RD Sharma Class 12 Exercise 10.8 Differentiation Solutions Maths

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RD Sharma Class 12 Solutions Chapter 10 Differentiation - Other Exercise
Differentiation Excercise: 10.8
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Also Read - RD Sharma Solution for Class 9 to 12 Maths
The RD Sharma class 12th exercise 10.8 is a top choice for hundreds of students who have excelled in their exams after practicing the book. The 10th chapter of the Class 12 maths book has the topic Differentiations. This part will discuss in detail the basic concept of differentiation of inverse trigonometric functions. It will also teach students about Differentiation of a function with respect to another function, Differentiation by using trigonometric substitutions, Differentiation of implicit functions, and the like. Exercise 10.8 includes 15 on linear trigonometric equations that need to be solved by students.

RD Sharma Class 12 Solutions Chapter 10 Differentiation - Other Exercise

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Differentiation Excercise: 10.8

Differentiation exercise 10.8 question 1

Answer: $2 \sec ^{2}\left(x^{2}\right) \tan \left(x^{2}\right)$
Hint:
$\begin{aligned} &\text { Let } u=\sec ^{2}\left(x^{2}\right) ; v=x^{2} \\\\ &\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}} \end{aligned}$

Given: $\sec ^{2}\left(x^{2}\right) \text { w.r.t } x^{2}$
Explanation:
$\begin{aligned} &\text { Let } u=\sec ^{2}\left(x^{2}\right), v=x^{2} \\\\ &\frac{d u}{d x}=2 \sec \left(x^{2}\right)\left[\sec x^{2} \tan x^{2} \times 2 x\right] \end{aligned}$
$=4 x \sec ^{2}\left(x^{2}\right) \tan \left(x^{2}\right)$ (chain rule)
$\begin{aligned} &\frac{d v}{d x}=2 x \\\\ &\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{4 x \sec ^{2}\left(x^{2}\right) \tan \left(x^{2}\right)}{2 x} \end{aligned}$
$=2 \sec ^{2}\left(x^{2}\right) \tan \left(x^{2}\right)$

Differentiation exercise 10.8 question 3

Answer: $x(\log x)^{x-1}\{1+\log x \cdot \log (\log x)\}$
Hint: $\text { Let } u=(\log x)^{x}, v=\log x$
$\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}$

Given: $(\log x)^{x} \text { w.r.t } \log x$
Explanation:
$\text { Let } u=(\log x)^{x}, v=\log x$
$u=(\log x)^{x}$
Taking log both sides,
$\begin{aligned} &\log u=\log (\log x)^{x} \\\\ &\log u=x \log (\log x) \end{aligned}$
Differentiate both sides w.r.t. $x$
$\begin{aligned} &\frac{1}{u} \frac{d u}{d x}=x\left[\frac{1}{\log x} \times \frac{1}{x}\right]+\log (\log x) \times 1 \\\\ &\frac{1}{u} \frac{d u}{d x}=(\log x)^{-1}+\log (\log x) \\\\ &\frac{d u}{d x}=u\left[(\log x)^{-1}+\log (\log x)\right] \end{aligned}$
$\begin{aligned} &\frac{d u}{d x}=u\left[\frac{1}{\log x}+\log (\log x)\right] \\\\ &\frac{d u}{d x}=\frac{u}{\log x}[1+\log x \log (\log x)] \\\\ &\frac{d u}{d x}=(\log x)^{x-1}[1+\log x \cdot \log (\log x)] \end{aligned}$
$\begin{aligned} &v=\log x \\\\ &\frac{d v}{d x}=\frac{1}{x} \end{aligned}$
$\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{(\log x)^{x-1}[1+\log x \cdot \log (\log x)]}{\frac{1}{x}}$
$=x(\log x)^{x-1}\{1+\log x \cdot \log (\log x)\}$

Differentiation exercise 10.8 question 4(i)

Answer: 1
Hint: $u=\sin ^{-1} \sqrt{1-x^{2}}, v=\cos ^{-1} x$
$\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}$

Given: $\sin ^{-1} \sqrt{1-x^{2}} \text { w.r.t } \cos ^{-1} x, x \in(0,1)$
Explanation:
$\text { Let } u=\sin ^{-1} \sqrt{1-x^{2}}, v=\cos ^{-1} x$
$\begin{aligned} &\text { Let } u=\sin ^{-1} \sqrt{1-x^{2}}, v=\cos ^{-1} x \\\\ &u=\sin ^{-1} \sqrt{1-x^{2}} \\\\ &\text { Let } x=\cos \theta \\\\ &u=\sin ^{-1} \sqrt{1-\cos ^{2} \theta} \\\\ &u=\sin ^{-1}(\sin \theta) \end{aligned}$
$\begin{aligned} &x \in(0,1) \\\\ &\cos \theta \in(0,1) \quad \theta \in\left(0, \frac{\pi}{2}\right) \\\\ &u=\sin ^{-1}(\sin \theta)=\theta \end{aligned}$
$\begin{aligned} &u=\cos ^{-1} x \\\\ &\frac{d u}{d x}=\frac{-1}{\sqrt{1-x^{2}}} \\\\ &v=\cos ^{-1} x \end{aligned}$
$\begin{aligned} &\frac{d v}{d x}=\frac{-1}{\sqrt{1-x^{2}}} \\\\ &\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{-1}{\frac{\sqrt{1-x^{2}}}{\sqrt{1-x^{2}}}}=1 \end{aligned}$

Differentiation exercise 10.8 question 4(ii)

Answer: $-1$
Hint: $\text { Let } u=\sin ^{-1} \sqrt{1-x^{2}}, v=\cos ^{-1} x$
$\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}$

Given: $\sin ^{-1} \sqrt{1-x^{2}} \text { w.r.t } \cos ^{-1} x, x \in(-1,0)$
Explanation:
$\text { Let } u=\sin ^{-1} \sqrt{1-x^{2}}, v=\cos ^{-1} x$
$\begin{aligned} &u=\sin ^{-1} \sqrt{1-x^{2}} \\\\ &\text { Let } x=\cos \theta \\\\ &u=\sin ^{-1} \sqrt{1-\cos ^{2} \theta} \end{aligned}$
$\begin{aligned} &u=\sin ^{-1}(\sin \theta) \\\\ &x \in(-1,0) \\\\ &\cos \theta \in(-1,0) \quad \theta \in\left(-\frac{\pi}{2}, 0\right) \end{aligned}$
$\begin{aligned} &u=\sin ^{-1}(\sin \theta)=-\theta \quad \theta \in\left(-\frac{\pi}{2}, 0\right) \\\\ &u=-\cos ^{-1} x \end{aligned}$
$\begin{aligned} &\frac{d u}{d x}=\frac{-1}{\sqrt{1-x^{2}}} \\\\ &v=\cos ^{-1} x \\\\ &\frac{d v}{d x}=\frac{1}{\sqrt{1-x^{2}}} \end{aligned}$
$\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{\frac{-1}{\sqrt{1-x^{2}}}}{\frac{1}{\sqrt{1-x^{2}}}}=-1$

Differentiation exercise 10.8 question 5(i)

Answer: $\frac{-1}{x}$
Hint: $\text { Let } u=\sin ^{-1}\left(4 x \sqrt{1-4 x^{2}}\right), v=\sqrt{1-4 x^{2}}$
Given: $\sin ^{-1}\left(4 x \sqrt{1-4 x^{2}}\right) \text { w.r.t } \sqrt{1-4 x^{2}}$
Explanation:
$\text { Let } 2 x=\cos \theta$
$\begin{aligned} &u=\sin ^{-1}\left(4 x \sqrt{1-4 x^{2}}\right) \\\\ &u=\sin ^{-1}\left(2 \cos \theta \sqrt{1-(\cos \theta)^{2}}\right) \end{aligned}$
$\begin{aligned} &=\sin ^{-1}\left(2 \cos \theta \sqrt{1-\cos ^{2} \theta}\right) \\\\ &=\sin ^{-1}(2 \cos \theta \sin \theta) \\\\ &=\sin ^{-1}(\sin 2 \theta) \end{aligned}$
Now we try to find range of $\theta$
$x \in\left(\frac{-1}{2 \sqrt{2}}, \frac{1}{2 \sqrt{2}}\right)$
$2 x \in\left(\frac{-1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$
$\cos \theta \in\left(\frac{-1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\; \; \; \; \; \; \; \; \; \left[\begin{array}{l} \cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4} \\ \& \cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)=\frac{3 \pi}{4} \end{array}\right]$
$\begin{aligned} &\theta \in\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right) \\\\ &2 \theta \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right) \end{aligned}$
$\begin{aligned} &u=\sin ^{-1}(\sin 2 \theta) \\\\ &=\pi-2 \theta \\\\ &=\pi-2 \cos ^{-1}(2 x) \end{aligned}$ $\left[\begin{array}{l} 2 x=\cos \theta \\ \theta=\cos ^{-1}(2 x) \end{array}\right]$
$\frac{d u}{d x}=0-2\left[\frac{-1}{\sqrt{1-(2 x)^{2}}} \times 2\right]$
$=\frac{4}{\sqrt{1-x^{2}}}$
$\begin{aligned} &v=\sqrt{1-4 x^{2}} \\\\ &\frac{d v}{d x}=\frac{1}{2 \sqrt{1-4 x^{2}}}(-8 x) \\\\ &=\frac{-4 x}{\sqrt{1-x^{2}}} \end{aligned}$
$\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{\frac{4}{\sqrt{1-x^{2}}}}{\frac{-4 x}{\sqrt{1-x^{2}}}}=\frac{-1}{x}$

Differentiation exercise 10.8 question 5(ii)

Answer: $\frac{1}{x}$
Hint: $\text { : Let } u=\sin ^{-1}\left(4 x \sqrt{1-4 x^{2}}\right), v=\sqrt{1-4 x^{2}}$

Given: $\sin ^{-1}\left(4 x \sqrt{1-4 x^{2}}\right) \text { w.r.t } \sqrt{1-4 x^{2}}$

Explanation: $\text { Let } 2 x=\cos \theta$
$\begin{aligned} &u=\sin ^{-1}\left(4 x \sqrt{1-4 x^{2}}\right) \\\\ &u=\sin ^{-1}\left(2 \cos \theta \sqrt{1-(\cos \theta)^{2}}\right) \end{aligned}$
$\begin{aligned} &=\sin ^{-1}\left(2 \cos \theta \sqrt{1-\cos ^{2} \theta}\right) \\\\ &=\sin ^{-1}(2 \cos \theta \sin \theta) \\\\ &=\sin ^{-1}(\sin 2 \theta) \end{aligned}$
Now we try to find the range of $\theta$
$\begin{aligned} &x \in\left(\frac{1}{2 \sqrt{2}}, \frac{1}{2}\right) \\\\ &2 x \in\left(\frac{1}{\sqrt{2}}, 1\right) \\\\ &\cos \theta \in\left(\frac{1}{\sqrt{2}}, 1\right) \end{aligned}$
$\begin{aligned} &\theta \in\left(0, \frac{\pi}{4}\right) \\\\ &2 \theta \in\left(0, \frac{\pi}{2}\right) \\\\ &u=\sin ^{-1}(\sin 2 \theta) \end{aligned}$
$=2 \theta \quad\left[\begin{array}{l} 2 x=\cos \theta \\ \theta=\cos ^{-1}(2 x) \end{array}\right]$
$\begin{aligned} &u=2 \cos ^{-1}(2 x) \\\\ &\frac{d u}{d x}=2\left[\frac{-1}{\sqrt{1-(2 x)^{2}}} \times 2\right] \end{aligned}$
$=\frac{-4}{\sqrt{1-4 x^{2}}}$
$\begin{aligned} &v=\sqrt{1-4 x^{2}} \\\\ &\frac{d v}{d x}=\frac{1}{2 \sqrt{1-4 x^{2}}}(-8 x) \end{aligned}$
$=\frac{-4 x}{\sqrt{1-4 x^{2}}}$
$\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{\frac{-4}{\sqrt{1-x^{2}}}}{\frac{-4 x}{\sqrt{1-4 x^{2}}}}=\frac{1}{x}$

Differentiation exercise 10.8 question 5(iii)

Answer: $-\frac{1}{x}$
Hint: $\text { Let } u=\sin ^{-1}\left(4 x \sqrt{1-4 x^{2}}\right), v=\sqrt{1-4 x^{2}}$

Given: $\sin ^{-1}\left(4 x \sqrt{1-4 x^{2}}\right) \text { w.r.t } \sqrt{1-4 x^{2}}$
Explanation: $\text { Let } 2 x=\cos \theta$
$\begin{aligned} &u=\sin ^{-1}\left(4 x \sqrt{1-4 x^{2}}\right) \\\\ &u=\sin ^{-1}\left(2 \cos \theta \sqrt{1-(\cos \theta)^{2}}\right) \end{aligned}$
$\begin{aligned} &=\sin ^{-1}\left(2 \cos \theta \sqrt{1-\cos ^{2} \theta}\right) \\\\ &=\sin ^{-1}(2 \cos \theta \sin \theta) \\\\ &=\sin ^{-1}(\sin 2 \theta) \end{aligned}$
$u=2 \theta \quad\left[\begin{array}{l} 2 x=\cos \theta \\\; \theta=\cos ^{-1}(2 x) \end{array}\right]$
$\begin{aligned} &=2 \cos ^{-1}(2 x) \\\\ &\frac{d u}{d x}=2\left[\frac{-1}{\sqrt{1-(2 x)^{2}}} \times 2\right] \end{aligned}$
$=\frac{4}{\sqrt{1-4 x^{2}}}$
$\begin{aligned} &v=\sqrt{1-4 x^{2}} \\\\ &\frac{d v}{d x}=\frac{1}{2 \sqrt{1-4 x^{2}}}(-8 x) \end{aligned}$
$=\frac{-4 x}{\sqrt{1-4 x^{2}}}$
$\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{\frac{4}{\sqrt{1-4 x^{2}}}}{\frac{-4 x}{\sqrt{1-4 x^{2}}}}=\frac{-1}{x}$

Differentiation exercise 10.8 question 6

Answer: $\frac{1}{4}$
Hint: $\text { Let } u=\tan ^{-1}\left[\frac{\sqrt{1+x^{2}}-1}{x}\right]$ ,
$v=\sin ^{-1}\left[\frac{2 x}{1+x^{2}}\right]$
Given: $\tan ^{-1}\left[\frac{\sqrt{1+x^{2}}-1}{x}\right] \text { w.r.t } \sin ^{-1}\left[\frac{2 x}{1+x^{2}}\right]$
$-1<x<1, x \neq 0$
Explanation: $\text { Let } x=\tan \theta$
$u=\tan ^{-1}\left[\frac{\sqrt{1+x^{2}}-1}{x}\right]$
$\begin{aligned} &=\tan ^{-1}\left[\frac{\sqrt{1+\tan ^{2} \theta}-1}{\tan \theta}\right] \\\\ &=\tan ^{-1}\left[\frac{\sqrt{\sec ^{2} \theta}-1}{\tan \theta}\right] \\\\ &=\tan ^{-1}\left[\frac{\sec \theta-1}{\tan \theta}\right] \end{aligned}$
$\begin{aligned} &=\tan ^{-1}\left[\frac{\frac{1}{\cos \theta}-1}{\frac{\sin \theta}{\cos \theta}}\right] \\\\ &=\tan ^{-1}\left[\frac{1-\cos \theta}{\sin \theta}\right] \end{aligned}$ .......................(1)
$\begin{aligned} &\text { Now } \cos 2 \theta=1-2 \sin ^{2} \theta \\\\ &2 \sin ^{2} \theta=1-\cos 2 \theta \\\\ &2 \sin ^{2} \frac{\theta}{2}=1-\cos \theta \end{aligned}$
$\begin{aligned} &\sin 2 \theta=2 \sin \theta \cos \theta \\\\ &\sin \theta=2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \end{aligned}$
Put in (1)
$\begin{aligned} &u=\tan ^{-1}\left[\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right] \\\\ &u=\tan ^{-1}\left[\tan \frac{\theta}{2}\right] \end{aligned}$
Now,
$\begin{aligned} &-1<x<1 \\\\ &-1<\tan \theta<1 \\\\ &-\frac{\pi}{4}<\theta<\frac{\pi}{4} \end{aligned}$ .........(2)
$\begin{aligned} &-\frac{\pi}{8}<\frac{\theta}{2}<\frac{\pi}{8} \\\\ &u=\frac{\theta}{2} \text { as } \frac{\theta}{2} \in\left(-\frac{\pi}{8}, \frac{\pi}{8}\right) \end{aligned}$
$\begin{aligned} &u=\frac{\tan ^{-1} x}{2} \\\\ &\frac{d u}{d x}=\frac{1}{2}\left[\frac{1}{1+x^{2}}\right] \end{aligned}$
$=\frac{1}{2\left(1+x^{2}\right)}$
$\begin{aligned} &v=\sin ^{-1}\left[\frac{2 x}{1+x^{2}}\right] \\\\ &x=\tan \theta \\\\ &v=\sin ^{-1}\left[\frac{2(\tan \theta)}{1+\tan ^{2} \theta}\right] \end{aligned}$
$=\sin ^{-1}(\sin 2 \theta)$
$-\frac{\pi}{4}<\theta<\frac{\pi}{4}$ ........From (2)
$-\frac{\pi}{2}<2 \theta<\frac{\pi}{2}$
$v=\sin ^{-1}(\sin 2 \theta) \; \; \; \; \; \quad 2 \theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
$=2 \theta$
$\begin{aligned} &v=2 \tan ^{-1} x \\\\ &\frac{d v}{d x}=\frac{2}{1+x^{2}} \end{aligned}$
$\begin{aligned} &\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{\frac{1}{2\left(1+x^{2}\right)}}{\frac{2}{1+x^{2}}} \\\\ &=\frac{1}{4} \end{aligned}$

Differentiation exercise 10.8 question 7(i)

Answer: 2
Hint: $\text { Let } u=\sin ^{-1}\left[2 x \sqrt{1-x^{2}}\right]$
$v=\sec ^{-1}\left[\frac{1}{\sqrt{1-x^{2}}}\right]$

Given: $\sin ^{-1}\left[2 x \sqrt{1-x^{2}}\right] \text { w.r.t } \sec ^{-1}\left[\frac{1}{\sqrt{1-x^{2}}}\right]$
$x \in\left(0, \frac{1}{\sqrt{2}}\right)$
Explanation:
$\text { Let } x=\sin \theta$
$\begin{aligned} &u=\sin ^{-1}\left[2 x \sqrt{1-x^{2}}\right] \\ &v=\sec ^{-1}\left[\frac{1}{\sqrt{1-x^{2}}}\right] \end{aligned}$
$u=\sin ^{-1}\left[2 \sin \theta \sqrt{1-\sin ^{2} \theta}\right]$
$\begin{aligned} &=\sin ^{-1}[2 \sin \theta \cos \theta] \\\\ &=\sin ^{-1}[\sin 2 \theta] \end{aligned}$
$\begin{aligned} &x \in\left(0, \frac{1}{\sqrt{2}}\right) \\\\ &\sin \theta \in\left(0, \frac{1}{\sqrt{2}}\right) \\\\ &\theta \in\left(0, \frac{\pi}{4}\right) \quad 2 \theta \in\left(0, \frac{\pi}{2}\right) \end{aligned}$
$\begin{aligned} &u=\sin ^{-1}[\sin 2 \theta]=2 \theta \quad \text { when } 2 \theta \in\left(0, \frac{\pi}{2}\right) \\\\ &u=2 \sin ^{-1} x \end{aligned}$
$=\frac{2}{\sqrt{1-x^{2}}}$
$\begin{aligned} &v=\sec ^{-1}\left[\frac{1}{\sqrt{1-\sin ^{2} \theta}}\right] \\\\ &\sec ^{-1}\left(\frac{1}{\cos \theta}\right)=\sec ^{-1}(\sec \theta) \\\\ &=\theta \quad \text { when } \theta \in\left(0, \frac{\pi}{4}\right) \end{aligned}$
$\begin{aligned} &=\sin ^{-1} x \\\\ &v=\sin ^{-1} x \\\\ &\frac{d v}{d x}=\frac{1}{\sqrt{1-x^{2}}} \end{aligned}$
$\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{\frac{2}{\sqrt{1-x^{2}}}}{\frac{1}{\sqrt{1-x^{2}}}}=2$

Differentiation exercise 10.8 question 7(ii)

Answer: $-2$
Hint: $\text { Let } u=\sin ^{-1}\left[2 x \sqrt{1-x^{2}}\right]$
$v=\sec ^{-1}\left[\frac{1}{\sqrt{1-x^{2}}}\right]$
Given: $\sin ^{-1}\left[2 x \sqrt{1-x^{2}}\right] \text { w.r.t } \sec ^{-1}\left[\frac{1}{\sqrt{1-x^{2}}}\right]$
$x \in\left(\frac{1}{\sqrt{2}}, 1\right)$
Explanation:
$\begin{aligned} &x \in\left(\frac{1}{\sqrt{2}}, 1\right) \\\\ &\sin \theta \in\left(\frac{1}{\sqrt{2}}, 1\right) \\\\ &\theta \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right) \quad 2 \theta \in\left(\frac{\pi}{2}, \pi\right) \end{aligned}$
$\begin{aligned} &\pi-2 \theta \in\left(0, \frac{\pi}{2}\right) \\\\ &u=\sin ^{-1}[\sin 2 \theta] \quad \text { when } \pi-2 \theta \in\left(0, \frac{\pi}{2}\right) \\\\ &u=\sin ^{-1}[\sin (\pi-2 \theta)] \end{aligned}$
$\begin{aligned} &=\pi-2 \theta \\\\ &u=\pi-2 \sin ^{-1} x \\\\ &\frac{d u}{d x}=\frac{-2}{\sqrt{1-x^{2}}} \\\\ &v=\sec ^{-1}\left[\frac{1}{\sqrt{1-\sin ^{2} \theta}}\right] \end{aligned}$
$\begin{aligned} &\sec ^{-1}\left(\frac{1}{\cos \theta}\right)=\sec ^{-1}(\sec \theta) \\\\ &=\theta \quad \text { when } \theta \in\left(0, \frac{\pi}{4}\right) \end{aligned}$
$\begin{aligned} &=\sin ^{-1} x \\\\ &v=\sin ^{-1} x \\\\ &\frac{d v}{d x}=\frac{1}{\sqrt{1-x^{2}}} \end{aligned}$
$\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{\frac{-2}{\sqrt{1-x^{2}}}}{\frac{1}{\sqrt{1-x^{2}}}}=-2$

Differentiation exercise 10.8 question 8

Answer: $\frac{(\cos x)^{\sin x}\{\cos x \cdot \log \cos x-\sin x \tan x\}}{(\sin x)^{\cos x}\{-\sin x \log \sin x+\cos x \cot x\}}$

Hint: $\text { Let } u=(\cos x)^{\operatorname{sin} x}, v=(\sin x)^{\cos x}$

Given: $(\cos x)^{\sin x} \text { w.r.t }(\sin x)^{\cos x}$
Explanation: Apply log on both sides
$\begin{aligned} &\log u=\log (\cos x)^{\sin x} \\\\ &\log u=\sin x \log (\cos x) \end{aligned}$
Differentiate both side w.r.t. $x$
$\frac{1}{u} \frac{d u}{d x}=\sin x\left[\frac{1}{\cos x}(-\sin x)\right]+\log (\cos x) \cos x$
$=-\sin x \tan x+\cos x \log (\cos x)$
$\begin{aligned} &\frac{d u}{d x}=(\cos x)^{\sin x}\{-\sin x \tan x+\cos x \cdot \log \cos x\} \\\\ &v=(\sin x)^{\cos x} \end{aligned}$
Apply log on both sides
$\begin{aligned} &\log v=\log (\sin x)^{\cos x} \\\\ &\log v=\cos x \cdot \log (\sin x) \end{aligned}$
Differentiate both side w.r.t. $x$
$\frac{1}{v} \frac{d v}{d x}=\cos x\left[-\frac{1}{\sin x}(\cos x)\right]+\log (\sin x)(-\sin x)$
$=\cos x \cot x-\sin x \log \sin x$
$\begin{aligned} &\frac{d v}{d x}=(\sin x)^{\cos x}\{-\sin x \log \sin x+\cos x \cot x\} \\\\ &\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}} \end{aligned}$
$=\frac{(\cos x)^{\sin x}\{\cos x \cdot \log \cos x-\sin x \tan x\}}{(\sin x)^{\cos x}\{-\sin x \log \sin x+\cos x \cot x\}}$

Differentiation exercise 10.8 question 9

Answer: 1
Hint: $\text { Let } u=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right), v=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$
Given: $\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) \text { w.r.t } \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$
$0<x<1$
Explanation: $\text { Let } u=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$
$\begin{aligned} &v=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) \\\\ &\text { Let } x=\tan \theta \\\\ &u=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right) \end{aligned}$
$\begin{aligned} &=\sin ^{-1}(\sin 2 \theta) \\\\ &0<x<1 \\\\ &0<\tan \theta<1 \\\\ &0<\theta<\frac{\pi}{4} \\ &0<2 \theta<\frac{\pi}{2} \end{aligned}$
$\begin{aligned} &u=\sin ^{-1}(\sin 2 \theta)=2 \theta\; \; \; \; \; \; \quad 2 \theta \in\left(0, \frac{\pi}{2}\right) \\ &u=2 \tan ^{-1} x \end{aligned}$
$\begin{aligned} &\frac{d u}{d x}=\frac{2}{1+x^{2}} \\\\ &v=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) \end{aligned}$
$\begin{aligned} &v=\cos ^{-1}\left(\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right) \\\\ &=\cos ^{-1}\left(\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right) \\\\ &=\cos ^{-1}(\cos 2 \theta) \end{aligned}$
$\begin{aligned} &=2 \theta\; \; \; \; \; \; \; \; \quad 2 \theta \in\left(0, \frac{\pi}{2}\right) \\\\ &v=2 \tan ^{-1} x \\\\ &\frac{d v}{d x}=\frac{2}{1+x^{2}} \end{aligned}$
$\begin{aligned} &\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}} \\\\ &=\frac{\frac{2}{1+x^{2}}}{\frac{2}{1+x^{2}}}=1 \end{aligned}$

Differentiation exercise 10.8 question 10

Answer: $\frac{1}{a x \sqrt{1+a^{2} x^{2}}}$
Hint: $\text { Let } u=\tan ^{-1}\left(\frac{1+a x}{1-a x}\right), v=\sqrt{1+a^{2} x^{2}}$
Given: $\tan ^{-1}\left(\frac{1+a x}{1-a x}\right) \text { w.r.t } \sqrt{1+a^{2} x^{2}}$
Explanation: $\text { Let } u=\tan ^{-1}\left(\frac{1+a x}{1-a x}\right)$
$\begin{aligned} &v=\sqrt{1+a^{2} x^{2}} \\\\ &\text { Let } a x=\tan \theta \\\\ &u=\tan ^{-1}\left(\frac{1+\tan \theta}{1-\tan \theta}\right) \end{aligned}$
$=\tan ^{-1}\left(\frac{\tan \frac{\pi}{4}+\tan \theta}{1-\tan \frac{\pi}{4} \tan \theta}\right) \quad\left[\tan \frac{\pi}{4}=1\right]$
$\begin{aligned} &=\tan ^{-1} \tan \left(\frac{\pi}{4}+\theta\right) \\\\ &=\frac{\pi}{4}+\theta \end{aligned}$
$u=\frac{\pi}{4}+\tan ^{-1}(a x) \; \; \; \; \; \; \; \quad\left[\begin{array}{l} a x=\tan \theta \\ \theta=\tan ^{-1} a x \end{array}\right]$
$\begin{aligned} &\frac{d u}{d x}=\frac{1}{1+(a x)^{2}}[a \times 1] \\\\ &=\frac{a}{1+a^{2} x^{2}} \\\\ &v=\sqrt{1+a^{2} x^{2}} \end{aligned}$
$\begin{aligned} &\frac{d v}{d x}=\frac{1}{2 \sqrt{1+a^{2} x^{2}}} \frac{d}{d x}\left(a^{2} x^{2}\right) \\\\ &=\frac{2 a^{2} x}{2 \sqrt{1+a^{2} x^{2}}}=\frac{a^{2} x}{\sqrt{1+a^{2} x^{2}}} \end{aligned}$
$\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}$
$=\frac{\frac{a}{1+a^{2} x^{2}}}{\frac{a^{2} x}{\sqrt{1+a^{2} x^{2}}}}=\frac{1}{a x \sqrt{1+a^{2} x^{2}}}$

Differentiation exercise 10.8 question 11

Answer: 2
Hint: $\text { Let } u=\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right), v=\tan ^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right)$

Given: $\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right) \text { w.r.t } \tan ^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right)$

Explanation:
$\text { Let } u=\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)$
$\begin{aligned} &v=\tan ^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right) \\\\ &\text { Let } x=\sin \theta \\\\ &u=\sin ^{-1}\left(2 \sin \theta \sqrt{1-\sin ^{2} \theta}\right) \end{aligned}$
$\begin{aligned} &=\sin ^{-1}(2 \cos \theta \sin \theta) \\\\ &=\sin ^{-1}(\sin 2 \theta) \end{aligned}$
$\begin{aligned} &v=\tan ^{-1}\left(\frac{\sin \theta}{\sqrt{1-\sin ^{2} \theta}}\right) \\\\ &=\tan ^{-1}\left(\frac{\sin \theta}{\cos \theta}\right)=\tan ^{-1}(\tan \theta) \end{aligned}$
Now,
$\begin{aligned} &\frac{-1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}} \\\\ &\frac{-1}{\sqrt{2}}<\sin \theta<\frac{1}{\sqrt{2}} \end{aligned}$
$\begin{aligned} &\frac{-\pi}{4}<\theta<\frac{\pi}{4} \\\\ &\frac{-\pi}{2}<2 \theta<\frac{\pi}{2} \end{aligned}$
$\begin{aligned} &u=\sin ^{-1}(\sin 2 \theta) \\\\ &=\sin ^{-1}(\sin 2 \theta) \\\\ &=2 \theta \\\\ &=2 \sin ^{-1} x \end{aligned}$
$\begin{aligned} &\frac{d u}{d x}=\frac{2}{\sqrt{1-x^{2}}} \\\\ &v=\tan ^{-1}(\tan \theta) \\\\ &=\theta \\\\ &v=\sin ^{-1} x \end{aligned}$
$\begin{aligned} &\frac{d v}{d x}=\frac{1}{\sqrt{1-x^{2}}} \\\\ &\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}} \end{aligned}$
$=\frac{\frac{2}{\sqrt{1-x^{2}}}}{\frac{1}{\sqrt{1-x^{2}}}}=2$

Differentiation exercise 10.8 question 12

Answer: 1
Hint: $\text { : Let } u=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right), v=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$

Given: $\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) \text { w.r.t } \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$
$0<x<1$
Explanation:
$\text { Let } u=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$
$\begin{aligned} &v=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) \\\\ &\text { Let } x=\tan \theta \\\\ &u=\tan ^{-1}\left(\frac{2 \tan \theta}{1-\tan ^{2} \theta}\right) \\\\ &=\tan ^{-1}(\tan 2 \theta) \end{aligned}$
$\begin{aligned} &v=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) \\\\ &v=\cos ^{-1}\left(\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right) \\\\ &=\cos ^{-1}\left(\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right) \end{aligned}$
$\begin{aligned} &=\cos ^{-1}(\cos 2 \theta)\\\\ &\text { Now, }\\\\ &0<x<1\\\\ &0<\tan \theta<1 \end{aligned}$
$\begin{aligned} &0<\theta<\frac{\pi}{4} \\\\ &0<2 \theta<\frac{\pi}{2} \\\\ &u=\tan ^{-1}(\tan 2 \theta) \end{aligned}$
$\begin{array}{ll} =2 \theta & 2 \theta \in\left(0, \frac{\pi}{2}\right) \\\\ =2 \tan ^{-1} x & \left(\begin{array}{l} \tan \theta=x \\ \theta=\tan ^{-1} x \end{array}\right) \end{array}$
$\begin{aligned} &v=\cos ^{-1}(\cos 2 \theta) \\\\ &=2 \theta \quad 2 \theta \in\left(0, \frac{\pi}{2}\right) \\\\ &=2 \tan ^{-1} x \end{aligned}$
$\begin{aligned} &\frac{d u}{d x}=\frac{2}{1+x^{2}} \\\\ &\frac{d v}{d x}=\frac{2}{1+x^{2}} \end{aligned}$
$\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}$
$=\frac{\frac{2}{1+x^{2}}}{\frac{2}{1+x^{2}}}=1$

Differentiation exercise 10.8 question 13

Answer: $\frac{\sqrt{1-x^{2}}}{3\left(1+x^{2}\right)}$

Hint: $\text { Let } u=\tan ^{-1}\left(\frac{x-1}{x+1}\right), v=\sin ^{-1}\left(3 x-4 x^{3}\right)$

Given: $\tan ^{-1}\left(\frac{x-1}{x+1}\right) \text { w.r.t } \sin ^{-1}\left(3 x-4 x^{3}\right)$
$\frac{-1}{2}<x<\frac{1}{2}$
Explanation: $\text { Let } u=\tan ^{-1}\left(\frac{x-1}{x+1}\right)$
$\begin{aligned} &v=\sin ^{-1}\left(3 x-4 x^{3}\right) \\\\ &u=\tan ^{-1}\left(\frac{x-1}{x+1}\right) \\\\ &=\tan ^{-1} x-\tan ^{-1} 1 \\\\ &=\tan ^{-1} x-\frac{\pi}{4} \end{aligned}$
$\begin{aligned} &\frac{d u}{d x}=\frac{1}{1+x^{2}} \\\\ &v=\sin ^{-1}\left(3 x-4 x^{3}\right) \\\\ &\text { Let } x=\sin \theta \end{aligned}$
$\begin{aligned} &\frac{-1}{2}<x<\frac{1}{2} \\\\ &\frac{-1}{2}<\sin \theta<\frac{1}{2} \\\\ &-\frac{\pi}{6}<\theta<\frac{\pi}{6} \end{aligned}$
$\begin{aligned} &v=\sin ^{-1}\left(3 \sin \theta-4 \sin ^{3} \theta\right) \\\\ &=\sin ^{-1}(\sin 3 \theta) \\\\ &=3 \theta \\\\ &v=3 \sin ^{-1} x \end{aligned}$
$\begin{aligned} &\frac{d v}{d x}=\frac{3}{\sqrt{1-x^{2}}} \\\\ &\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{\frac{1}{1+x^{2}}}{\frac{3}{\sqrt{1-x^{2}}}}=\frac{\sqrt{1-x^{2}}}{3\left(1+x^{2}\right)} \end{aligned}$

Differentiation exercise 10.8 question 14

Answer: $\frac{-x \sqrt{x^{2}-1}}{2}$

Hint: $\text { Let } u=\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right), v=\sec ^{-1} x$

Given: $\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right) \text { w.r.t } \sec ^{-1} x$
Explanation:
$\text { Let } u=\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)$
$\begin{aligned} &v=\sec ^{-1} x \\\\ &u=\tan ^{-1}\left(\frac{\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}}{\cos ^{2} \frac{x}{2}+\sin ^{2} \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}}\right) \end{aligned}$

$\left[\begin{array}{l} \cos 2 x=\cos ^{2} x-\sin ^{2} x \\ \sin 2 x=2 \sin x \cos x \\ 1=\cos ^{2} x+\sin ^{2} x \end{array}\right]$
$u=\tan ^{-1}\left(\frac{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^{2}}\right)$
$u=\tan ^{-1}\left(\frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}\right)$
Divide numerator and denominator by $\cos \frac{x}{2}$
$u=\tan ^{-1}\left(\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}\right)$
$u=\tan ^{-1}\left(\frac{\tan \frac{\pi}{4}-\tan \frac{x}{2}}{1+\tan \frac{\pi}{4} \tan \frac{x}{2}}\right)$
$\begin{aligned} &u=\tan ^{-1}\left(\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right) \\\\ &u=\frac{\pi}{4}-\frac{x}{2} \\\\ &\frac{d u}{d x}=\frac{-1}{2} \end{aligned}$
$\begin{aligned} &v=\sec ^{-1} x \\\\ &\frac{d v}{d x}=\frac{1}{x \sqrt{x^{2}-1}} \\\\ &\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{\frac{-1}{2}}{\frac{1}{x\sqrt{x^{2}-1}}}=\frac{-x \sqrt{x^{2}-1}}{2} \end{aligned}$

Differentiation exercise 10.8 question 15

Answer: 1
Hint: $\text { Let } u=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right), v=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$

Given: $\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) \text { w.r.t } \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$
$-1<x<1$
Explanation:
$\begin{aligned} &\text { Let } u=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) \\\\ &v=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) \\\\ &\text { Let } x=\tan \theta \end{aligned}$
$\begin{aligned} &u=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right), v=\tan ^{-1}\left(\frac{2 \tan \theta}{1-\tan ^{2} \theta}\right) \\\\ &u=\sin ^{-1}(\sin 2 \theta), v=\tan ^{-1}(\tan 2 \theta) \end{aligned}$
$\begin{aligned} &-1<x<1 \\\\ &-1<\tan \theta<1 \end{aligned}$
$\begin{aligned} &-\frac{\pi}{4}<\theta<\frac{\pi}{4} \\\\ &-\frac{\pi}{2}<2 \theta<\frac{\pi}{2} \end{aligned}$
$\begin{aligned} &u=\sin ^{-1}(\sin 2 \theta)=2 \theta \\\\ &v=\tan ^{-1}(\tan 2 \theta)=2 \theta \\\\ &u=2 \tan ^{-1} x \end{aligned}$
$\begin{aligned} &\frac{d u}{d x}=\frac{2}{1+x^{2}} \\\\ &v=2 \tan ^{-1} x \end{aligned}$
$\begin{aligned} &\frac{d v}{d x}=\frac{2}{1+x^{2}} \\\\ &\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{\frac{2}{1+x^{2}}}{\frac{2}{1+x^{2}}}=1 \end{aligned}$

Differentiation exercise 10.8 question 16

Answer: 3
Hint: $\text { Let } u=\cos ^{-1}\left(4 x^{3}-3 x\right), v=\tan ^{-1}\left(\frac{\sqrt{1-x^{2}}}{x}\right)$

Given: $\cos ^{-1}\left(4 x^{3}-3 x\right) \text { w.r.t } \tan ^{-1}\left(\frac{\sqrt{1-x^{2}}}{x}\right)$
$\frac{1}{2}<x<1$
Explanation: $u=\cos ^{-1}\left(4 x^{3}-3 x\right)$
$\begin{aligned} &v=\tan ^{-1}\left(\frac{\sqrt{1-x^{2}}}{x}\right) \\\\ &\text { Let } x=\cos \theta \end{aligned}$
$\begin{aligned} &u=\cos ^{-1}\left(4 \cos ^{3} \theta-3 \cos \theta\right) \\\\ &u=\cos ^{-1}(\cos 3 \theta) \end{aligned}$ $\begin{aligned} &v=\tan ^{-1}\left(\frac{\sqrt{1-\cos ^{2} \theta}}{\cos \theta}\right) \\ &v=\tan ^{-1}\left(\frac{\sin \theta}{\cos \theta}\right)=\tan ^{-1}(\tan \theta) \end{aligned}$
$\begin{aligned} &\frac{1}{2}<x<1 \\\\ &\frac{1}{2}<\cos \theta<1 \\\\ &0<\theta<\frac{\pi}{3} \end{aligned}$
$\begin{array}{ll} u=\cos ^{-1}(\cos 3 \theta)=3 \theta & \theta \in\left(0, \frac{\pi}{3}\right) \\\\ v=\tan ^{-1}(\tan \theta)=\theta & \theta \in\left(0, \frac{\pi}{3}\right) \end{array}$
$\begin{aligned} &u=3 \cos ^{-1} x \\\\ &\frac{d u}{d x}=\frac{-3}{\sqrt{1-x^{2}}} \\\\ &v=\cos ^{-1} x \end{aligned}$
$\begin{aligned} &\frac{d v}{d x}=\frac{-1}{\sqrt{1-x^{2}}} \\\\ &\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{\frac{-3}{\sqrt{1-x^{2}}}}{\frac{-1}{\sqrt{1-x^{2}}}}=3 \end{aligned}$

Differentiation exercise 10.8 question 17

Answer: $\frac{1}{2}$
Hint: $\text { Let } u=\tan ^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right), v=\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)$

Given: $\tan ^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right) \text { w.r.t } \sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)$
$-\frac{1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}$
Explanation:
$u=\tan ^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right)$
$\begin{aligned} &v=\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right) \\\\ &\text { Let } x=\sin \theta \end{aligned}$
$\begin{aligned} &u=\tan ^{-1}\left(\frac{\sin \theta}{\sqrt{1-\sin ^{2} \theta}}\right), \\\\ &u=\tan ^{-1}\left(\frac{\sin \theta}{\sqrt{\cos ^{2} \theta}}\right) \\\\ &u=\tan ^{-1}(\tan \theta) \end{aligned}$ $\begin{aligned} &v=\sin ^{-1}\left(2 \sin \theta \sqrt{1-\sin ^{2} \theta}\right) \\\\ &v=\sin ^{-1}(2 \sin \theta \cos \theta) \\\\ &v=\sin ^{-1}(\sin 2 \theta) \end{aligned}$
Now
$\begin{aligned} &-\frac{1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}} \\\\ &-\frac{1}{\sqrt{2}}<\sin \theta<\frac{1}{\sqrt{2}} \end{aligned}$

$\begin{aligned} &-\frac{\pi}{4}<\theta<\frac{\pi}{4} \\\\ &-\frac{\pi}{2}<2 \theta<\frac{\pi}{2} \end{aligned}$
$\begin{array}{ll} u=\tan ^{-1}(\tan \theta)=\theta & \theta \in\left(-\frac{\pi}{4}, \frac{\pi}{4}\right) \\\\ v=\sin ^{-1}(\sin 2 \theta)=2 \theta & 2 \theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \end{array}$
$\begin{aligned} &u=\sin ^{-1} x \\\\ &\frac{d u}{d x}=\frac{1}{\sqrt{1-x^{2}}} \\\\ &v=2 \sin ^{-1} x \end{aligned}$
$\frac{d v}{d x}=\frac{2}{\sqrt{1-x^{2}}}$
$\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{\frac{1}{\sqrt{1-x^{2}}}}{\frac{2}{\sqrt{1-x^{2}}}}=\frac{1}{2}$
$\frac{d u}{d v}=\frac{1}{2}$

Differentiation exercise 10.8 question 18

Answer: 1
Hint: $\text { Let } x=\cos \theta$

Given: $\sin ^{-1} \sqrt{1-x^{2}} \text { w.r.t } \cot ^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right)$
Explanation:
$\text { Let } x=\cos \theta$
$\begin{aligned} &u=\sin ^{-1} \sqrt{1-x^{2}} \\\\ &=\sin ^{-1} \sqrt{1-\cos ^{2} \theta} \\\\ &=\sin ^{-1}(\sin \theta) \end{aligned}$
$\begin{aligned} &v=\cot ^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right) \\\\ &=\cot ^{-1}\left(\frac{\cos \theta}{\sqrt{1-\cos ^{2} \theta}}\right) \end{aligned}$
$=\cot ^{-1}\left(\frac{\cos \theta}{\sin \theta}\right)=\cot ^{-1}(\cot \theta)$
Now
$\begin{aligned} &0<x<1 \\\\ &0<\cos \theta<1 \\\\ &0<\theta<\frac{\pi}{2} \end{aligned}$
$\begin{array}{ll} u=\sin ^{-1}(\sin \theta)=\theta & \quad \theta \in\left(0, \frac{\pi}{2}\right) \\\\ v=\cot ^{-1}(\cot \theta)=\theta & 2 \theta \in\left(0, \frac{\pi}{2}\right) \end{array}$
$\begin{aligned} &u=\cos ^{-1} x \quad\left[\begin{array}{l} x=\cos \theta \\ \theta=\cos ^{-1} x \end{array}\right] \\ &\frac{d u}{d x}=\frac{-1}{\sqrt{1-x^{2}}} \end{aligned}$
$\begin{aligned} &v=\cos ^{-1} x \\\\ &\frac{d v}{d x}=\frac{-1}{\sqrt{1-x^{2}}} \end{aligned}$
$\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{\frac{-1}{\sqrt{1-x^{2}}}}{\frac{-1}{\sqrt{1-x^{2}}}}=1$
$\frac{d u}{d v}=1$

Differentiation exercise 10.8 question 19

Answer: $\frac{-2}{ax}$
Hint: $\text { Let } a x=\sin \theta$

Given: $\sin ^{-1}\left(2 a x \sqrt{1-a^{2} x^{2}}\right) \text { w.r.t } \sqrt{1-a^{2} x^{2}}$
$\frac{-1}{\sqrt{2}}<a x<\frac{1}{\sqrt{2}}$
Explanation:
$\text { Let } \mathrm{a} x=\sin \theta$
$\begin{aligned} &u=\sin ^{-1}\left(2 a x \sqrt{1-a^{2} x^{2}}\right) \\\\ &=\sin ^{-1}\left(2 \sin \theta \sqrt{1-\sin ^{2} \theta}\right) \\\\ &=\sin ^{-1}(2 \sin \theta \cos \theta) \\\\ &=\sin ^{-1}(\sin 2 \theta) \end{aligned}$ $\begin{aligned} &u=\sin ^{-1}\left(2 a x \sqrt{1-a^{2} x^{2}}\right) \\\\ &=\sin ^{-1}\left(2 \sin \theta \sqrt{1-\sin ^{2} \theta}\right) \\\\ &=\sin ^{-1}(2 \sin \theta \cos \theta) \\ &=\sin ^{-1}(\sin 2 \theta) \end{aligned}$
Now
$\begin{aligned} &\frac{-1}{\sqrt{2}}<a x<\frac{1}{\sqrt{2}} \\\\ &\frac{-1}{\sqrt{2}}<\sin \theta<\frac{1}{\sqrt{2}} \end{aligned}$
$\begin{aligned} &-\frac{\pi}{4}<\theta<\frac{\pi}{4} \\\\ &-\frac{\pi}{2}<2 \theta<\frac{\pi}{2} \end{aligned}$
$\begin{aligned} &u=\sin ^{-1}(\sin 2 \theta)=2 \theta \quad 2 \theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \\ &=2 \sin ^{-1} a x \end{aligned}$
$\begin{aligned} &\frac{d u}{d x}=\frac{2}{\sqrt{1-a^{2} x^{2}}}(a) \\\\ &=\frac{2 a}{\sqrt{1-a^{2} x^{2}}} \end{aligned}$
$\begin{aligned} &v=\sqrt{1-a^{2} x^{2}} \\\\ &\frac{d v}{d x}=\frac{1}{2 \sqrt{1-a^{2} x^{2}}}\left(-a^{2} 2 x\right) \\\\ &\frac{d v}{d x}=\frac{-a^{2} x}{\sqrt{1-a^{2} x^{2}}} \end{aligned}$ $\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{\frac{2 a}{\sqrt{1-a^{2} x^{2}}}}{\frac{-a^{2} x}{\sqrt{1-a^{2} x^{2}}}}=\frac{-2}{a x}$

Differentiation exercise 10.8 question 20

Answer: $\frac{\sqrt{1-x^{2}}}{x\left(1+x^{2}\right)}$
Hint: $\tan ^{-1}\left(\frac{a-b}{a+b}\right)=\tan ^{-1} a-\tan ^{-1} b$
Given: $\tan ^{-1}\left(\frac{1-x}{1+x}\right) \text { w.r.t } \sqrt{1-x^{2}}$
$-1<x<1$
Explanation:
$\tan ^{-1}\left(\frac{1-x}{1+x}\right)=u$
$\begin{aligned} &u=\tan ^{-1}(1)-\tan ^{-1} x \\\\ &u=\frac{\pi}{4}-\tan ^{-1} x \\\\ &\frac{d u}{d x}=0-\frac{1}{1+x^{2}}=-\frac{1}{1+x^{2}} \end{aligned}$
$\begin{aligned} &v=\sqrt{1-x^{2}} \\\\ &\frac{d v}{d x}=\frac{1}{2 \sqrt{1-x^{2}}}(-2 x) \\\\ &\frac{d v}{d x}=\frac{-x}{\sqrt{1-x^{2}}} \end{aligned}$
$\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{-\frac{1}{1+x^{2}}}{\frac{-x}{\sqrt{1-x^{2}}}}=\frac{\sqrt{1-x^{2}}}{x\left(1+x^{2}\right)}$

Differentiation exercise 10.8 question 21

Answer: $-2 e^{-\cos x} \cos x$
Hint: $\text { let } u=\sin ^{2} x, v=e^{\cos x}$

Given: $\sin ^{2} x \text { w.r.t } e^{\cos x}$
$-1<x<1$
Explanation:
$u=\sin ^{2} x$
$\begin{aligned} &\frac{d u}{d x}=2 \sin x \cos x \\\\ &v=e^{\cos x} \\\\ &\frac{d v}{d x}=e^{\cos x}(-\sin x)=-\sin x e^{\cos x} \end{aligned}$
$\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{2 \sin x \cos x}{-\sin x e^{\cos x}}=-2 \cos x e^{-\cos x}$

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