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

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, $f^{\prime}\left(\frac{\pi^{2}}{16}\right)=\frac{\sec ^{2} \sqrt{\frac{\pi^{2}}{16}}}{4 \sqrt{\frac{\pi^{2}}{16} \tan \left(\sqrt{\frac{\pi^{2}}{16}}\right)}}$
$=\frac{\sec ^{2}\left(\frac{\pi}{4}\right)}{4\left(\frac{\pi}{4}\right) \sqrt{\tan \left(\frac{\pi}{4}\right)}}$
$\Rightarrow \frac{\sec ^{2}\left(\frac{\pi}{4}\right)}{\pi \times(1)} \quad\left[B u t \tan \frac{\pi}{4} \quad \Rightarrow 1\right]$
$\begin{aligned} &\Rightarrow \frac{\sec ^{2}\left(\frac{\pi}{4}\right)}{\pi} \\\\ &\Rightarrow \frac{\left(\sec \left(\frac{\pi}{4}\right)\right)^{2}}{\pi} \end{aligned}$
$\begin{aligned} &=\frac{(\sqrt{2})^{2}}{\pi} \\\\ &f^{\prime}\left(\frac{\pi^{2}}{16}\right)=\frac{2}{\pi} \end{aligned}$