴ Proved

Differentiation exercise 10.2 question 67

Answer:Proved
Hint: you must know the rules of derivative of exponential and trigonometric functions.
Given: $y=e^{x} \cos x$
Prove: $\frac{d y}{d x}=\sqrt{2} e^{x} \cdot \cos \left(x+\frac{\pi}{4}\right)$

Solution:
Let $y=e^{x} \cos x$

Differentiate with respect to x,use product rule
$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(e^{x} \cos x\right) \\\\ &\frac{d y}{d x}=e^{x} \frac{d}{d x}(\cos x)+\cos x \frac{d}{d x} e^{x} \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}=e^{x}(-\sin x)+e^{x}(\cos x) \\\\ &\frac{d y}{d x}=e^{x}(\cos x-\sin x) \end{aligned}$

Multiply and divide by $\sqrt{2}$
$\begin{aligned} &\frac{d y}{d x}=\sqrt{2} e^{x}\left(\frac{\cos x}{\sqrt{2}}-\frac{\sin x}{\sqrt{2}}\right) \\\\ &\frac{d y}{d x}=\sqrt{2} e^{x}\left(\cos \frac{\pi}{4} \cos x-\sin \frac{\pi}{4} \sin x\right) \end{aligned}$
$\frac{d y}{d x}=\sqrt{2} e^{x} \cos \left(x+\frac{\pi}{4}\right)$ [ ∴using property $\operatorname{cos \; a\; cos} b-\operatorname{sin\: a\; sin} b=\cos (a+b)$ ]
∴ Proved

Differentiation exercise 10.2 question 68

Answer:Proved
Hint: you must know the rules of derivative of logarithm and trigonometric functions.
Given: $y=\frac{1}{2} \log \left(\frac{1-\cos 2 x}{1+\cos 2 x}\right)$

Prove: $\frac{d y}{d x}=2 \operatorname{cosec} 2 x$

Solution:
$y=\frac{1}{2} \log \left(\frac{1-\cos 2 x}{1+\cos 2 x}\right)$ $\left[\therefore 1-\cos 2 x=2 \sin ^{2} x ; 1+\cos 2 x=2 \cos ^{2} x\right]$
$\begin{aligned} &y=\frac{1}{2} \log \left(\frac{2 \sin ^{2} x}{2 \cos ^{2} x}\right) \\\\ &y=\frac{1}{2} \log \tan ^{2} x \end{aligned}$
$\begin{aligned} &y=\frac{1}{2} \times 2 \log \tan x \\\\ &y=\log \tan x \end{aligned}$

Differentiate with respect to x,use chain rule
$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}(\log \tan x) \\\\ &\frac{d y}{d x}=\frac{1}{\tan x} \frac{d}{d x}(\tan x) \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}=\frac{1}{\tan x} \times \sec ^{2} x \\\\ &\frac{d y}{d x}=\frac{\cos x}{\sin x} \times \frac{1}{\cos ^{2} x} \\\\ &\frac{d y}{d x}=\frac{1}{\sin x \cos x} \end{aligned}$

Multiply and divide by 2
$\frac{d y}{d x}=\frac{2}{2 \sin x \cos x}$ $[\therefore 2 \sin x \cos x=\sin 2 x]$
$\frac{d y}{d x}=\frac{2}{\sin 2 x}$ $\left[\therefore \frac{1}{\sin x}=\operatorname{cosec} x\right]$
$\frac{d y}{d x}=2 \operatorname{cosec} 2 x$
∴ Proved

Differentiation exercise 10.2 question 69

Answer: Proved
Hint: you must know the rules of solving derivative of inverse trigonometric functions.
Given: $y=x \sin ^{-1} x+\sqrt{1-x^{2}}$

Prove: $\frac{d y}{d x}=\sin ^{-1} x$

Solution:
$y=x \sin ^{-1} x+\sqrt{1-x^{2}}$
Differentiate with respect to x,use product rule
$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(x \sin ^{-1} x\right)+\frac{d}{d x}\left(\sqrt{1-x^{2}}\right) \\\\ &\frac{d y}{d x}=(x) \frac{d}{d x} \sin ^{-1} x+\sin ^{-1} x \frac{d}{d x}(x)+\frac{d}{d x}\left(\sqrt{1-x^{2}}\right) \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}=\left(\frac{x}{\sqrt{1-x^{2}}}+\sin ^{-1} x\right)-\frac{2 x}{2 \sqrt{1-x^{2}}} \\\\ &\frac{d y}{d x}=\frac{x}{\sqrt{1-x^{2}}}+\sin ^{-1} x-\frac{x}{\sqrt{1-x^{2}}} \\\\ &\frac{d y}{d x}=\sin ^{-1} x \end{aligned}$
∴ Proved

Differentiation exercise 10.2 question 70

Answer:Proved
Hint: you must know the rules of solving derivatives.
Given: $y=\sqrt{x^{2}+a^{2}}$

Prove: $y \frac{d y}{d x}-x=0$

Solution:
$y=\sqrt{x^{2}+a^{2}}$
Squaring both sides,
$y^{2}=x^{2}+a^{2}$
Differentiate both sides,
$\begin{aligned} &2 y \frac{d y}{d x}=\frac{d}{d x}\left(x^{2}+a^{2}\right) \\\\ &2 y \frac{d y}{d x}=2 x+0 \end{aligned}$
$\begin{aligned} &y \frac{d y}{d x}=x\\\\ &\text { Or }\\\\ &y \frac{d y}{d x}-x=0 \end{aligned}$

∴ Proved

Differentiation exercise 10.2 question 71

Answer:Proved
Hint: you must know the rules of solving derivatives of exponential functions.
Given: $y=e^{x}+e^{-x}$
Prove: $\frac{d y}{d x}=\sqrt{y^{2}-4}$

Solution:
$y=e^{x}+e^{-x}$
Differentiate with respect to x,
$\frac{d y}{d x}=\frac{d}{d x}\left(e^{x}+e^{-x}\right)$ $\left[\therefore \frac{d}{d x} e^{x}=e^{x} ; \frac{d}{d x} e^{-x}=-e^{-x}\right]$
$\begin{aligned} &\frac{d y}{d x}=e^{x}-e^{-x} \\\\ &\frac{d y}{d x}=\sqrt{\left(e^{x}-e^{-x}\right)^{2}-4 e^{x} \times e^{-x}} \end{aligned}$ $\left[\therefore(a-b)=\sqrt{\left(a^{2}+b^{2}\right)-2 a b}=\sqrt{(a+b)^{2}-4 a b}\right]$
$\frac{d y}{d x}=\sqrt{y^{2}-4}$ $[\because \left.e^{x}+e^{-x}=y\right]$
∴ Proved

Differentiation exercise 10.2 question 72

Answer:Proved
Hint: you must know the rules of solving derivatives.
Given: $y=\sqrt{a^{2}-x^{2}}$

Prove: $y \frac{d y}{d x}+x=0$

Solution:
$y=\sqrt{a^{2}-x^{2}}$
Squaring both sides,
$y^{2}=a^{2}-x^{2}$
Differentiate with respect to x,
$\begin{aligned} &2 y \frac{d y}{d x}=0-2 x \\\\ &2 y \frac{d y}{d x}=-2 x \end{aligned}$
$\begin{aligned} &y \frac{d y}{d x}=-x \\\\ &y \frac{d y}{d x}+x=0 \end{aligned}$

∴ Proved

Differentiation exercise 10.2 question 73

Answer:Proved
Hint: you must know the rules of solving derivatives.
Given: $x y=4$

Prove: $x\left(\frac{d y}{d x}+y^{2}\right)=3 y$

Solution:
$\begin{gathered} x y=4 \\\\ y=\frac{4}{x} \end{gathered}$
Differentiate with respect to x,
$\begin{aligned} &\frac{d y}{d x}=4 \frac{d}{d x}\left(x^{-1}\right) \\\\ &\frac{d y}{d x}=4(-1) \times\left(x^{-1-1}\right) \end{aligned}$
$\frac{d y}{d x}=-\frac{4}{x^{2}}$
$\frac{d y}{d x}=-\frac{y^{2}}{4}$ ( multiplying by 4 in num. & den.)

$\begin{aligned} &4 \frac{d y}{d x}=-y^{2} \\\\ &4 \frac{d y}{d x}=3 y^{2}-4 y^{2} \\\\ &4 \frac{d y}{d x}+4 y^{2}=3 y^{2} \\\\ &4\left(\frac{d y}{d x}+4 y^{2}\right)=3 y^{2} \end{aligned}$
Dividing both sides with x,
$\begin{aligned} &\frac{4}{x}\left(\frac{d y}{d x}+y^{2}\right)=\frac{3 y^{2}}{x} \\\\ &y\left(\frac{d y}{d x}+y^{2}\right)=\frac{3 y^{2}}{x} \\\\ &x\left(\frac{d y}{d x}+y^{2}\right)=\frac{3 y^{2}}{y} \\ &x\left(\frac{d y}{d x}+y^{2}\right)=3 y \end{aligned}$
∴ Proved

Differentiation exercise 10.2 question 74

Answer:Proved
Hint: you must know the rules of solving derivatives.
Given: Prove that
$\frac{d}{d x}\left\{\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}\right\}=\sqrt{a^{2}-x^{2}}$
Solution:
$\text { L.H.S } \Rightarrow \frac{d}{d x}\left\{\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}\right\}$
$\Rightarrow \frac{d}{d x}\left(\frac{x}{2} \sqrt{a^{2}-x^{2}}\right)+\frac{d}{d x}\left(\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}\right)$
$\Rightarrow \frac{1}{2}\left[x \frac{d}{d x} \sqrt{a^{2}-x^{2}}+\sqrt{a^{2}-x^{2}} \frac{d}{d x}(x)\right]+\frac{a^{2}}{2} \times \frac{1}{\sqrt{1-\left(\frac{x}{a}\right)^{2}}}\left(\frac{1}{a}\right)$
$\Rightarrow \frac{1}{2}\left[x \times \frac{1}{2 \sqrt{a^{2}-x^{2}}} \frac{d}{d x}\left(a^{2}-x^{2}\right)+\sqrt{a^{2}-x^{2}}\right]+\frac{a^{2}}{2} \times \frac{1}{\sqrt{\frac{a^{2}-x^{2}}{a^{2}}}}\left(\frac{1}{a}\right)$
$\Rightarrow \frac{1}{2}\left[\frac{-2 x^{2}}{2 \sqrt{a^{2}-x^{2}}}+\sqrt{a^{2}-x^{2}}\right]+\left(\frac{a^{2}}{2}\right) \frac{a}{\sqrt{a^{2}-x^{2}}} \times\left(\frac{1}{a}\right)$
$\Rightarrow \frac{1}{2}\left[\frac{-2 x^{2}+2\left|a^{2}-x^{2}\right|}{2 \sqrt{a^{2}-x^{2}}}\right]+\frac{a^{2}}{2 \sqrt{a^{2}-x^{2}}}$
$\Rightarrow \frac{a^{2}-x^{2}-x^{2}}{2 \sqrt{a^{2}-x^{2}}}+\frac{a^{2}}{2 \sqrt{a^{2}-x^{2}}}$
$\Rightarrow \frac{2 a^{2}-2 x^{2}}{2 \sqrt{a^{2}-x^{2}}} \quad[\therefore x=\sqrt{x} \times \sqrt{x}]$
$\begin{aligned} &\Rightarrow \frac{\left(a^{2}-x^{2}\right)}{\sqrt{a^{2}-x^{2}}} \\\\ &\Rightarrow \sqrt{a^{2}-x^{2}} \quad \Rightarrow \text { R.H.S } \end{aligned}$

∴ Proved

Differentiation exercise 10.2 question 75

Answer: $\frac{2}{3}$
Hint: you must know the rules of solving derivatives of trigonometric function
Given: $f(x)=\sqrt{\frac{\sec x-1}{\sec x+1}}$

Find : $f^{\prime}\left(\frac{\pi}{3}\right)$

Solution:
$f(x)=\sqrt{\frac{\sec x-1}{\sec x+1}}$
$=\sqrt{\frac{1-\cos x}{1+\cos x}} \quad\left[\therefore \sec x=\frac{1}{\cos x}\right]$

Now rationalize
$=\sqrt{\frac{1-\cos x}{1+\cos x} \times \frac{1-\cos x}{1-\cos x}}$
$f(x)=\frac{1-\cos x}{\sin x}$
$=\frac{1}{\sin x}-\frac{\cos x}{\sin x}$
$f(x)=\operatorname{cosec} x-\cot x$

Differentiate with respect to x,
$\begin{aligned} &f^{\prime}(x)=-\operatorname{cosec} x \cot x-\left(-\operatorname{cosec}^{2} x\right)\\ \\ &f^{\prime}\left(\frac{\pi}{3}\right)=-\operatorname{cosec}\left(\frac{\pi}{3}\right) \cot \left(\frac{\pi}{3}\right)+\operatorname{cosec}^{2}\left(\frac{\pi}{3}\right) \end{aligned}$
$\begin{aligned} &=\frac{-2}{\sqrt{3}} \times \frac{1}{\sqrt{3}}+\left(\frac{2}{\sqrt{3}}\right)^{2} \\\\ &=\frac{-2}{3}+\frac{4}{3} \\\\ &\Rightarrow \frac{-2+4}{3} \Rightarrow \frac{2}{3} \end{aligned}$

Differentiation exercise 10.2 question 76

Answer: $\frac{2}{\pi }$
Hint: you must know the rules of solving derivative of trigonometric functions
Given: $f(x)=\sqrt{\tan \sqrt{x}}$

Find: $f^{'}\left(\frac{\pi^{2}}{16}\right)$

Solution:
$f(x)=\sqrt{\tan (\sqrt{x})}$
Differentiate with respect to x,
$f^{\prime}(x)=\frac{1}{2 \sqrt{\tan (\sqrt{x})}} \frac{d}{d x} \tan \sqrt{x}$
$=\frac{1}{2 \sqrt{\tan (\sqrt{x})}} \sec ^{2} \sqrt{x} \times \frac{1}{2 \sqrt{x}}$
$f^{\prime}(x) \Rightarrow \frac{\sec ^{2} \sqrt{x}}{4 \sqrt{x \tan (\sqrt{x})}}$
Now, $f^{\prime}\left(\frac{\pi^{2}}{16}\right)=\frac{\sec ^{2} \sqrt{\frac{\pi^{2}}{16}}}{4 \sqrt{\frac{\pi^{2}}{16} \tan \left(\sqrt{\frac{\pi^{2}}{16}}\right)}}$
$=\frac{\sec ^{2}\left(\frac{\pi}{4}\right)}{4\left(\frac{\pi}{4}\right) \sqrt{\tan \left(\frac{\pi}{4}\right)}}$
$\Rightarrow \frac{\sec ^{2}\left(\frac{\pi}{4}\right)}{\pi \times(1)} \quad\left[B u t \tan \frac{\pi}{4} \quad \Rightarrow 1\right]$
$\begin{aligned} &\Rightarrow \frac{\sec ^{2}\left(\frac{\pi}{4}\right)}{\pi} \\\\ &\Rightarrow \frac{\left(\sec \left(\frac{\pi}{4}\right)\right)^{2}}{\pi} \end{aligned}$
$\begin{aligned} &=\frac{(\sqrt{2})^{2}}{\pi} \\\\ &f^{\prime}\left(\frac{\pi^{2}}{16}\right)=\frac{2}{\pi} \end{aligned}$

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