RD Sharma Class 12 Exercise 10.3 Differentiation Solutions Maths-Download PDF Free Online

Differentiation exercise 10.3 question 2

Answer : $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2}\left(\frac{-1}{\sqrt{1-\mathrm{x}^{2}}}\right)$
Hint:
$\begin {array} {ll}\cos ^{-1}(\cos \theta)=\theta \quad if \ \ \theta \varepsilon[0, \mathrm{\pi}] \\\\ \cos ^{-1}(\cos \theta)=-\theta \quad if \ \ \ \theta \varepsilon[-\mathrm{\pi}, 0] \end{array}$
Given:
$\begin {array} {ll}\cos ^{-1}\left\{\sqrt{\frac{1+x}{2}}\right\},-1<x<1\end{array}$
Solution:
$\begin {array} {ll}Let, y=\cos ^{-1}\left\{\sqrt{\frac{1+x}{2}}\right\} \\\\ Let \: \,x=\cos 2 \theta, \theta=\frac{1}{2} \cos ^{-1} x \\\\ Now, y=\cos ^{-1}\left\{\sqrt{\frac{1+\cos 2 \theta}{2}}\right\}\end{array}$
$\begin {array} {ll}By \ using\ \cos 2 \theta=2 \cos ^{2} \theta-1\\\\ y=\cos ^{-1}\left\{\sqrt{\frac{2 \cos ^{2} \theta}{2}}\right\}\\\\ y=\cos ^{-1}(\cos \theta)\\\\ \therefore \cos ^{-1}(\cos \theta)=\theta\\\\\end{array}$
$\begin {array} {ll} Now\\\\ y=\cos ^{-1}(\cos \theta)\\\\ y=\theta\\\\ y=\frac{1}{2} \cos ^{-1} x \ \ \ \ \ \ \ \therefore \cos ^{-1}(\cos \theta)=\theta \quad if \theta \varepsilon[0, \pi] \end{array}$
Differentiating with respect to x, we get
$\begin {array} {ll} \frac{\mathrm{dy}}{\mathrm{dx}} \Rightarrow \frac{1}{2}\left(\frac{-1}{\sqrt{1-\mathrm{x}^{2}}}\right) \end{array}$
As we know,
$\begin {array} {ll} \frac{\mathrm{d}}{\mathrm{dx}} \ (constants) =0 \\\\ \frac{d}{d x}\left(\cos ^{-1} x\right)=\frac{-1}{\sqrt{1-x^{2}}} \end{array}$
Hence, final answer is $\begin {array} {ll} \frac{1}{2}\left(\frac{-1}{\sqrt{1-x^{2}}}\right) \end{array}$

Differentiation exercise 10.3 question 3

Answer : -$\frac{-1}{2 \sqrt{1-x^{2}}}$
Hint:
Use substitution method to differentiate this function
Given:
$\sin ^{-1}\left\{\sqrt{ \left.\frac{1-x}{2}\right\}}\right\}, 0<x<1$
Solution:
$\begin{array}{l} \operatorname{Let} y=\sin ^{-1}\left\{\sqrt{\frac{1-x}{2}}\right\}\\\\ \end{array}$
$\begin{array}{l} \text { let } x =\cos 2 \theta \\\\ \left[\theta=\frac{1}{2} \cos ^{-1} x\right] \ \ - (1) \\ \end{array}$
$\begin{array}{l} \therefore y=\sin ^{-1}\left\{\sqrt{\frac{1-\cos 2 \theta}{2}}\right\} \\\\ =\sin ^{-1}\left\{\sqrt{\frac{2 \sin ^{2} \theta}{2}}\right\} \\\\ y =\sin ^{-1}\left\{\sqrt{\sin ^{2} \theta}\right\} \ \ \ \ \left[\because 1-\cos 2 \theta=2 \sin ^{2} \theta\right] \\\\ =\sin ^{-1}(\sin \theta) \\\\ y =\theta=\frac{1}{2} \cos ^{-1} x[\text { using (i)] } \end{array}$
Now differentiating y w.r.t x then

$\begin{array}{l} \frac{d y}{d x}=\frac{1}{2} \frac{d\left(\cos ^{-1} x\right)}{d x} \\\\ \quad=\frac{1}{2} \cdot \frac{-1}{\sqrt{1-x^{2}}} \\\\ \quad=\frac{-1}{2 \sqrt{1-x^{2}}} \quad\left[\because \frac{d}{d x} \cos ^{-1} x=\frac{-1}{\sqrt{1-x^{2}}}\right] \\\\ \therefore \frac{d y}{d x}=\frac{-1}{2 \sqrt{1-x^{2}}} \end{array}$

Differentiation exercise 10.3 question 4

Answer :
$\frac{d y}{d x}=-\frac{1}{\sqrt{1-x^{2}}}$
Hint:
$\frac{d}{d x}\left(x^{n}\right)=n x^{n-1} ; \frac{d}{d x} (constant) = 0$
Given:
$\sin ^{-1}\left\{\sqrt{1-\mathrm{x}^{2}}\right\}, 0<\mathrm{x}<1$
Solution:
$\begin {array}{l} Let \ \ x=\cos \theta\\\\ \theta=\cos ^{-1} x\\\\ y=\sin ^{-1}\left\{\sqrt{ \left.1-\cos ^{2} \theta\right\}}\right.\\\\ \therefore \cos ^{2} \theta+\sin ^{2} \theta=1\\\\ y=\sin ^{-1}\left\{\sqrt{\sin ^{2} \theta}\right\}\\\\ \mathrm{y}=\sin ^{-1}\{\sin \theta\} \end{}$
Considering the limits,
$\begin {array}{l} 0<\cos \theta<1 \\\\ 0<\theta<\frac{\pi}{2} \ \ \ \ \ \ \ \ \ \ \left\{\cos \frac{1}{2}=1\right\}\\\\ Now, y=\sin ^{-1}(\sin \theta)\\\\ \mathrm{y}=\theta\\\\ \left\{\sin ^{-1}(\sin \theta)=\theta\right\}\\\\ if \theta \varepsilon\left[\frac{-\pi}{2}, \frac{\pi}{2}\right] \\\\ y=\cos ^{-1} x \end{}$
differentiating with respects to x, we get
$\begin {array}{l} \frac{d y}{d x}=\frac{\mathrm{d} (\cos^{-1}x)}{\mathrm{d} x}\\\\ \therefore \frac{\mathrm{d}}{\mathrm{dx}}\left(\cos ^{-1} \mathrm{x}\right) \Rightarrow \frac{-1}{\sqrt{1-\mathrm{x}^{2}}}\\\\ \frac{\mathrm{d} y}{\mathrm{dx}}=\frac{-1}{\sqrt{1-\mathrm{x}^{2}}} \end{}$

Differentiation exercise 10.3 question 6

Answer :$\frac{d y}{d x}=\frac{a}{a^{2}+x^{2}}$
Hint:

Given:
$\sin ^{-1}\left\{\frac{\mathrm{x}}{\sqrt{\mathrm{x}^{2}+\mathrm{a}^{2}}}\right\}$
Solution:
$\begin {array} {ll} Let \ y=\sin ^{-1}\left\{-\frac{x}{\sqrt{x^{2}+a^{2}}}\right\}\\\\ Let \ \mathrm{x}=\operatorname{atan} \theta\\\\ Now,\\ \mathrm{y}=\sin ^{-1}\left\{\frac{\operatorname{atan} \theta}{\sqrt{a^{2} \tan ^{2} \theta+a^{2}}}\right\}\\\\ \mathrm{y}=\sin ^{-1}\left\{\frac{\text { atsan } \theta}{\sqrt{a^{2}\left(\tan ^{2} \theta+1\right)}}\right\}\\\\\ \mathrm{y}=\sin ^{-1}\left\{\frac{\operatorname{atan} \theta}{\mathrm{a}\left(\tan ^{2} \theta+1\right)}\right\}\\\\ \left.\mathrm{y}=\sin ^{-1} \int \frac{\operatorname{atan} \theta}{\mathrm{a} \sqrt{\sec ^{2} \theta}}\right\} \end{array}$
$\begin{array}{l} \text { Using }:\left\{\tan \theta=\frac{\sin \theta}{\cos \theta}, \cos \theta=\frac{1}{\sec \theta}\right\}\\\\ \mathrm{y}=\sin ^{-1}\{\sin \theta\}\\\\ \mathrm{y}=\theta \quad \ \ \ \ \ \ \ \ \ \therefore\left\{\sin ^{-1} \theta(\sin \theta)=\theta\right\} \quad \text { if } \theta \varepsilon\left[\frac{-\pi}{2}, \frac{\mathrm{\pi}}{2}\right]\\\\ \mathrm{x}=\operatorname{atan} \theta\\\\ \mathrm{x}=\operatorname{atan} \theta\\\\ \theta=\tan ^{-1}\left(\frac{x}{a}\right)\\\\ y=\tan ^{-1}\left(\frac{x}{a}\right)\\\\ \end{array}$
differentiating with respects to x, we get
$\begin{array}{l} \frac{d y}{d x}=\frac{d}{d x}\left(\tan ^{-1}\left(\frac{x}{a}\right)\right) \\\\ \frac{\partial}{\partial x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^{2}} \\\\ \frac{d y}{d x}=\left(\frac{1}{1+\left(\frac{x}{a}\right)^{2}}\right) \times \frac{1}{a} \\\\ \end{array}$
$\begin{array}{l} \frac{d y}{d x}=\left(\frac{1}{1+\frac{x^{2}}{a^{2}}}\right) \times \frac{1}{a} \\\\ \frac{d y}{d x}=\left(\frac{1}{\frac{a^{2}+x^{2}}{a^2}}\right) \times \frac{1}{a} \\\\ \frac{d y}{d x}=\frac{a^{2}}{a^{2}+x^{2}} \times \frac{1}{a} \\\\ \frac{d y}{d x}=\frac{a}{a^{2}+x^{2}} \end{array}$

Differentiation exercise 10.3 question 7

Answer:
$\frac{d y}{d x}=\frac{2}{\sqrt{1-x^{2}}}$
Hint:
$\frac{\mathrm{d}}{\mathrm{dx}}( Constant )=0 ; \frac{\partial}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}-\mathbf{1}}$
Given:
$\sin ^{-1}\left(2 x^{2}-1\right), 0<x<1$
Solution:
$\begin {array}{l} Let,\\ y=\sin ^{-1}\left(2 x^{2}-1\right)\\\\ Let \\ \mathrm{x}=\cos \theta\\\\ Now,\\ y=\sin ^{-1}\left\{2 \cos ^{2} \theta-1\right\}\\\\ Using\\\\ 2 \cos ^{2} \theta-1=\cos 2 \theta\\\\ y=\sin ^{-1}(\cos 2 \theta)\\\\ y=\sin ^{-1}\left\{\sin \left(\frac{n}{2}-2 \theta\right)\right\} \end{}$
As we know,
$\sin \left(\frac{n}{2}-\theta\right)=\cos \theta$
Considering the limits, $0<\mathrm{x}<1$
$\begin {array}{l} 0<\cos \theta<1\\\\ 0<\theta<\frac{n}{2} \ \ \ \ \ \ \left\{\cos \frac{\pi}{2}=0, \cos 0=1\right\}\\\\ 0>-2 \theta>-\pi\\\\ \frac{n}{2}>\frac{n}{2}-2 \theta>-\frac{\pi}{2}\\\\ \text { Now, } \mathrm{y}=\sin ^{-1}\left\{\sin \left(\frac{\mathrm{n}-20}{2}\right)\right\}\\\\ \mathrm{y}=\frac{\mathrm{n}}{2}-2 \theta \ \ \ \ \ \ \ \ \ \left\{\sin ^{-1}(\sin \theta)=\theta\right\} \text { if } \theta \in\left[-\frac{n}{2}, \frac{n}{2}\right]\\\\ \therefore \mathrm{x}=\cos \theta, \theta=\cos ^{-1} \mathrm{x}\\\\ y=\frac{n}{2}-2 \cos ^{-1} x \end{}$
Differentiating with respect ts? x, weget
$\frac{d y}{d x}=\frac{d}{\partial x}\left(\frac{n}{2}-2 \cos ^{-1} x\right)$
As we know
$\begin {array}{l} \frac{d}{d x}(constant)=0, \frac{\partial}{d x}\left(\cos ^{-1} x\right)=\frac{-1}{\sqrt{1-x^{2}}}\\\\ \frac{\mathrm{dy}}{\mathrm{dx}}=0-2\left(-\frac{1}{\sqrt{1-\mathrm{x}^{2}}}\right)\\\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2}{\sqrt{1-\mathrm{x}^{2}}} \end{}$

Differentiation exercise 10.3 question 8

Answer :$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-2}{\sqrt{1-\mathrm{x}^{2}}}$

Hint:
$\frac{\partial}{\mathrm{d} \mathrm{x}}( Constant )=0 ; \frac{\partial}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}-\mathbf{1}}$

Given:
$\sin ^{-1}\left(1-2 x^{2}\right), 0<x<1$
Solution:
$\begin {array} {ll}Let \ y=\sin ^{-1}\left\{1-2 x^{2}\right\}\\\\ Let \ x=\sin \theta\\\\ Then,\\ 0=\sin ^{-1} x\\\\ y=\sin ^{-1}\left\{-2 \sin ^{2} \theta\right\}\\\\ Using,\\ 1=2 \sin ^{2} \theta=\cos 2 \theta\\\\ \mathrm{y}=\sin ^{-1}\{\cos 2 \theta\}\\\\ \mathrm{y}=\sin ^{-1}\left\{\sin \left(\frac{\mathrm{n}}{2}-2 \theta\right)\right\} \end{}$
As we know,
$\begin {array} {ll}\sin \left(\frac{n-\theta}{2}-\theta\right)=\cos \theta \end{}$
Considering the limits,
$\begin {array} {ll}0<x<1\\\\ 0<\sin \theta<1\\\\ 0<\theta<\frac{n}{2}\ \ \ \ \ \ \ \left\{\sin \frac{1}{2}=1, \sin \theta=0\right\}\\\\ 0 \angle 2 \theta=\pi\\\\ 0>-2 \theta>-\pi\\\\ \frac{1}{2}>\frac{\pi}{2}-2 \theta>-\frac{n}{2}\\\\ \end{}$
Now,
$\begin {array} {ll}\mathrm{y}=\sin ^{-1}\left\{\sin \left(\frac{\mathrm{n}}{2}-2 \theta\right)\right\} \\\\ y=\frac{n}{2}-2 \theta \\\\ y=\frac{n}{2}=2 \sin ^{-1} x \end{}$
Differentiating with respect to x,
$\begin{array}{l} \frac{d y}{d x}=\frac{d}{d x}\left(\frac{n}{2}-2 \sin ^{-1} x\right) \\\\ \text { As we know } \\\\ \frac{d}{d x}(\text { constant })=0 \\\\ \frac{d}{d x}\left(\sin ^{-1} x\right)=\frac{1}{\sqrt{1-x^{2}}} \\\\ \frac{d y}{d x}=0-2 \frac{1}{\sqrt{1-x^{2}}} \\\\ \frac{d y}{d x}=\frac{-2}{\sqrt{1-x^{2}}} \end{array}$

Differentiation exercise 10.3 question 9

Answer :

$\frac{d y}{d x}=\frac{-a}{a^{2}+x^{2}}$
Hint:
$\frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}(Constant )=0 \frac{d}{dx}\left(x^{n}\right)=n x^{n-1}$
Given:
$\cos ^{-1}\left\{\frac{x}{\sqrt{x^{2}+a^{2}}}\right\}$
Solution:
$Let,\\\\$
$\mathrm{y}=\cos ^{-1}\left\{\frac{\mathrm{x}}{\sqrt{\mathrm{x}^{2}+\mathrm{a}^{2}}}\right)\\\\$
$Let\\\\$
$\mathrm{x}=\operatorname{acot} \theta\\\\$
$\theta=\cot ^{-1}\left(\frac{x}{a}\right)\\\\$
$\begin{array}{l} \text { Now }\\ y=\cos ^{-1}\left\{\frac{\operatorname{acot} \theta}{\sqrt{a^{2} \cot ^{2} \theta+a^{2}}}\right\}\\\\ y=\cos ^{-1}\left\{\frac{\operatorname{acot} \theta}{\sqrt{a^{2}\left(\cot ^{2} \theta+1\right)}}\right\}\\\\ y=\cos ^{-1}\left\{\frac{\operatorname{acot} \theta}{a \sqrt{\cot ^{2} \theta+1}}\right\}\\\\ \text { Using } 1+\cot ^{2} \theta=\operatorname{cosec}^{2} \theta\\\\ y=\cos ^{-1}\left\{\frac{\operatorname{acot} \theta}{a \sqrt{\operatorname{cosec}^{2} \theta}}\right\}\\\\ \mathrm{y}=\cos ^{-1}\left\{\frac{\operatorname{acot} \theta}{\mathrm{acosec} \theta}\right\}\\\\ \mathrm{y}=\cos ^{-1}\left\{\frac{\cot \theta}{\operatorname{cosec} \theta}\right\} \end{array}$
As we Know,
$\begin{array}{l} \cot \theta=\frac{\cos \theta}{\sin \theta} \text { , }\\\\ \operatorname{cosec} \theta=\frac{1}{\sin \theta}\\\\ \mathrm{y}=\cos ^{-1}\left\{\frac{\cos \theta}{\sin \theta} \times \sin \theta\right\}\\\\ y=\cos ^{-1}(\cos \theta)\\\\ \mathrm{y}=\theta\\\\ \mathrm{y}=\cot ^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)\\\\ \text { Differentiating with respect to } x \text { We get }\\\\ \frac{d y}{d x}=\frac{d}{d x}\left(\cot ^{-1}\left(\frac{x}{a}\right)\right)\\\\ \therefore \frac{\partial}{d x}\left(\cot ^{-1} x\right)=\frac{-1}{1+x^{2}}\\\\ \frac{\mathrm{d} y}{\mathrm{dx}}=\frac{-1}{1+\left(\frac{\mathrm{x}}{\mathrm{a}}\right)^{2}} \times \frac{1}{\mathrm{a}} \end{array}$
$\begin{array}{l} \frac{d y}{d x}=\frac{-1}{1+\frac{x^{2}}{a^{2}}} \times \frac{1}{a} \\\\ \frac{d y}{d x}=\frac{-1}{\frac{a^{2}+x^{2}}{a^{2}}} \times \frac{1}{a} \\\\ \frac{d y}{d x}=\frac{a^{2}}{a^{2}+x^{2}} \times \frac{1}{a} \\\\ \frac{d y}{d x}=\frac{-a}{a^{2}+x^{2}} \end{array}$

Differentiation Exercise 10.3 Question 10

Answer: $\frac{d y}{d x}=1$
Hint:
$\frac{\mathrm{d}}{\mathrm{d} \mathrm{x}} \text { (constants) }=0$
$\frac{d}{d\mathrm{x}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{nx}^{\mathrm{n}-1}$
Given:
$\sin ^{-1}\left\{\frac{\sin x+\cos x}{\sqrt{2}}\right\}$
$\frac{-3 \mathrm{\pi}}{4}<\mathrm{x}<\frac{\pi}{4}$
Soluton:
Let,$y=\sin ^{-1}\left\{\frac{\sin x+\cos x}{\sqrt{2}}\right\}$
Now,
$\mathrm{y}=\sin ^{-1}\left\{\sin \mathrm{x} \frac{1}{\sqrt{2}}+\cos \mathrm{x} \frac{1}{\sqrt{2}}\right\}$
$\mathrm{y}=\sin ^{-1}\left\{\sin \mathrm{x} \cos \left(\frac{\mathrm{\pi}}{4}\right)+\cos \mathrm{xsin}\left(\frac{\mathrm{\pi}}{4}\right)\right\}$
$\sin \left(\frac{\mathrm{\pi}}{4}\right)=\frac{1}{\sqrt{2}} \&$
$\cos \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}$
Using,
$\sin (A+B)=\sin A \cos B+\cos A \sin B$
$\mathrm{y}=\sin ^{-1}\left\{\sin \left(\mathrm{x}+\frac{\pi}{4}\right)\right\}$
Considering the limits,
$-\frac{3 \pi}{4}<x<\frac{\pi}{4}$
$-\frac{3 \pi}{4}+\frac{\pi}{4}<x+\frac{\pi}{4}<\frac{\pi}{4}+\frac{1}{4}$
$-\frac{\mathrm{\pi}}{2}<\mathrm{x}+\frac{\mathrm{\pi}}{4}<\frac{\pi}{2}$
Now,
$\mathrm{y}=\sin ^{-1}\left\{\sin \left(\mathrm{x}+\frac{\mathrm{\pi}}{4}\right)\right\}$
$y=x+\frac{\pi}{4}$ $\left\{\sin ^{-1}(\sin \theta)=\theta, \text { if } \theta \varepsilon\left[\frac{\pi}{2}, \frac{\pi}{2}\right]\right\}$
Differentiating with respect to x , We get
$\frac{d y}{d x}=1+0$
$\frac{\mathrm{d}}{\mathrm{dx}} \text { (constant) }=0$
$\frac{d y}{d x}=1$

Differentiation Exercise 10.3 Question 11

Answer: $\frac{\mathrm{d} y}{\mathrm{~d} \mathrm{x}}=-1$
Hint:
$\frac{\mathrm{d}}{\mathrm{dx}} \text { (constants) }=0$
$\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}$
Given:
$\cos ^{-1}\left\{\frac{\cos x+\sin x}{\sqrt{2}}\right\}$
$-\frac{\mathrm{\pi}}{4}<\mathrm{x}<\frac{\mathrm{\pi}}{4}$
Solution:
Let,
$y=\cos ^{-1}\left\{\frac{\cos x+\sin x}{\sqrt{2}}\right\}$
Now
$\mathrm{y}=\cos ^{-1}\left\{\cos \mathrm{x} \frac{1}{\sqrt{2}}+\sin \mathrm{x} \frac{1}{\sqrt{2}}\right\}$
$y=\cos ^{-1}\left\{\cos x \cos \left(\frac{\pi}{4}\right)+\sin x \sin \left(\frac{\pi}{4}\right)\right\}$
$\sin \left(\frac{\mathrm{\pi}}{4}\right)=\frac{1}{\sqrt{2}} \&$
$\cos \left(\frac{\mathrm{\pi}}{4}\right)=\frac{1}{\sqrt{2}}$
Using,
$\cos (A-B)=\cos A \cos B+\sin A \sin B$
$y=\cos ^{-1}\left\{\cos \left(x-\frac{\pi}{4}\right)\right\}$
Considering the limits,

$\frac{-\pi}{4}<\mathrm{x}<\frac{\mathrm{\pi}}{4}$
$-\frac{\pi}{4}-\frac{\pi}{4}<x-\frac{\pi}{4}<\frac{\pi}{4}-\frac{\pi}{4}$
$-\frac{2 \pi}{4}<x-\frac{\pi}{4}<0$
$\frac{-\pi}{2}<x-\frac{\pi}{4}<0$
Now,
$y=\cos ^{-1}\left\{\cos \left(x-\frac{\pi}{4}\right)\right\}$
$\mathrm{y}=-\left(\mathrm{x}-\frac{\mathrm{\pi}}{4}\right)$ $\left\{\begin{array}{l} \cos ^{-1}(\cos \theta)=-\theta \\ \text { if } \theta \in[-\pi, 0] \end{array}\right\}$
Differentiating with respect to x , We get

$\frac{d y}{d x}=-1$

$\frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}(\text { constants })=0$

Differentiation Exercise 10.3 Question 12 (i)

Answer:
$\frac{dy}{dx}=\frac{1}{2\sqrt{1-x^{2}}}$
Hint:
$\frac{d}{dx}(constant)=0$
$\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}$
Given:
$\tan ^{-1}\left \{ \frac{x}{1+\sqrt{1-x^{2}}} \right \}$
$-1< x< 1$
Solution:
Let,
$\begin{aligned} &\mathrm{y}=\tan ^{-1}\left\{\frac{\mathrm{x}}{1+\sqrt{1-\mathrm{x}^{2}}}\right\} \\ &\mathrm{x}=\sin \theta \\ &\theta=\sin ^{-1} \mathrm{x} \end{aligned}$
Now,$y=\tan ^{-1}\left \{ \frac{\sin \theta }{1+\sqrt{1\sin ^{2}\theta }} \right \}$
Using $\sin ^{2}\theta +\cos ^{2}\theta =1$
$\begin{aligned} &\mathrm{y}=\tan ^{-1}\left\{\frac{\sin \theta}{1+\sqrt{\cos ^{2} \theta}}\right\} \\ &\mathrm{y}=\operatorname{tran}^{-1}\left\{\frac{\sin \theta}{1+\cos \theta}\right\} \\ &\text { Using } 2 \cos ^{2} \theta=1+\cos 2 \theta \\ &2 \sin \theta \cos \theta=\sin 2 \theta \end{aligned}$$\begin{aligned} &\mathrm{y}=\tan ^{-1}\left\{\frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}}\right\} \\ &\mathrm{y}=\tan ^{-1}\left\{\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}\right\} \\ &\tan \theta=\frac{\sin \theta}{\cos \theta} \\ &\mathrm{y}=\tan ^{-1}\left\{\tan \frac{\theta}{2}\right\} \end{aligned}$

Considering the limits,

$\begin{aligned} &-1<x<1 \\ &-1<\sin \theta<1 \\ &-\frac{\pi}{2}<\frac{\theta}{2}<\frac{\pi}{2} \\ &\frac{-\pi}{4}<\frac{\theta}{2}<\frac{\pi}{4} \end{aligned}$

Now,

$\begin{aligned} &\mathrm{y}=\tan ^{-1}\left\{\tan \frac{\theta}{2}\right\} \\ &\mathrm{y}=\frac{\theta}{2} \end{aligned}$ $\tan ^{-1}(\tan \theta)=0 \text { if } \theta \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$

$y=\frac{1}{2}\sin ^{-1}x$

Differentiating with respect to x , We get

$\begin{aligned} &\frac{d y}{d x}=-1 \\ &\frac{d}{d x}(\text { constants })=0 \\ &\frac{d y}{d x}=\frac{1}{2}\left(\frac{1}{2} \frac{d}{d x}\left(\sin ^{-1} x\right)\right. \\ &\frac{d}{d x}\left(\sin ^{-1} x\right)=\frac{1}{\sqrt{1-x^{2}}} \\ &\frac{d y}{d x}=\frac{1}{2} \frac{1}{\sqrt{1-x^{2}}} \\ &\frac{d y}{d x}=\frac{1}{2 \sqrt{1-x^{2}}} \end{aligned}$

$-1< x< 1$

Differentiation Exercise 10.3 Question 12 (ii)

Answer:
$\frac{dy}{dx}=\frac{1}{2}$
Hint:
$\cos 2x=2\cos ^{2}x-1$
$\tan \left ( \frac{\pi }{2} -\theta \right )=\cot \theta$
Given:
$\tan ^{-1}\left \{ \frac{1+\cos x}{sinx} \right \}$
Solution:
$\begin{aligned} &y=\tan ^{-1}\left\{\frac{1+\cos x}{\sin x}\right\} \\ &y=\tan ^{-1}\left\{\frac{2 \cos ^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}\right\} \\ &y=\tan ^{-1}\left(\cot \frac{x}{2}\right) \\ &y=\tan ^{-1}\left(\tan \left(\frac{\pi}{2}-\frac{x}{2}\right)\right) \end{aligned}$ $\left [ \left ( \frac{\pi }{2} -\theta \right )=\cot \theta \right ]$
$y=\frac{x}{2}$
Differentiating w.r.t x
$\frac{dy}{dx}=\frac{1}{2}$

Differentiation Exercise 10.3 Question 13

Answer:$\frac{dy}{dx}=\frac{1}{2\sqrt{a^{2}-x^{2}}}$
Hint:
$\begin{aligned} &\frac{\mathrm{d}}{\mathrm{dx}}(\text { constants })=0 \\ &\frac{d}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}-1} \end{aligned}$
Given:
$\begin{aligned} &\tan ^{-1}\left\{\frac{\mathrm{x}}{\mathrm{a}+\sqrt{\mathrm{a}^{2}-\mathrm{x}^{2}}}\right\} \\ &-\mathrm{a}<\mathrm{x}<-\mathrm{a} \end{aligned}$
Solution:
Let,
$x=a\sin \theta$
$\theta =\sin ^{-1}\frac{x}{a}$
Now,
$\begin{aligned} &y=\tan ^{-1}\left\{\frac{\operatorname{asin} \theta}{a+\sqrt{a^{2}-a^{2} \sin ^{2} \theta}}\right\} \\ &y=\tan ^{-1}\left\{\frac{\operatorname{asin} \theta}{a+a \sqrt{1-\sin ^{2} \theta}}\right\} \end{aligned}$
Using
$\begin{aligned} &\cos ^{2} \theta+\sin ^{2} \theta=1 \\ &y=\tan ^{-1}\left\{\frac{\operatorname{asin} \theta}{a+a \sqrt{\cos \theta}}\right\} \\ &y=\tan ^{-1}\left\{\frac{\operatorname{asin} \theta}{a+\operatorname{acos} \theta}\right\} \\ &y=\tan ^{-1}\left\{\frac{\operatorname{asin} \theta}{a(1+\operatorname{los} \theta)}\right\} \\ &U \operatorname{sing} 2 \cos ^{2} \theta=1+\cos \theta \\ &y=\tan ^{-1}\left\{\frac{\sin \theta}{1+\cos \theta}\right\} \text { and } \\ &2 \sin \theta \cos \theta=\sin 2 \theta \\ &y=\tan ^{-1}\left\{\frac{\sin \theta}{1+\cos \theta}\right\} \end{aligned}$
$\begin{aligned} &y=\tan ^{-1}\left\{\frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}}\right\} \\ &y=\tan ^{-1}\left\{\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}\right\} \\ &y=\tan ^{-1}\left(\tan \frac{\theta}{2}\right) \end{aligned}$ $\therefore \frac{\sin \theta }{\cos \theta }=\tan \theta$
Considering the limit

$\begin{aligned} &-a<x<a \\ &-1<\sin \theta<1 \\ &\frac{-\pi}{2}<\theta<\frac{\pi}{2} \\ &-\frac{\pi}{4}<\frac{\theta}{2}<\frac{\pi}{4} \end{aligned}$ $\left \{ \sin \frac{\pi}{2}=1 \right \}$
Now,$\begin{aligned} &y==\tan ^{-1}\left\{\tan \frac{\theta}{2}\right\} \\ &y=\frac{\theta}{2} \\ &y=\frac{1}{2} \sin ^{-1} \frac{x}{a} \end{aligned}$

Differentiating with respect to x , We get

$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(\frac{1}{2} \sin ^{-1} \frac{x}{a}\right) \\ &\therefore \frac{d}{d x}\left(\sin ^{-1} x\right)=\frac{1}{\sqrt{1-x^{2}}} \\ &\frac{d y}{d x}=\frac{1}{2} \frac{1}{\sqrt{1-(x)^{2}}} \times \frac{1}{a} \\ &\Rightarrow \frac{1}{2} \times \frac{1}{\sqrt{1-\frac{x^{2}}{a^{2}}}} \times \frac{1}{a} \\ &\Rightarrow \frac{d y}{d x}=\frac{1}{2} \times \frac{1}{\sqrt{\frac{a^{2}-x^{2}}{a^{2}}}} \times \frac{1}{a} \\ &\frac{d y}{d x}=\frac{1}{2 \sqrt{a^{2}-x^{2}}} \end{aligned}$

Differentiation Exercise 10.3 Question 14

Answer:$\frac{dy}{dx}=\frac{1}{\sqrt{1-x^{2}}}$
Hint:
$\begin{aligned} &\frac{\mathrm{d}}{\mathrm{dx}}(\text { constants })=0 \\ &\frac{d}{d x}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{nx}^{\mathrm{n}-1} \end{aligned}$
Given:
$\begin{aligned} &\sin ^{-1}\left\{\frac{\mathrm{x}+\sqrt{1-\mathrm{x}^{2}}}{\sqrt{2}}\right\} \\ &-1<\mathrm{x}<1 \end{aligned}$
Solution:
Let
$y=\sin ^{-1}\left\{\frac{\mathrm{x}+\sqrt{1-\mathrm{x}^{2}}}{\sqrt{2}}\right\}$
Let
$x=\sin \theta$
$\theta =\sin ^{-1}x$
Now,
$\mathrm{y}=\sin ^{-1}\left\{\frac{\sin \theta+\sqrt{1-\sin ^{2} \theta}}{\sqrt{2}}\right\}$
Using
$\begin{aligned} &\sin ^{2} \theta+\cos ^{2} \theta=1 \\ &y=\sin ^{-1}\left\{\frac{\sin \theta+\sqrt{\cos ^{2} \theta}}{\sqrt{2}}\right] \\ &y=\sin ^{-1}\left\{\frac{\sin \theta+\cos \theta}{\sqrt{2}}\right\} \end{aligned}$
Now,
$\begin{aligned} &\mathrm{y}=\sin ^{-1}\left\{\sin \theta \frac{1}{\sqrt{2}}+\cos \theta \frac{1}{\sqrt{2}}\right\} \\ &\mathrm{y}=\sin ^{-1}\left\{\sin \theta \cos \left(\frac{\mathrm{\pi}}{4}\right)+\cos \theta \sin \left(\frac{\mathrm{\pi}}{4}\right)\right\} \\ &\sin \left(\frac{\mathrm{\pi}}{4}\right)=\frac{1}{\sqrt{2}} \& \cos \left(\frac{\mathrm{\pi}}{4}\right) \end{aligned}$
$y=\sin ^{-1}\left\{\sin \left(\theta+\frac{\pi}{4}\right)\right\}$

Considering the limits,

$\begin{aligned} &-1<x<1 \\ &-1<\sin \theta<1 \\ &-\frac{\pi}{2}<\theta<\frac{\pi}{2} \end{aligned}$ $\left \{ \sin \frac{\pi}{2}=1 \right \}$

$\begin{aligned} &\frac{-\pi}{2}+\frac{\pi}{4}<\theta+\frac{\pi}{4}<\frac{\pi}{2}+\frac{\pi}{4} \\ &-\frac{\pi}{4}<\theta+\frac{\pi}{4}<\frac{3 \pi}{4} \end{aligned}$

Now,

$\begin{aligned} &\mathrm{y}=\sin ^{-1}\left\{\sin \left(\theta+\frac{\mathrm{\pi}}{4}\right)\right\} \\ &\mathrm{y}=\theta+\frac{\mathrm{\pi}}{4} \\ &\mathrm{y}=\sin ^{-1} \mathrm{x}+\frac{\mathrm{\pi}}{4} \end{aligned}$ $sin^{-}\left ( \sin \theta \right )=\theta if\theta \varepsilon \left [ \frac{-\pi}{2} ,\frac{\pi}{2}\right ]$Differentiating with respect to x , We get
$\begin{aligned} &\frac{\mathrm{d}}{\mathrm{dx}}\left(\sin ^{-1} \mathrm{x}\right)=\frac{1}{\sqrt{1-\mathrm{x}^{2}}} \\ &\frac{\mathrm{d}}{d \mathrm{x}} \text { (constant) }=0 \\ &\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\sqrt{1-\mathrm{x}^{2}}} \end{aligned}$

Differentiation Exercise 10.3 Question 15

Answer:$\frac{dy}{dx}=\frac{-1}{\sqrt{1-x^{2}}}$
Hint:
$\begin{aligned} &\frac{\mathrm{d}}{\mathrm{dx}}(\text { constants })=0 \\ &\frac{d}{d \mathrm{x}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}-1} \end{aligned}$
Given:
$\begin{aligned} &\cos ^{-1}\left\{\frac{x+\sqrt{1-x^{2}}}{\sqrt{2}}\right\} \\ &-1<x<1 \end{aligned}$
Solution:
$y=\cos ^{-1}\left\{\frac{x+\sqrt{1-x^{2}}}{\sqrt{2}}\right\} \\$
Let,
$x=\sin \theta$
$\theta =\sin ^{-1}x$
$\begin{aligned} &\mathrm{y}=\cos ^{-1}\left\{\frac{\sin \theta+\sqrt{1-\sin ^{2} \theta}}{\sqrt{2}}\right\} \\ &\mathrm{y}=\cos ^{-1}\left\{\frac{\sin \theta+\sqrt{\cos ^{2} \theta}}{\sqrt{2}}\right\} \\ &\text { Using } \sin ^{2} \theta+\cos ^{2} \theta=1 \\ &\mathrm{y}=\cos ^{-1}\left\{\frac{\sin \theta+\cos \theta}{\sqrt{2}}\right\} \\ &\mathrm{y}=\cos ^{-1}\left\{\sin \theta \frac{1}{\sqrt{2}}+\cos \theta \frac{1}{\sqrt{2}}\right\} \\ &\mathrm{y}=\cos ^{-1}\left\{\sin \theta \sin \left(\frac{\mathrm{\pi}}{4}\right)++\cos \theta \cos \left(\frac{\mathrm{\pi}}{4}\right)\right\} \end{aligned}$
$\therefore \sin \left ( \frac{\pi}{4} \right )=\frac{1}{\sqrt{2}}$
$\cos \left ( \frac{\pi}{4} \right )=\frac{1}{\sqrt{2}}$
Using,
$\begin{aligned} &\cos (A-B)=\cos A \cos B+\sin A \sin B \\ &y=\cos ^{-1}\left\{\cos \left(\theta-\frac{\pi}{4}\right)\right\} \end{aligned}$
Considering the limits,
$\begin{aligned} &-1<x<1 \\ &-1<\sin \theta<1 \\ &-\frac{\pi}{2}<\theta<\frac{\pi}{2} \\ &-\frac{\pi}{2}-\frac{\pi}{4}<\theta-\frac{\pi}{4}<\frac{1}{2}-\frac{\pi}{4} \\ &-\frac{3 \pi}{4}<8-\frac{\pi}{4}<\frac{\pi}{4} \end{aligned}$
Now,
$\begin{aligned} &y=\cos ^{-1}\left\{\cos \left(\theta-\frac{\pi}{4}\right)\right\} \\ &y=-\left(\theta-\frac{\pi}{4}\right) \end{aligned}$ $\left\{\therefore \cos ^{-1}(\cos \theta)=-0 \text { if } \theta \varepsilon[-\pi, 0]\right\}$
$\begin{aligned} &y=-\left(\theta-\frac{\pi}{4}\right) \\ &y=-\sin ^{-1} x+\frac{\pi}{4} \end{aligned}$ $\left [ since\: x=\sin \theta \right ]$
Differentiating with respect to x , We get
$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(-\sin ^{-1} x+\frac{\pi}{4}\right) \\ &\therefore \frac{d}{d x}(\text { constants })=0 \\ &\frac{d}{d x}\left(\sin ^{-1} x\right)=\frac{1}{\sqrt{1-x^{2}}} \\ &\frac{d y}{d x}=-\frac{1}{\sqrt{1-x^{2}}}+0 \\ &\frac{d y}{d x}=\frac{-1}{\sqrt{1-x^{2}}} \end{aligned}$

Differentiation Exercise 10.3 Question 16

Answer: $\frac{dy}{dx}=\frac{4}{1+4x^{2}}$
Hint:
$\begin{aligned} &\frac{d}{d x}(\text { constants })=0 \\ &\frac{d}{d x}\left(x^{n}\right)=n x^{n-1} \end{aligned}$
Given:
$\begin{aligned} &\tan ^{-1}\left\{\frac{4 x}{1-4 x^{2}}\right\} \\ &\frac{-1}{2}<x<\frac{1}{2} \end{aligned}$
Solution:
Let,
$y=\tan ^{-1}\left\{\frac{4 x}{1-4 x^{2}}\right\}$
Let,
$2x=\tan \theta$
$\theta =\tan ^{-1}2x$
$y =\tan ^{-1}\left \{ \frac{2\tan \theta }{1-\tan ^{2}\theta } \right \}$
Using $\tan 2\theta =\frac{2\tan \theta }{1-\tan ^{2}\theta }$$y=\tan ^{-1}\left ( \tan 2\theta \right )$

Considering the limits,

$\begin{aligned} &\frac{-1}{2}<x<\frac{1}{2} \\ &-1<2 x<1 \\ &-1<\tan \theta<1 \\ &-\frac{\pi}{4}<\theta<\frac{\pi}{4} \\ &-\frac{\pi}{2}<2 \theta<\frac{\pi}{2} \end{aligned}$ $\left \{ \tan 45^{\circ}=1 \right \}$

Now,

$y=\tan ^{-1}\left ( \tan 2\theta \right )$

$y=2\theta$

$y=2\tan ^{-1}\left ( 2x \right )$ $\left [ since,2x=\tan \theta \right ]$

Differentiating with respect to x , We get

$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(2 \tan ^{-1} 2 x\right) \\ &\frac{d y}{d x}=2 \times \frac{2}{1+(2 x)^{2}} \\ &\frac{d y}{d x}=\frac{4}{1+4 x^{2}} \end{aligned}$ $\frac{d\left ( tan^{-1}x \right )}{dx}=\frac{1}{1+x^{2}}$

Diffrentiation Exercise 10.3 Question 17

Answer: $\frac{dy}{dx}=\frac{2^{x+1}log2}{1+4x}$
Hint:
$\begin{aligned} &\frac{\mathrm{d}}{\mathrm{dx}}(\text { constants })=0 \\ &\frac{d}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}-1} \end{aligned}$
Given:
$\begin{aligned} &\tan ^{-1}\left[\frac{2^{x+1}}{1-4^{x}}\right] \\ &-\infty<x<0 \end{aligned}$
Solution:
$y=\tan ^{-1}\left[\frac{2^{x+1}}{1-4^{x}}\right]$
Let,
$2^{x}=\tan \theta$
$y=\tan ^{-1}\left \{ \frac{2x2^{x}}{1-\left ( 2x \right )^{2}} \right \}$
As we Know
$\begin{aligned} &a^{m+n} \rightarrow a^{m} \cdot a^{n} \\ &y=\tan ^{-1}\left\{\frac{2 \tan \theta}{1-\tan ^{2} \theta}\right\} \end{aligned}$
Using
$\tan 2\theta =\frac{2\tan \theta }{1-\tan ^{2}\theta }$$y=\tan ^{-1}\left ( \tan 2\theta \right )$

Considering the limits,

$\begin{aligned} &-\infty<x<0 \\ &2^{-\infty}<2^{x}<2^{0} \\ &0<\tan \theta<1 \\ &0<\frac{\pi}{4} \end{aligned}$ $\left\{2^{0}=1,2^{-\infty}=0\right\}$

$0< 2\theta < \frac{\pi}{2}$ $\left \{ \tan \frac{\pi}{4} =1\right \}$

$0< 2\theta < \frac{\pi}{2}$

Now

$y=\tan ^{-1}\left ( \tan \theta \right )$

$y==2\theta$

$y=2\tan ^{-1}\left ( 2^{x} \right )$ $\left\{\begin{array}{r} \sin c e 2^{x}=\tan \theta \\ \theta=\tan ^{-1}\left(2^{x}\right) \end{array}\right\}$

Differentiating with respect to x , We get

$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(2 \tan ^{-1}\left(2^{x}\right)\right) \\ &\frac{d y}{d x}=2 \times \frac{2^{x} \log 2}{1+\left(2^{x}\right)^{2}} \\ &\frac{d y}{d x}=\frac{2^{x+1} \log 2}{1+4^{x}} \end{aligned}$ $\begin{gathered} \therefore \frac{\mathrm{d}\left(\tan ^{-1} \mathbf{x}\right)}{\mathrm{dx}}=\frac{1}{1+\mathrm{x}^{2}} \\ \frac{\mathrm{d}}{\mathrm{dx}}\left(2^{\mathrm{x}}\right)=\log 2 \\ \left\{\mathrm{a}^{\mathrm{m}} \cdot \mathbf{a}^{\mathrm{n}} \Rightarrow \mathbf{a}^{\mathrm{m}+\mathrm{h}}\right\} \end{gathered}$

Differentiation Exercise 10.3 Question 18

Answer:$\frac{d y}{d x}=\frac{2 a^{x} l o g a}{1+a^{2 x}}$
Hint:
$\frac{d c}{d x}=0;\frac{d}{dx}(x^{n})=nx^{n-1}$
Given:
$\tan ^{-1}\left\{\frac{2 \mathrm{a}^{x}}{1-\mathrm{a}^{2 x}}\right\}, \mathrm{a}>1,-\infty<\mathrm{x}<0$
Solution:
Let
$y=\tan ^{-1}\left\{\frac{2 \mathrm{a}^{x}}{1-\mathrm{a}^{2 x}}\right\},$
Let,
$\begin{aligned} &\begin{aligned} &\mathrm{a}^{\mathrm{x}}=\tan \theta \\ &\mathrm{y}=\tan ^{-1}\left\{\frac{2 \tan \theta}{1-\tan ^{2} \theta}\right\} \\ &\text { Using } \tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^{2} \theta} \\ &\mathrm{y}=\tan ^{-1}(\tan 2 \theta) \end{aligned}\\ &\text { Considering the limits, }\\ &-\infty<\mathrm{x}<0\\ &a^{-\infty}<a^{x}<a^{0} \end{aligned}$
$0< \tan \theta < 1$ $\left \{ a^{o} =1\right \}$
$0< \theta < \frac{\pi}{4}$ $\left \{ \tan \left ( \frac{\pi}{4} \right )=1 \right \}$
Now,$y=\tan ^{-1}\left ( \tan 2\theta \right )$
$y=2\theta$
$y=2\tan ^{-1}\left ( a^{x} \right )$ $since\: a^{x}=\tan \theta ,\theta =\tan ^{-1}\left ( a^{x} \right )$
Differentiating with respect to $x$, we get
$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(2 \tan ^{-1}\left(a^{x}\right)\right) \\ &\frac{d y}{d x}=2 \times \frac{a^{x}}{1+\left(a^{x}\right)^{2}} \log a \\ &\frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^{2}} \\ &\frac{d y}{d x}=\frac{2 a^{x} \log a}{1+a^{2 x}} \end{aligned}$

Differentiation Exercise 10.3 Question 19

Answer:$\frac{dy}{dx}=\frac{-1}{2\sqrt{1-x^{2}}}$
Hint:
$\begin{aligned} &\frac{\mathrm{d}}{\mathrm{dx}}(\text { constants })=0 \\ &\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{nx}^{\mathrm{n}-1} \end{aligned}$
Given:
$\sin ^{-1}\left\{\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right\}, 0<x<1$
Solution:
Let
$y=\sin ^{-1}\left\{\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right\}$
Let,$x=\cos 2\theta$
$\begin{aligned} &\text { Now, } \sin ^{-1}\left\{\frac{\sqrt{1+\cos 2 \theta}+\sqrt{1-\cos 2 \theta}}{2}\right\} \\ &\text { Using } 1-2 \sin ^{2} \theta=\cos 2 \theta \\ &2 \cos ^{2} \theta-1=\cos 2 \theta \\ &y=\sin ^{-1}\left\{\frac{\sqrt{2 \cos ^{2} \theta}+\sqrt{2 \sin ^{2} \theta}}{2}\right\} \end{aligned}$
Now,
$\begin{aligned} &\mathrm{y}=\sin ^{-1}\left\{\frac{\sqrt{2} \cos \theta+\sqrt{2} \sin \theta}{2}\right\} \\ &\mathrm{y}=\sin ^{-1}\left\{ \cos \theta \frac{1}{\sqrt{2}}+\sin \theta \frac{1}{\sqrt{2}}\right\} \\ &\mathrm{y}=\sin ^{-1}\left\{\sin \theta \cos \left(\frac{\mathrm{\pi}}{4}\right)+\cos \theta \sin \left(\frac{\mathrm{\pi}}{4}\right)\right\} \\ &\therefore \sin \left(\frac{\mathrm{\pi}}{4}\right)=\frac{1}{\sqrt{2}} \& \cos \left(\frac{\mathrm{\pi}}{4}\right)=\frac{1}{\sqrt{2}} \\ &\mathrm{U} \sin \mathrm{sin}(\mathrm{A}+\mathrm{B})=\sin \mathrm{A} \cos \mathrm{B}+\cos \mathrm{Asin} \mathrm{B} \\ &\mathrm{y}=\sin ^{-1}\left\{\sin \left(\theta+\frac{\mathrm{\pi}}{4}\right)\right\} \end{aligned}$
Considering the limits,
$\begin{aligned} &0<\mathrm{x}<1 \\ &0<\cos 2 \theta<1 \\ &0<2 \theta<\frac{\mathrm{\pi}}{2} \\ &0<\theta<\frac{\mathrm{\pi}}{4} \\ &0+\frac{\mathrm{\pi}}{4}<\left(\theta+\frac{1}{4}\right)<\frac{\mathrm{\pi}}{2}+\frac{\pi}{4} \\ &\frac{\mathrm{\pi}}{4}<\left(\theta+\frac{\pi}{4}\right)<\frac{\mathrm{\pi}}{2} \end{aligned}$
Now,$y=\sin ^{-1}$ $\left \{ \sin \left ( \theta +\frac{\pi}{4} \right ) \right \}$
$y=\theta +\frac{\pi }{4}$ $\sin ^{-1}\left ( \sin \theta \right )=\theta ,\: if\: \theta \varepsilon \left [ -\frac{\pi}{2},\frac{\pi}{2} \right ]$
$y=\frac{1}{2} \cos ^{-1} x+\frac{\pi}{4}$ $\left \{ since\: x=\cos 2\theta \right \}$
$\frac{d y}{d x}=\frac{d}{d x}\left(\frac{1}{2} \cos ^{-1} x+\frac{\pi}{4}\right)$
$\begin{aligned} &\frac{\mathrm{d}}{\mathrm{dx}}(\text { constant })=0 \ \\ &\frac{d}{\mathrm{dx}}\left(\cos ^{-1} \mathrm{x}\right)=\frac{-1}{\sqrt{1-\mathrm{x}^{2}}} \\ &\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2}\left(\frac{-1}{\sqrt{1-\mathrm{x}^{2}}}\right)+0 \\ &\frac{\mathrm{dy}}{\mathrm{dx}} \Rightarrow \frac{-1}{2 \sqrt{1-\mathrm{x}^{2}}} \end{aligned}$

Differentiation Exercise 10.3 Question 20

Answer: $\frac{dy}{dx}=\frac{a}{2\left ( 1+a^{2}x^{2} \right )}$
Hint:
$\begin{aligned} &\frac{d}{d \mathrm{x}}(\text { constant })=0 \\ &\frac{d}{d \mathrm{x}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}-1} \end{aligned}$
Given:
$\begin{aligned} &\tan ^{-1}\left\{\frac{\sqrt{1+a^{2} x^{2}}-1}{a x}\right\}, \\ &x \neq 0 \end{aligned}$
Solution:
Let,
$y=\tan ^{-1}\left\{\frac{\sqrt{1+a^{2} x^{2}}-1}{a x}\right\}$
$Let\: ax=\tan \theta$
Now
$y=\tan ^{-1}\left \{ \frac{\sqrt{1+\tan ^{2}\theta -1}}{\tan \theta } \right \}$
Using
$\begin{aligned} &\sec ^{2} \theta=1+\tan ^{2} \theta \\ &\mathrm{y}=\tan ^{-1}\left\{\frac{\sqrt{\sec ^{2} \theta}-1}{\tan \theta}\right\} \\ &\mathrm{y}=\tan ^{-1}\left\{\frac{\sec \theta-1}{\tan \theta}\right\} \\ &\mathrm{y}=\tan ^{-1}\left\{\frac{\frac{1}{\cos \theta}-1}{\frac{\sin \theta}{\cos \theta}}\right\} \end{aligned}$
$\begin{aligned} &\text { Using } \tan \theta=\frac{\sin \theta}{\cos \theta}, \sec \theta=\frac{1}{\cos \theta} \\ &y=\tan ^{-1}\left\{\frac{\frac{1-\cos \theta}{\cos \theta}}{\frac{\sin d}{\cos \theta}}\right\} \\ &y=\tan ^{-1}\left\{\frac{1-\cos \theta}{\cos \theta} \times \frac{\cos \theta}{\sin \theta}\right\} \\ &y=\tan ^{-1}\left\{\frac{1-\cos \theta}{\sin \theta}\right\} \end{aligned}$
$\begin{aligned} &\text { Using } 2 \sin ^{2} \theta=1-\cos 2 \theta \\ &\text { and } 2 \sin \theta \cos \theta=\sin 2 \theta \\ &y=\tan ^{-1}\left\{\frac{2 \sin ^{2} \theta / 2}{2 \sin \theta / 2 \cos \theta / 2}\right\} \\ &y=\tan ^{-1}\left\{\tan \frac{\theta}{2}\right\} \end{aligned}$
$y=\frac{\theta }{2}$
$y=\frac{1}{2}tan^{-1}ax$
Differentiating with respect to x , We get
$\begin{aligned} &\frac{dy}{d x}=\frac{d}{d x}\left(\frac{1}{2} \operatorname{tan}^{-1} x\right)\\ &\text { Using }\\ &\frac{d}{dx}\left(\operatorname{tan}^{-1} x\right) \Rightarrow \frac{1}{1+x^{2}}\\ &\frac{d y}{d x}=\frac{1}{2} \times \frac{1}{1+(a x)^{2}} \times \frac{1}{a}\\ &\frac{d y}{d x}=\frac{a}{2\left(1+a^{2} x^{2}\right)} \end{aligned}$

Differentiation Exercise 10.3 Question 21

Answer:$\frac{dy}{dx}=\frac{1}{2}$
Hint:
$\frac{dy}{dx}\left ( x^{n} \right )=nx^{n-1}$
$\frac{dy}{dx}\left (constant\right )=0$
Given:
$\tan ^{-1}\left\{\frac{\sin x}{1+\cos x}\right\},-\pi<x<\pi$
Solution:
Let,
$y=\tan ^{-1}\left\{\frac{\sin x}{1+\cos x}\right\}$
Function y is defined for all the real numbers where $\cos x\neq -1$
$\begin{aligned} &\text { Using } 2 \cos ^{2} \theta=1+\cos 2 \theta \\ &2 \sin \theta \cos \theta=\sin 2 \theta \\ &y=\tan ^{-1}\left\{\frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}\right\} \\ &y=\tan ^{-1}\left\{-\frac{\sin x / 2}{\cos x / 2}\right\} \\ &y=\tan ^{-1}\left\{\tan \frac{x}{2}\right\} \\ &y=\frac{x}{2} \end{aligned}$
Differentiating with respect to x , We get
$\frac{dy}{dx}=\frac{d}{dx}\left ( \frac{x}{2} \right )$
$\frac{dy}{dx}=\frac{1}{2}$

Differentiation Exercise 10.3 Question 22

Answer: $\frac{dy}{dx}=\frac{-1}{1+x^{2}}$
Hint:
$\begin{aligned} &\frac{d}{d x}(\text { Constant })=0 \\ &\frac{d}{d x}\left(\mathrm{x}^{n}\right)=n x^{n-1} \end{aligned}$
Given:
$\sin ^{-1}\left \{ \frac{1}{\sqrt{1+x^{2}}} \right \}$
Solution:
Let,
$y=\sin ^{-1}\left \{ \frac{1}{\sqrt{1+x^{2}}} \right \}$
Let
$x=\cot \theta$
$\theta =\cot ^{-1}x$
Now,
$y=\sin ^{-1}\left \{ \frac{1}{\sqrt{1+\cot ^{2}\theta }} \right \}$
Using,
$\begin{aligned} &1+\cot ^{2} \theta=\operatorname{cosec}^{2} \theta \\ &y=\sin ^{-1}\left\{\frac{1}{\sqrt{\operatorname{cosec}^{2} \theta}}\right\} \\ &y=\sin ^{-1}\left\{\frac{1}{\operatorname{cosec} \theta}\right\} \end{aligned}$ $\left \{ \frac{1}{cosec\theta }=\sin \theta \right \}$
$y=\sin ^{-1}\left ( \sin \theta \right )$ $\sin ^{-1}(\sin \theta)=\sin \theta \text {,if } \theta \varepsilon\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$
$y=\theta$$y=\cot ^{-1}x$

Differentiating with respect to x , We get

$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(\cot ^{-1} x\right) \\ &\therefore \frac{d}{d x}\left(\cot ^{-1} x\right) \Rightarrow \frac{-1}{1+x^{2}} \\ &\frac{d y}{d x}=-\frac{1}{1+x^{2}} \end{aligned}$

Differentiation Exercise 10.3 Question 23

Answer:
$\frac{dy}{dx}=\frac{2nx^{n-1}}{1+x^{2n}}$
Hint:
$\begin{aligned} &\frac{\mathrm{d}}{\mathrm{dx}}(\text { constsant })=0 \\ &\frac{d}{d \mathrm{x}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}-1} \end{aligned}$
Given:
$\begin{aligned} &\cos ^{-1}\left\{\frac{1-x^{2 n}}{1+x^{2 n}}\right\} \\ &0<x<\infty \end{aligned}$
Solution:
$y=cos ^{-1}\left\{\frac{1-x^{2 n}}{1+x^{2 n}}\right\}$
Let,
$x^{n}=\tan \theta$
$\theta =\tan ^{-1}\left ( x^{n} \right )$
$y=\cos ^{-1}\left \{ \frac{1-\tan ^{2}\theta }{1+\tan ^{2}\theta } \right \}$$y=\cos ^{-1}\left \{ \cos 2\theta \right \}$

Using,

$\frac{1-\tan ^{2}\theta }{1+\tan ^{2}\theta }=\cos 2\theta$

Considering the limits,

$\begin{aligned} &0<x<\infty \\ &0<x^{n}<\infty \\ &0<\theta<\frac{\pi}{2} \end{aligned}$

Now, $\left \{ \cos ^{-1}\left ( \cos \theta \right )=\theta ,if\: \theta \varepsilon \left [ 0,\pi \right ] \right \}$

$y=\cos ^{-1}\left ( \cos 2\theta \right )$

$y=2\theta$ $Since\hspace{0.2cm}x^{n}=\tan \theta$
Differentiating with respect to x we get
$\frac{dy}{dx}=\frac{d}{dx}\left ( 2\tan ^{-1}\left ( x^{n} \right ) \right )$
Using
$\begin{aligned} &\frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^{2}} \\ &\frac{d y}{d x}=\frac{2 \times 1}{1+\left(x^{n}\right)^{2}} \times n x^{n-1} \\ &\frac{d y}{d x}=\frac{2 n x^{n-1}}{1+x^{2 n}} \end{aligned}$

Differentiation Exercise 10.3 Question 24

Answer:$\frac{dy}{dx}=0$
Hint:
$\begin{aligned} &\frac{\mathrm{d}}{\mathrm{dx}}(\text { constsant })=0 \\ &\frac{d}{d \mathrm{x}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}-1} \end{aligned}$
Given:
$\sin ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)+\sec ^{-1}\left(\frac{1+x^{2}}{1-x^{2}}\right)$
Solution:
Let,
$y=\sin ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)+\sec ^{-1}\left(\frac{1+x^{2}}{1-x^{2}}\right)$
Using
$\begin{aligned} &\sec ^{-1} x=\cos ^{-1}\left(\frac{1}{x}\right) \\ &y=\sin ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)+\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) \end{aligned}$
Using$\begin{aligned} &\sin ^{-1} \mathrm{x}+\cos ^{-1} \mathrm{x}=\frac{\pi}{2} \\ &\mathrm{y}=\frac{\pi}{2} \end{aligned}$

Differentiating with respect to x we get

$\frac{dy}{dx}=\frac{d}{dx}\left ( \frac{\pi }{2} \right )$

$\frac{dy}{dx}=0$ $\left \{ \frac{d}{dx}\left ( constant \right )=0 \right \}$

Differentiation Exercise 10.3 Question 26

Answer:$\frac{1}{2\sqrt{x}\left ( 1+x \right )}$
Hint:
$\frac{d}{dx}\left ( constant \right )=0;$
$\frac{d}{d x}\left ( x^{n} \right )=nx^{n-1}$
Given:
$\tan ^{-1}\left ( \frac{\sqrt{x}+\sqrt{a}}{1-\sqrt{xa}} \right )$
Solution:
Let,
$y=\tan ^{-1}\left ( \frac{\sqrt{x}+\sqrt{a}}{1-\sqrt{xa}} \right )$
Since,$\tan ^{-1} \mathrm{x}+\tan ^{-1} \mathrm{y}=\tan ^{-1}\left(\frac{\mathrm{x}+\mathrm{y}}{1-\mathrm{xy}}\right)$
$y=\tan ^{-1}\sqrt{x}+\tan ^{-1}\sqrt{a}$
Differentiating it with respect to x using chain qule.
$\frac{dy}{dx}=\frac{d}{dx}\left ( \tan ^{-1}\sqrt{x} \right )+\frac{d}{dx}\left ( \tan ^{-1}\sqrt{a} \right )$
Since ,
$\begin{aligned} &\frac{\mathrm{d}}{\mathrm{dx}}\left(\tan ^{-1} \mathrm{x}\right)=\frac{1}{1+\mathrm{x}^{2}} ; \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}-1} \\ &\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{1+(\sqrt{\mathrm{x}})^{2}} \frac{\mathrm{d}}{\mathrm{dx}}(\sqrt{\mathrm{x}})+0 \\ &\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{1+\left(\mathrm{x}^{1 / 2}\right)^{2}} \times \frac{1}{2}(\mathrm{x})^{\frac{1}{2}-1} \\ &\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{1+\mathrm{x}} \times \frac{1}{2}(\mathrm{x})^{-\frac{1}{2}} \\ &\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{1+\mathrm{x}} \times \frac{1}{2 \sqrt{\mathrm{x}}} \\ &\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2 \sqrt{\mathrm{x}}(1+\mathrm{x})} \end{aligned}$

Differentiation Exercise 10.3 Question 27

Answer:$\frac{dy}{dx}=1$
Hint:
$\frac{d}{dx}\left ( constant \right )=0$
$\frac{d }{dx}\left (x^{n} \right )=nx^{n-1}$
Given:
$\tan ^{-1}\left[\frac{a+\operatorname{btan} x}{b-\operatorname{atan} x}\right]$
Solution:
Let,
$\begin{aligned} &y=\tan ^{-1}\left[\frac{a+b \tan x}{b-\operatorname{atan} x}\right] \\ &y=\tan ^{-1}\left[\frac{\frac{a+b \tan x}{b}}{\frac{b-a t a n x}{b}}\right] \\ &y=\tan ^{-1}\left[\frac{\frac{a}{b}+\tan x}{1-\frac{a}{b} \tan x}\right] \end{aligned}$
$\begin{aligned} &y=\tan ^{-1}\left[\frac{\tan \left(\tan ^{-1} \frac{a}{b}\right)+\tan x}{1-\tan \left(\tan ^{-1} \cdot \frac{g}{b}\right) \tan x}\right] \\ &y=\tan ^{-1}\left[\tan \left(\tan ^{-1} \frac{a}{b}+x\right)\right] \\ &\therefore \tan (A+B)=\frac{\tan A+\tan \vec{B}}{1-\tan A \tan B} \\ &y=\tan ^{-1}\left(\frac{a}{b}\right)+x \end{aligned}$
Differentiating it with respect to x , We get
$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(\tan ^{-1}\left(\frac{a}{b}\right)\right)+\frac{d}{d x}(x) \\ &\frac{d}{d x}(\text { constant })=0 \\ &\frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^{2}} \\ &\frac{d y}{d x}=\frac{1}{1+\left(\frac{a}{b}\right)^{2}} \times 0+1 \\ &\frac{d y}{d x}=0+1 \\ &\frac{d y}{d x}=1 \end{aligned}$

Differentiation Exercise 10.3 Question 28

Answer:$\frac{dy}{dx}=\frac{1}{1+x^{2}}$
Hint:
$\begin{aligned} &\frac{d}{d x}(\text { constant })=0 ; \\ &\frac{d}{d x}\left(x^{n}\right)=n x^{n-1} \end{aligned}$
Given:
$\tan ^{-1}\left ( \frac{a+bx}{b-ax} \right )$
Solution:
Let,
$\begin{aligned} &y=\tan ^{-1}\left(\frac{a+b x}{b-a x}\right) \\ &y=\tan ^{-1}\left[\frac{\frac{a+b x}{b}}{\frac{b-a x}{b}}\right] \end{aligned}$

Since,$\tan ^{-1}x+\tan ^{-1}y=\tan ^{-1}\left ( \frac{x+y}{1-xy} \right )$

Differentiating it with respect to x , We get

$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(\tan ^{-1}\left(\frac{a}{b}\right)+\tan ^{-1} x\right. \\ &\text { Since, } \frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^{2}} \\ &\frac{d y}{d x}=\frac{1}{1+\left(\frac{a}{b}\right)^{2}} \times 0+\frac{1}{1+x^{2}} \\ &\frac{d y}{d x}=0+\frac{1}{1+x^{2}} \\ &\frac{d y}{d x}=\frac{1}{1+x^{2}} \end{aligned}$

Differentiation Exercise 10.3 Question 29

Answer:$\frac{dy}{dx}=\frac{a}{a^{2}+x^{2}}$
Hint:
$\begin{aligned} &\frac{d}{d x}(\text { constan } t)=0 ; \\ &\frac{d}{d x}\left(x^{n}\right)=n x^{n-1} \end{aligned}$
Given:
$\tan ^{-1}\left ( \frac{x-a}{x+a} \right )$
Solution:
Let,
$\begin{aligned} &y=\tan ^{-1}\left(\frac{x-a}{x+a}\right) \\ &y=\tan ^{-1}\left(\frac{\frac{x-a}{x}}{\frac{x+a}{x}}\right) \end{aligned}$$\begin{aligned} &y=\tan ^{-1}\left(\frac{\frac{x}{x}-\frac{a}{x}}{\frac{x}{x}+\frac{a}{x}}\right) \\ &y=\tan ^{-1}\left(\frac{1-\frac{a}{x}}{1+1 \times \frac{a}{x}}\right) \\ &y=\tan ^{-1}(1)+\tan ^{-1}\left(\frac{a}{x}\right) \end{aligned}$

Using,

$\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$

Differentiating its with respect to $x$using chain tule.

$\begin{aligned} &\frac{d y}{d x}=0-\frac{1}{1+\left(\frac{a}{x}\right)^{2}} \frac{d}{d x}\left(\frac{a}{x}\right) \\ &\frac{d y}{d x}=\frac{-1}{1+\frac{a^{2}}{x^{2}}}\left(-\frac{a}{x^{2}}\right) \\ &\frac{d y}{d x}=\frac{-1}{\frac{x^{2}+a^{2}}{x^{2}}} \times\left(\frac{-a}{x^{2}}\right) \\ &\frac{d y}{d x}=\frac{-x^{2}}{x^{2}+a^{2}} \times\left(\frac{-a}{x^{2}}\right) \end{aligned}$ Using $\frac{d}{dx}\left ( x^{n} \right )=nx^{n-1}$

$\frac{d}{dx}=\frac{a}{x^{2}+a^{2}}$ $\frac{d}{dx}\left ( constant \right )=0$

Differentiation Exercise 10.3 Question 30

Answer: $\frac{dy}{dx}=\frac{3}{1+9x^{2}}-\frac{2}{1+4x^{2}}$
Hint:
$\begin{aligned} &\frac{\mathrm{d}}{\mathrm{dx}}(\text { constan } \mathrm{t})=0 ; \\ &\frac{d}{d \mathrm{x}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}-1} \end{aligned}$
Given:
$\tan ^{-1}\left(\frac{x}{1+6 x^{2}}\right)$
Solution:
Let, $y=\tan ^{-1}\left(\frac{x}{1+6 x^{2}}\right)$
$y=\tan ^{-1}\left(\frac{3 x-2 x}{1+(3 x)(2 x)}\right)$
Since,$3x-2x=x$
$\begin{aligned} &y=\tan ^{-1} 3 x-\tan ^{-1} 2 x \\ &{\left[\text { Since }+\tan x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)\right]} \end{aligned}$
Differentiating it with respect tox using chain rule,
$\begin{aligned} &\frac{d y}{d x}=\frac{1}{1+\left(3 x^{2}\right)} \frac{d}{d x}(3 x)-\frac{1}{1+(2 x)^{2}} \frac{d}{d x}(2 x) \\ &\left\{\text { Since } \Rightarrow \frac{\mathrm{d} }{\mathrm{d} x}\tan ^{-1}(x)=\frac{1}{1+x^{2}}\right\} \\ &\frac{d y}{d x}=\frac{1}{1+9 x^{2}}(3)-\frac{1}{1+4 x^{2}}(2) \\ &\frac{d y}{d x}=\frac{3}{1+9 x^{2}}-\frac{2}{1+4 x^{2}} \end{aligned}$

Differentiation Exercise 10.3 Question 31

Answer:$\frac{dy}{dx}=\frac{3}{1+9x^{2}}+\frac{2}{1+4x^{2}}$
Hint:
$\frac{d}{dx}\left ( constant \right )=0;$
$\frac{d }{dx}\left (x^{n}\right )=nx^{n-1}$
Given:
$\tan ^{-1}\left ( \frac{5x}{1-6x^{2}} \right )$
Solution:
Let,
$\begin{aligned} &y=\tan ^{-1}\left(\frac{5 x}{1-6 x^{2}}\right) \\ &y=\tan ^{-1}\left(\frac{3 x+2 x}{1-(3 x)(2 x)}\right) \end{aligned}$
$\begin{aligned} &\text { By using } 5 x=3 x+2 x \\ &y=\tan ^{-1}(3 x)+\tan ^{-1}(2 x) \end{aligned}$
$\text { Since, } \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$
Differentiating it with respect to x using chain rule,
$\frac{d y}{d x}=\frac{d}{d x}\left(\tan ^{-1}(3 x)\right)+\frac{d}{d x}\left(\tan ^{-1}(2 x)\right)$
Using
$\begin{aligned} &\frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^{2}} \\ &\frac{d y}{d x}=\frac{1}{1+(3 x)^{2}} \frac{d}{d x}(3 x)+\frac{1}{1+(2 x)^{2}} \frac{d}{d x}(2 x) \\ &\frac{d y}{d x}=\frac{1}{1+9 x^{2}}(3)+\frac{1}{1+4 x^{2}}(2) \\ &\frac{d y}{d x}=\frac{3}{1+9 x^{2}}+\frac{2}{1+4 x^{2}} \end{aligned}$

Differentiation Exercise 10.3 Question 32

Answer: $\frac{dy}{dx}=1$
Hint:
$\begin{aligned} &\frac{\mathrm{d}}{\mathrm{dx}}(\text { constan } \mathrm{t})=0 ; \\ &\frac{d}{d \mathrm{x}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}-1} \end{aligned}$
Given:
$\tan ^{-1}\left[\frac{\cos x+\sin x}{\cos x-\sin x}\right]$
Solution:
$\begin{aligned} &y=\tan ^{-1}\left[\frac{\cos x+\sin x}{\cos x-\sin x}\right] \\ &y=\tan ^{-1}\left[\frac{\frac{\cos x+\sin x}{\cos x}}{\frac{\cos x-\sin x}{\cos x}}\right] \\ &y=\tan ^{-1}\left[\frac{1+\tan x}{1-\tan x}\right] \\ &\text { since, } \frac{\sin x}{\cos x}=\operatorname{tian} x \end{aligned}$
$\begin{aligned} &y=\tan ^{-1}\left[\frac{\tan \frac{\pi}{4}+\tan x}{1-\frac{\tan \pi}{4} \tan x}\right] \\ &y=\tan ^{-1}\left[\tan \left(\frac{\pi}{4}+x\right)\right] \\ &\text { Since } \tan (A+B)=\frac{\tan A+\tan B}{1-\tan \tan B} \\ &y=\frac{\pi}{4}+x \end{aligned}$
Differentiating it with respect to x,
$\frac{dy}{dx}=0+1$ $\left\{\begin{array}{l} \frac{\mathrm{d}}{\mathrm{dx}}(\text { constant })=0 \\ \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})=1 \end{array}\right\}$
$\frac{dy}{dx}=1$

Differentiation Exercise 10.3 Question 33

Answer: $\frac{dy}{dx}=\frac{1}{3x^{2}\left ( 1+x^{2}3 \right )}$
Hint:
$\begin{aligned} &\frac{d}{d x}(\text { constant })=0 ; \\ &\frac{d}{d x}\left(x^{n}\right)=n x^{n-1} \end{aligned}$
Given:
$\tan ^{-1}\left[\frac{\mathrm{x}^{1 / 3}+\mathrm{a}^{1 / 3}}{1-(\mathrm{ax})^{1 / 3}}\right]$
Solution:
Let,
$\begin{aligned} &y=\tan ^{-1}\left[\frac{x^{13}+a^{1 / 3}}{1-(a x)^{1 / 3}}\right] \\ &y=\tan ^{-1}\left(x^{1 / 3}\right)+\tan ^{-1}\left(a^{1 / 3}\right) \\ &\text { Since, } \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right) \end{aligned}$
Differentiating it with respect to x using chain rule,
$\begin{aligned} &\frac{d y}{d x}=\frac{1}{1+\left(x^{1 / 3}\right)^{2}} \frac{d}{d x}\left(x^{1 / 3}\right)+\frac{1}{1+\left(a^{1 / 3}\right)^{2}} \frac{d}{d x}\left(a^{1 / 3}\right) \\ &\frac{d y}{d x}=\frac{1}{1+x^{2 / 3}} \times\left(\frac{1}{3} x^{\frac{1}{3}-1}\right)+\frac{1}{1+\left(a^{1 / 3}\right)^{2}} \times 0 \\ &\frac{d y}{d x}=\frac{1}{1+x^{2 / 3}} \times \frac{1}{3} x^{-2 / 3} \\ &\frac{d y}{d x}=\frac{1}{3 x^{2 / 3}\left(1+x^{2 / 3}\right)} \end{aligned}$
Using
$\begin{aligned} &\frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^{2}} \\ &\frac{d}{d x}\left(x^{n}\right)=n x^{n-1} \\ &\frac{d}{d x}(\text { constant })=0 \end{aligned}$

Differentiation Exercise 10.3 Question 34

Answer:$\frac{dy}{dx}=\frac{2^{x+1}log2}{1+4^{x}}$
Hint:
$\frac{d}{dx}\left ( constant \right )=0$
$\frac{\partial }{\partial x}\left ( x^{n} \right )=nx^{n-1}$
Given:
$\sin ^{-1}\left ( \frac{2^{x+1}}{1+4^{x}} \right )$
Solution:
Let
$y=\sin ^{-1}\left ( \frac{2^{x+1}}{1+4^{x}} \right )$
To find the domain we need to find all $x$ such that
$-1 \leq \frac{2^{x+1}}{1+4 x} \leq 1$
Since the quantity in the middle is always positive, we need to find
all such that $\frac{2^{x+1}}{1+4 x} \leq 1$
ie, all $x$ such that $2^{x+1}=1+4^{x}$
$2\leq \frac{1}{2^{x}}+2^{x}$, which is true for a $x$
Hence the function is defined at all real numbers
Putting $2^{x}=\tan \theta$
$\begin{aligned} &y=\sin ^{-1}\left(\frac{2^{x+1}}{1+4 x}\right) \\ &y=\sin ^{-1}\left(\frac{2^{x} \cdot 2}{1+\left(2^{x}\right)^{2}}\right) \\ &y=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right) \\ &\text { Using } \sec ^{2} \theta=1+\tan ^{2} \theta \end{aligned}$
$\begin{aligned} &\mathrm{y}=\sin ^{-1}\left(\frac{2 \tan \theta}{\sec ^{2} \theta}\right) \\ &\mathrm{y}=\sin ^{-1}\left(2 \frac{\sin \theta}{\cos \theta} \times \cos ^{2} \theta\right) \quad \text \; \; \; \; \; \; \; \; \; \; \; { Since, } \tan \theta=\frac{\sin \theta}{\cos \theta^{\prime} \sec \theta}, \frac{1}{\sec }=\cos \theta \\ &\mathrm{y}=\sin ^{-1}(2 \sin \theta \cos \theta) \\ &\mathrm{U} \sin \mathrm{sin}=\sin 2 \theta=2 \sin \theta \cos \theta \end{aligned}$
$y=\sin ^{-1}\left ( \sin \theta \right )$ $\left\{\begin{array}{l} \sin ^{-1}(\sin \theta)=\theta \\ \text { if } \theta \in\left[\frac{-1}{2}, \frac{0}{2}\right] \end{array}\right\}$
$y=2\theta$
$y=2\tan ^{-1}\left ( 2^{x} \right )$ $\left\{\begin{array}{l} \text { since } 2^{\mathrm{x}} \quad=\tan \theta \\ \boldsymbol{\theta}=\tan ^{-1}\left(2^{\mathrm{x}}\right) \end{array}\right\}$
$\begin{aligned} &\frac{d y}{d x} \Rightarrow \frac{d}{d x}\left(2 \tan ^{-1}\left(2^{x}\right)\right) \\ &\frac{d y}{d x}=2 \times \frac{1}{1+\left(2^{x}\right)^{2}} \frac{\partial}{d x}\left(2^{x}\right) \\ &\frac{d y}{d x}=2 \times \frac{1}{1+4^{x}} \cdot\left(2^{x}\right) \log 2 \\ &\frac{d y}{d x} \Rightarrow \frac{2^{x+1} \log 2}{1+4^{x}} \end{aligned}$ $\text { Since } \frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^{2}}$

Differentiation Exercise 10.3 Question 35

Answer: Hence Prove$\frac{dy}{dx}=\frac{4}{1+x^{2}}$
Hint:
$\begin{aligned} &\frac{\mathrm{d}}{d \mathrm{x}}(\text { constan } \mathrm{t})=0 ; \\ &\frac{d}{d \mathrm{x}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{nx}^{\mathrm{n}-1} \end{aligned}$
Given:
$\begin{aligned} &y=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)+\sec ^{-1}\left(\frac{1+x^{2}}{1-x^{2}}\right) \\ &0<x<1 \end{aligned}$
Solution:
Prove :$\frac{dy}{dx}=\frac{4}{1+x^{2}}$
$\begin{aligned} &y=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)+\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) \\ &\text { Since } \sec ^{-1} x=\cos ^{-1}\left(\frac{1}{x}\right) \\ &y=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)+\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) \end{aligned}$
Let,
$x=\tan \theta$
$\theta =\tan ^{-1}x$
$\begin{aligned} &y=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right)+\cos ^{-1}\left(\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right) \\ &y=\sin ^{-1}\left(\frac{2 \tan \theta}{\sec ^{2} \theta}\right)+\cos ^{-1}(\cos 2 \theta) \end{aligned}$
Using
$\begin{aligned} &\sec ^{2} \theta=1+\tan ^{2} \theta \\ &\frac{1}{\sec \theta}=\cos \theta \\ &\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}=\cos 2 \theta \end{aligned}$
$\begin{aligned} &\mathrm{y}=\sin ^{-1}\left(2 \frac{\sin \theta}{\cos \theta} \times \cos ^{2} \theta\right)+\cos ^{-1}(\operatorname{Cos} 2 \theta) \\ &\mathrm{y}=\sin ^{-1}(2 \sin \theta \cos \theta)+\cos ^{-1}(\cos 20) \\ &\mathrm{y}=\sin ^{-1}(\sin 2 \theta)+\cos ^{-1}(\cos 2 \theta)-(\mathrm{i}) \\ &\mathrm{u} \operatorname{sing} 2 \sin \theta \cos \theta=\sin 2 \theta \end{aligned}$
Considering Limits,
$\begin{aligned} &0<\mathrm{x}<1 \\ &0<\tan \theta<1 \\ &0<\theta<\frac{\mathrm{\pi}}{4} \\ &0<(2 \theta)<\frac{\mathrm{\pi}}{2} \end{aligned}$
so from $e_{i}\: \: \left ( i \right )$
$y=2\theta +2\theta$
$y=4\theta$
$\begin{aligned} &\cos ^{-1}(\cos \theta) \Rightarrow \text { if } \theta \in[0, \pi] \\ &y=4 \tan ^{-1} x \end{aligned}$ $since,x=\tan \theta ,\theta =\tan ^{-1}x$
Differentiating it with respect to x
$\begin{aligned} &\frac{d y}{d x}=4 \frac{d}{d x}\left(\tan ^{-1} x\right) \\ &\text { Using } \frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^{2}} \\ &\frac{d y}{d x}=4\left(-\frac{1}{1+x^{2}}\right) \\ &\frac{d y}{d x}=\frac{4}{1+x^{2}} \end{aligned}$

Differentiation Exercise 10.3 Question 36

Answer: Hence Prove ,$\frac{dy}{dx}=\frac{2}{1+x^{2}}$
Hint:
$\begin{aligned} &\frac{d}{d x}(\text { constan } t)=0 ; \\ &\frac{d}{d x}\left(x^{n}\right)=n x^{n-1} \end{aligned}$
Given:
$\begin{aligned} &\mathrm{y}=\sin ^{-1}\left(\frac{\mathrm{x}}{\sqrt{1+\mathrm{x}^{2}}}\right)+\cos ^{-1}\left(\frac{1}{\sqrt{1+\mathrm{x}^{2}}}\right) \\ &0<\mathrm{x}<\infty \end{aligned}$
Solution:
Let,
$x=\tan \theta ,$
$\theta =\tan ^{-1}x$
$\begin{aligned} &y=\sin ^{-1}\left(\frac{\tan \theta}{\sqrt{1+\tan ^{2} \theta}}\right)+\cos ^{-1}\left(\frac{1}{\sqrt{1+\tan ^{2} \theta}}\right) \\ &y=\sin ^{-1}\left(\frac{\tan \theta}{\sqrt{\sec ^{2} \theta}}\right)+\cos ^{-1}\left(\frac{1}{\sqrt{\sec ^{2} \theta}}\right) \\ &\text { Using } \sec ^{2} \theta=1+\tan ^{2} \theta \\ &\frac{\sin \theta}{\cos \theta}=\tan \theta \\ &\frac{1}{\sec \theta}=\theta \cos \theta \end{aligned}$
$\begin{aligned} &y=\sin ^{-1}\left(\frac{\tan \theta}{\sec \theta}\right)+\cos ^{-1}\left(\frac{1}{\sec \theta}\right) \\ &y=\sin ^{-1}\left(\frac{\sin \theta}{\cos \theta} \times \cos \theta\right)+\cos ^{-1}(\cos \theta) \\ &y=\sin ^{-1}(\sin \theta)+\cos ^{-1}(\cos \theta)-(i) \end{aligned}$
Considering limit

$\begin{aligned} &0<\mathrm{x}<\infty \\ &0<\tan \theta<\infty \\ &0<\theta<\frac{\mathrm{\pi}}{2} \end{aligned}$
So From eq (i)
$y=\theta +\theta$ $\text { Since, } \sin ^{-1}(\sin \theta)=\theta \text { if } \theta \varepsilon\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$
$y=2\theta$ $\cos ^{-1}(\cos \theta)=\theta \text { if } \theta \varepsilon[0, \pi]$
$y=2\tan ^{-1}x$ $\left [ since x=\tan \theta \right ]$

Differentiating it with respect to x

$\begin{aligned} &\frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^{2}} \\ &\frac{d y}{d x}=2 \times \frac{1}{1+x^{2}} \\ &\frac{d y}{d x}=\frac{2}{1+x^{2}} \end{aligned}$

Differentiation Exercise 10.3 Question 37(i)

Answer: $\frac{dy}{dx}=-1$
Hint:
$\begin{aligned} &\frac{\mathrm{d}}{\mathrm{dx}}(\text { constan } \mathrm{t})=0 \\ &\frac{d}{d \mathrm{x}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}-1} \end{aligned}$
Given:
$\cos ^{-1}\left ( \sin x \right )$
Solution:
Let,
$y=\cos ^{-1}\left ( \sin x \right )$
We observe that this function is defined for all real numbers
$\begin{aligned} &y=\cos ^{-1}(\sin x) \\ &y=\cos ^{-1}\left[\cos \left(\frac{\pi}{2}-x\right)\right] \\ &y=\frac{\pi}{2}-x \end{aligned}$ $\left\{\begin{array}{r} \operatorname{since} \cos ^{-1}(\cos \theta)=\theta \\ \text { if } \theta \in[\theta, \mathbf{\pi}] \end{array}\right\}$
Differentiating it with respect to x,
$\begin{aligned} &\frac{d y}{d x}=0-1 \\ &\frac{d y}{d x}=-1 \end{aligned}$ $\left \{ Since,\frac{d\left ( constant \right )}{dx}=0 \right \}$

Differentiation Exercise 10.3 Question 37 (ii)

Answer: $\frac{dy}{dx}=\frac{1}{1+x^{2}}$
Hint:
$\begin{aligned} &\frac{d}{d x}(\text { constan } t)=0 \\ &\frac{d}{d x}\left(x^{n}\right)=n x^{n-1} \end{aligned}$
Given:
$\cos ^{-1}\left ( \frac{1-x}{1+x} \right )$
Solution:
$y=\cos ^{-1}\left ( \frac{1-x}{1+x} \right )$
Let,
$\begin{aligned} &\mathrm{x}=\tan \theta \\ &\theta=\tan ^{-1} \mathrm{x} \\ &\mathrm{y}=\cot ^{-1}\left(\frac{1-\tan \theta}{1+\tan \theta}\right) \\ &\mathrm{y}=\cot ^{-1}\left(\frac{\tan \frac{\mathrm{n}}{4}-\tan \theta}{1+\tan \frac{\mathrm{n}}{4} \tan \theta}\right) \end{aligned}$
$\begin{aligned} &y=\cot ^{-1}\left[\tan \left(\frac{n}{4}-\theta\right)\right]\\ &\text { Using, }\\ &\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}\\ &y=\cot ^{-1}\left[\tan \left(\frac{\pi}{4}-\theta\right)\right]\\ &\mathrm{y}=\cot ^{-1}\left[\operatorname{cots}\left(\frac{\mathrm{\pi}}{2}-\left(\frac{\mathrm{\pi}-\theta}{4}-\theta\right)\right]\right. \end{aligned}$ $\therefore since\: \cot \left ( \frac{\pi}{2}-\theta \right )=\tan \theta$
$\begin{aligned} &y=\cot ^{-1}\left[\cot \left(\frac{\pi}{2}-\frac{\pi}{4}+\theta\right)\right] \\ &y=\cot ^{-1}\left[\cot \left(\frac{\pi}{4}+\theta\right)\right] \\ &y=\frac{\pi}{4}+\theta \end{aligned}$ $\left\{\cot ^{-1}(\cot \theta)=\theta, \text { if } \theta<\left[-\frac{\mathbf{\pi}}{2}, \frac{\pi}{2}\right]\right\}$
$y=\frac{\pi}{4}+\tan ^{-1}x$ $\left\{\begin{array}{c} \text { Since } \mathrm{x}=\tan \theta \\ \theta=\tan ^{-1} \mathrm{x} \end{array}\right\}$
Differentiating it with respect to x,
$\begin{aligned} &\frac{d y}{d x}=0+\frac{1}{1+x^{2}} \\ &\frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^{2}} \\ &\frac{\partial}{d x}(\text { constant })=0 \\ &\frac{d y}{d x}=\frac{1}{1+x^{2}} \end{aligned}$

Differentiation Exercise 10.3 Question 38

Answer:$\frac{dy}{dx}=\frac{1}{2},$ Hence,$\frac{dy}{dx}$ is independent of x
Hint:
$\begin{aligned} &\frac{d}{d x} \text { (constant) }=0 ; \\ &\frac{d}{dx}\left(x^{n}\right)=n x^{n-1} \end{aligned}$
Given:
$y=\cot ^{-1}\left\{\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right\}$
Solution:
$\begin{aligned} &y=\cot ^{-1}\left\{\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right\} \\ &\text { Then, } \frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}} \\ &\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}} \times \frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}+\sqrt{1-\sin x}} \end{aligned}$
$\begin{aligned} &\frac{(\sqrt{1+\sin x}+\sqrt{1+\sin x})^{2}}{(1+\sin x)-(1-\sin x)} \\ &\begin{array}{l} \text { Using }(a-b)(a+b)=a^{2}-b^{2} \\ (a+b)^{2}=a^{2}+b^{2}+2 a b \\ \frac{(1+\sin x)+(00(1-\sin x)+2 \sqrt{(1-\sin x)(1+\sin x)}}{(1+\sin x)-(1-\sin x)} \\ \frac{1+\sin x+1-\sin x+2 \sqrt{(1-\sin x)(1+\sin x)}}{1+\sin x-1+\sin x} \end{array} \end{aligned}$
$\begin{aligned} &\Rightarrow \frac{2+2 \sqrt{1+\sin x-\sin x-\sin ^{2} x}}{2 \sin x} \\ &\Rightarrow \frac{2+2 \sqrt{1-\sin ^{2} x}}{2 \sin x} \\ &\Rightarrow \frac{2\left(1+\sqrt{1-\sin ^{2} x}\right)}{2 \sin x} \\ &\Rightarrow \frac{1+\sqrt{1 \sin ^{2} x}}{\sin x} \end{aligned}$
Using $\cos ^{2}x+\sin ^{2}x=1$
$\begin{aligned} &\Rightarrow \frac{1+\sqrt{\cos ^{2} x}}{\sin x} \\ &\Rightarrow \frac{1+\cos x}{\sin x} \\ &\text { Using } \sin 2 \theta=2 \sin \theta \cos \theta \\ &1+\operatorname{Cos} 2 \theta=2 \cos ^{2} \theta \\ &\Rightarrow \frac{2 \cos ^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}} \\ &\Rightarrow \cot \frac{x}{2} \end{aligned}$ $\left \{ \frac{\cos x}{\sin x}=\cot x\right \}$

Therefore, equation (1) becomes

$\begin{aligned} &y=\cot ^{-1}\left(\cot \frac{x}{2}\right) \\ &y=\frac{x}{2} \end{aligned}$ $\left\{\cot ^{-1}(\cot \theta)=\theta \operatorname{if} \theta^{\varepsilon}\left[\frac{-\pi}{2}, \frac{\mathbf{\pi}}{2}\right]\right\}$

Differentiating it with respect to $x$

$\frac{dy}{dx}=\frac{1}{2}\frac{d}{dx}\left ( x \right )$

$\frac{dy}{dx}=\frac{1}{2}$

Differentiation Exercise 10.3 Question 39

Answer: $\frac{dy}{dx}=\frac{4}{1+x^{2}}$
Hint:
$\begin{aligned} &\frac{\mathrm{d}}{\mathrm{dx}}(\text { constiant })=0 \\ &\frac{d}{d \mathrm{x}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{nx}^{\mathrm{n}-1} \end{aligned}$
Given:
$\begin{aligned} &y=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)+\sec ^{-1}\left(\frac{1+x^{2}}{1-x^{2}}\right) \\ &x=0 \end{aligned}$
Prove
$\frac{dy}{dx}=\frac{1}{1+x^{2}}$
Solution:
Here,$y=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)+\sec ^{-1}\left(\frac{1+x^{2}}{1-x^{2}}\right) \$
Using $\sec ^{-1}$
$\begin{aligned} &\mathrm{x}=\cos ^{-1}\left(\frac{1}{\mathrm{x}}\right) \\ &\mathrm{y}=\tan ^{-1}\left(\frac{2 \mathrm{x}}{1-\mathrm{x}^{2}}\right)+\cos ^{-1}\left(\frac{1-\mathrm{x}^{2}}{1+\mathrm{x}^{2}}\right) \end{aligned}$
Put $x=\tan \theta$$\begin{aligned} &y=\tan ^{-1}\left(\frac{2 \tan \theta}{1-\tan ^{2} \theta}\right)+\cos ^{-1}\left(\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right) \\ &y=\tan ^{-1}(\tan 2 \theta)+\cos ^{-1}(\cos 2 \theta)-(1) \end{aligned}$

Using,

$\tan 2\theta =\frac{2\tan \theta }{1-\tan ^{2}\theta }$

$\begin{aligned} &\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}=\cos 2 \theta \\ &\text { Here, } 0<\mathrm{x}<\infty \\ &0<\tan \theta<\infty \\ &0<\theta<\frac{\mathrm{\pi}}{2} \\ &0<2 \theta<\mathrm{\pi} \end{aligned}$

So From eq(i)

$y=2\theta +2\theta$

$\left\{\text { Since, } \tan ^{-1}(\tan \theta)=\theta \text { if } \theta \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right] \text { and } \cos ^{-1}(\cos \theta)=\theta \text { if } \theta[0, \pi]\right\}$

$y=4\theta$

$y=4\tan ^{-1}x$ $\left [ Using\: x=\tan \theta ,\theta =\tan ^{-1}x \right ]$

Differentiating it with respect t0 $x$

$\begin{aligned} &\frac{d y}{d x}=\frac{4}{1+ x^{2}} \\ &\Rightarrow \frac{4}{1+x^{2}} \\ &\text { Using } \frac{d\left(\tan ^{-1} x\right)}{d x}=\frac{1}{1+x^{2}} \end{aligned}$

Differentiation Exercise 10.3 Question 40

Answer: $\frac{dy}{dx}=0$
Hint:
$\frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}-1} ; \frac{d}{\mathrm{dx}}(\text { constant })=0$
Given:
$y=\sec ^{-1}\left(\frac{x+1}{x-1}\right)+\sin ^{-1}\left(\frac{x-1}{x+1}\right) x>0$
Solution:
$\begin{aligned} &\mathrm{y}=\sec ^{-1}\left(\frac{\mathrm{x}+1}{\mathrm{x}-1}\right)+\sin ^{-1}\left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right) \mathrm{x}>0 \\ &\mathrm{y}=\cos ^{-1}\left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)+\sin ^{-1}\left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right) \end{aligned}$
Using $\sec ^{-1}x=\cos \left ( \frac{1}{x} \right )$
Since,$\cos ^{-1}x+\sin ^{-1}x=\frac{\pi}{2}$
$y=\frac{\pi}{2}$
Differentiating with respect to $x$
$\frac{dy}{dx}=0$ $\left [ \frac{d }{dx} \left ( constant \right )=a]\right ]$

Differentiation Exercise 10.3 Question 41

Answer:$\frac{dy}{dx}=\frac{-x}{\sqrt{1-x^{2}}}$
Hint:
$\begin{aligned} &\frac{\mathrm{d}}{\mathrm{dx}}(\text { constant })=0 ; \\ &\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}-1} \end{aligned}$
Given:
$\mathrm{y}=\sin \left[2 \tan ^{-1}\left\{\sqrt{\frac{\mathrm{x}}{1+\mathrm{x}}}\right\}\right]$
Solution:
Put $x=\cos 2\theta$
$\begin{aligned} &y=\sin \left[2 \tan ^{-1} \sqrt{\frac{1-\cos 2 \theta}{1+\cos 2 \theta}}\right] \\ &y=\sin \left[2 \tan ^{-1} \sqrt{\frac{2 \sin ^{2} \theta}{2 \cos ^{2} \theta}}\right] \\ &\text { Since } 1-\cos 2 \theta=2 \sin ^{2} \theta \\ &1+\cos 2 \theta=2 \cos ^{2} \theta \end{aligned}$
$\begin{aligned} &1+\cos 2 \theta=2 \cos ^{2} \theta \\ &y=\sin \left[2 \tan ^{-1} \sqrt{\tan ^{2} \theta}\right] \\ &\text { Since } \frac{\sin \theta}{\cos \theta}=\tan \theta \\ &y=\sin \left[2 \tan ^{-1}(\tan \theta)\right] \\ &y=\sin [2 \theta] \end{aligned}$ $\left \{ \tan ^{-1}\left ( \tan \theta \right )=\theta \: if\: \theta \varepsilon \left [ \frac{-\pi}{2},\frac{\pi}{2} \right ] \right \}$
$y=\sin \left(2 \times \frac{1}{2} \cos ^{-1} x\right)$ $\left [ Since,x=\cos 2\theta ,\theta =\frac{1}{2}\cos ^{-1}x \right ]$
$\begin{aligned} &y=\sin \left(\cos ^{-1} x\right) \\ &y=\sin \left(\sin ^{-1} \sqrt{1-x^{2}}\right) \\ &y=\sqrt{1-x^{2}} \end{aligned}$ $\left[\sin ^{-1}(\sin \theta) \Rightarrow 0 \text { if } \theta \in\left[\frac{-\pi}{2},\frac{\mathrm{\pi}}{2}\right]\right]$
Differentiating it with respect to $x$
$\begin{aligned} \frac{d y}{d x} &=\frac{d}{d x}\left(\sqrt{1-x^{2}}\right) \\ &=\frac{d}{d x}\left(1-x^{2}\right)^{1 / 2} \end{aligned}$
As we know,
$\begin{aligned} &\frac{d}{d x}\left(x^{n}\right)=n x^{n-1} \\ &\qquad \begin{aligned} \frac{d y}{d x} &=\frac{1}{2}\left(1-x^{2}\right)^{\frac{1}{2}} \frac{d}{\partial x}\left(1-x^{2}\right) \\ d y &=\frac{1}{d x} \\ &=\frac{1}{2 \sqrt{1-x^{2}}}(-2 x) \\ \frac{d y}{d x} &=\frac{-x}{\sqrt{1-x^{2}}} \end{aligned} \end{aligned}$

Differentiation Exercise 10.3 Question 42

Answer: $\frac{dy}{dx}=\frac{2}{\sqrt{1-4x^{2}}}$
Hint:
$\begin{aligned} &\frac{d}{d \mathrm{x}} \text { (constant) }=0 \\ &\frac{d}{d\mathrm{x}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{nx}^{-1} \end{aligned}$
Given:
$y=\cos ^{-1}(2 x)+2 \cos ^{-1} \sqrt{1+4 x^{2}}, 0<x<\frac{1}{2}$
Solution:
Put $2x=\cos \theta$
So,
$\begin{aligned} &y=\cos ^{-1}(\cos \theta)+2 \cos ^{-1} \sqrt{1-\cos ^{2} \theta} \\ &\text { Since } \cos ^{2} \theta+\sin ^{2} \theta=1 \\ &y=\cos ^{-1}(\cos \theta)+2 \cos ^{-1} \sqrt{\sin ^{2} \theta} \\ &y=\cos ^{-1}(\cos \theta)+2 \cos ^{-1}(\sin \theta) \\ &y=\cos (\cos \theta)+2 \cos ^{-1}\left(\cos \left(\frac{\pi}{2}-\theta\right)\right)-(i) \end{aligned}$
$\begin{aligned} &0<x=1 \\ &0<2 x<1 \\ &0<\cos \theta=1 \\ &0<\theta<\frac{\pi}{2} \\ &\text { And } 0>-\theta>-\frac{\pi}{2} \\ &\frac{n}{2}>\left(\frac{n}{2}-\theta\right)>0 \end{aligned}$
So From eq (i)
$y=\theta +2\left ( \frac{\pi}{2}-\theta \right )$ $\left \{ Since \cos ^{-1}\left ( \cos \theta \right )=0\: if\: \theta \varepsilon \left [ 0,n \right ] \right \}$
$\begin{aligned} &y \Rightarrow \theta+\pi-2 \theta \\ &y \Rightarrow \pi-C \\ &y=\pi-\cos ^{-1}(2 x) \\ &\text { Since, } 2 x=\cos \theta \end{aligned}$
Differentiating it with respect to $x$
$\frac{d y}{d x}=0-\left[\frac{-1}{\sqrt{1-(2 x)^{2}}}\right] \frac{d}{d x}(2 x)$
Since,$\frac{d}{dx}\left ( constant \right )=0;$
$\begin{aligned} &\frac{d}{d x}\left(\cos ^{-1} x\right)=\frac{1}{\sqrt{1-x^{2}}} \\ &\frac{d y}{d x}=\frac{1}{\sqrt{1-4 x^{2}}}(2) \\ &\frac{d y}{d x}=\frac{2}{\sqrt{1-4 x^{2}}} \end{aligned}$

Differentiation Exercise 10.3 Question 43

Answer: Hence Prove ,$1+a^{2}=b$
Hint:
$\begin{aligned} &\frac{d}{d x}\left(x^{n}\right)=n x^{n-1} \\ &\frac{d(\text { constant) }}{d x}=0 \end{aligned}$
Given:
$\frac{d}{\mathrm{dx}}\left[\tan ^{-1}(\mathrm{a}+\mathrm{bx})\right]=1 \text { at } \mathrm{x}=0$
Solution:
So, using chain rule,
we know $\frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^{2}}$
$\begin{aligned} &{\left[\left\{\frac{1}{1+(a+b x)^{2}}\right\} \frac{d}{dx}(a+b x)\right]_{x=0}=1} \\ &{\left[\frac{1}{1+(a+b x)^{2}} \times(b)\right]_{x=0}=1} \\ &{\left[\frac{b}{\left[1+(a+b x)^{2}\right.}\right]_{x=0}=1} \\ &\therefore(a+b)^{2}=a^{2}+b^{2}+2 a b \end{aligned}$
$\begin{aligned} &{\left[\frac{b}{1+\left(a^{2}+b^{2} x^{2}+2 a b x\right)}\right]_{x=0}=1} \\ &\frac{b}{1+\left(a^{2}+0+0\right)}=1 \\ &\frac{b}{1+a^{2}}=1 \\ &b=1+a^{2} \end{aligned}$
Hence Proved,

Differentiation Exercise 10.3 Question 44

Answer:$\frac{dy}{dx}=\frac{-6}{\sqrt{1-4x^{2}}}$
Hint:
$\begin{aligned} &\frac{d}{d x}\left(x^{n}\right)=n x^{n-1} \\ &\frac{d}{d x}(\text { Constant })=0 \end{aligned}$
Given:

Solution:
$\begin{aligned} &y=\cos ^{-1}(2 x)+2 \cos ^{-1} \sqrt{1-4 x^{2}}, \\ &-\frac{1}{2}<x<0 \end{aligned}$
Put$2x=\cos \theta$
So
$y=\cos ^{-1}(\operatorname{coc} \theta)+2 \cos ^{-1} \sqrt{1-\cos ^{2} \theta}$
Using,$\sin ^{2}\theta +\cos ^{2}\theta =1$
$\begin{aligned} &y=\cos ^{-1}(\cos \theta)+2 \cos \sqrt{\sin ^{2} \theta} \\ &y=\cos ^{-1}(\cos \theta)+2 \cos ^{-}(\sin \theta) \\ &y=\cos ^{-1}(\cos \theta)+2 \cos \left(\cos ^{-1}\left(\frac{\pi}{2}-\theta\right)\right)-(i) \end{aligned}$
Considering limit,
$\begin{aligned} &-\frac{1}{2}<\mathrm{x}<0 \\ &-1<2 \mathrm{x}<0 \\ &-1-\cos \theta<0 \\ &\frac{\mathrm{\pi}}{2}<\theta<\mathrm{\pi} \end{aligned}$
And,
$\begin{aligned} &-\frac{\mathrm{\pi}}{2}>-\theta>-\mathrm{\pi} \\ &\left(\frac{\pi}{2}-\frac{\pi}{2}\right)>\left(\frac{\pi}{2}-\theta\right)>\left(\frac{\mathrm{\pi}}{2}-\pi\right) \\ &0>\left(\frac{\pi}{2}-\theta\right)>-\frac{\pi}{2} \end{aligned}$
So, from equation (i)

$\mathrm{y}=\theta+2\left[-\left(\frac{\pi}{2}-\theta\right)\right]$
$\left\{\operatorname{Since}, \cos ^{-1}(\cos \theta)=0 \text { if } \theta \varepsilon[0, \pi], \cos ^{-1} \cos (\theta)=-\theta, \text { if } \theta \varepsilon[-\pi \cdot 0]\right\}$$\begin{aligned} &\mathrm{y}=\theta-2 \times \frac{\pi}{2}+2 \theta \\ &\mathrm{y}=-\pi+3 \theta \\ &\mathrm{y}=-\pi+3 \cos ^{-1}(2 \mathrm{x}) \end{aligned}$

Differentiating its with respect to $x$using chain rule

$\frac{d y}{d x}=0+3\left[\frac{-1}{\sqrt{1-(2 x)^{2}}}\right] \frac{d}{d x}(2 x)$

As we Know,

$\begin{aligned} &\frac{\mathrm{d}}{\mathrm{dx}}(\text { constant })=0 \\ &\frac{\mathrm{d}}{\mathrm{dx}}\left(\cos ^{-1} \mathrm{x}\right)=\frac{-1}{\sqrt{1-\mathrm{x}^{2}}} \\ &\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-3}{\sqrt{1-4 \mathrm{x}^{2}}}-(2) \\ &\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{6}{\sqrt{1-4 \mathrm{x}^{2}}} \end{aligned}$

Differentiation Exercise 10.3 Question 45

Answer:$\frac{dy}{dx}=\frac{1}{2\sqrt{1-x^{3}}}$
Hint:
$\begin{aligned} &\frac{d}{d x}(\text { constant })=0 \\ &\frac{d}{d x}\left(x^{n}\right)=n x^{n-1} \end{aligned}$
Given:
$y=\tan ^{-1}\left\{\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right\}$
Solution:
Put,
$x=\cos 2\theta$
So,
$\begin{aligned} y &=\tan ^{-1}\left[\frac{\sqrt{1+\cos 2 \theta}-\sqrt{1-\cos 2 \theta}}{\sqrt{1+\cos 2 \theta}+\sqrt{1-\cos 2 \theta}}\right] \\ &=\tan ^{-1}\left(\frac{\sqrt{2 \cos ^{2} \theta}-\sqrt{2 \sin ^{2} \theta}}{\sqrt{2 \cos ^{2} \theta}+\sqrt{2 \sin ^{2} \theta}}\right) \end{aligned}$
Using$\begin{aligned} &1+\cos 2 \theta=2 \cos ^{2} \theta \\ &y=\tan ^{-1}\left(\frac{\sqrt{2} \cos \theta-\sqrt{2} \sin \theta}{\sqrt{2} \cos \theta+\sqrt{2} \sin \theta}\right) \\ &y=\tan ^{-1}\left(\frac{\sqrt{2}(\cos \theta-\sin \theta)}{\sqrt{2}(\cos \theta+\sin \theta)}\right) \end{aligned}$

Dividing numerator and denominator by $\cos \theta$

$\begin{aligned} &y=\tan ^{-1}\left(\frac{\frac{\cos \theta}{\cos \theta}-\frac{\sin \theta}{\cos \theta}}{\frac{\cos \theta}{\cos \theta}+\frac{\sin \theta}{\cos \theta}}\right) \\ &y=\tan ^{-1}\left(\frac{\cos \theta-\sin \theta}{\frac{\cos \theta}{\cos \theta+\sin \theta}} \cos \theta\right) \end{aligned}$

Using

$\begin{aligned} &\tan \theta=\frac{\sin \theta}{\cos \theta} \\ &y=\tan ^{-1}\left(\frac{1-\tan \theta}{1+\tan \theta}\right) \\ &y=\tan ^{-1}\left(\frac{\tan \frac{n}{4}-\tan \theta}{1+\tan \frac{\pi}{4} \tan \theta}\right) \\ &y=\tan ^{-1}\left(\tan \left(\frac{n}{4}-\theta\right)\right) \\ &\text { Using } \tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B} \end{aligned}$

$y=\frac{\pi-0}{4}-\theta$ $\left [ Using\: \tan ^{-1}\left ( \tan \theta \right ) =\theta \: if\: \theta \varepsilon \left [ -\frac{\pi}{2},\frac{\pi}{2} \right ]\right ]$

$y=\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\text { Using } x=\cos 2 \theta]$
Differentiating it with respect to x
$\frac{\mathrm{dy}}{\mathrm{d} x}=0-\frac{1}{2}\left(\frac{-1}{\sqrt{1-\mathrm{x}^{2}}}\right) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left\{\frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}\left(\cos ^{-1} \mathrm{x}\right)=\frac{-1}{\sqrt{1-\mathrm{x}}^{2}}\right\}$
$\begin{aligned} &\frac{d y}{d x}=\frac{1}{2 \sqrt{1-x^{2}}} \\ &\frac{d y}{d x}=\frac{-1}{\sqrt{1-x^{2}}} \end{aligned}$

Differentiation Exercise 10.3 Question 47

Answer: $\frac{dy}{dx}=\frac{-1}{\sqrt{1-x^{2}}}$
Hint:
$\begin{aligned} &\frac{d}{\mathrm{dx}}(\text { constsant })=0 \\ &\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}}-1 \end{aligned}$
Given:
$y=\cos ^{-1}\left\{\frac{2 x-3 \sqrt{1-x^{2}}}{\sqrt{13}}\right)$
Solution:
Let,$x=\cos \theta$
$\begin{aligned} &\mathrm{y}=\cos ^{-1}\left\{\frac{2 \cos \theta-3 \sqrt{1-\cos ^{2} \theta}}{\sqrt{4}}\right\} \\ &\text { Using } \cos ^{2} \theta+\sin ^{2} \theta=1 \\ &\mathrm{y}=\cos ^{-1}\left\{\frac{2}{\sqrt{13}} \cos \theta-\frac{3}{\sqrt{13}} \sqrt{\sin ^{2} \theta}\right] \\ &\mathrm{y}=\cos ^{-1}\left\{\frac{2}{\sqrt{13}} \cos \theta-\frac{3}{\sqrt{13}} \sin \theta\right] \end{aligned}$
$\cos \phi=\frac{2}{\sqrt{13}}$
Let,$\sin \phi=\sqrt{1-\cos ^{2}\phi}$
$\begin{aligned} &=\sqrt{1-\left(\frac{2}{\sqrt{13}}\right)^{2}}+\sqrt{1-\frac{4}{13}} \\ &=\sqrt{\frac{13-4}{13}}=\sqrt{\frac{9}{13}} \end{aligned}$
$\sin \phi=\frac{3}{\sqrt{13}}$
So,
$\begin{aligned} &\begin{aligned} \mathrm{y} &=\cos ^{-1}\{\cos \phi \cos \theta-\sin \phi \sin \theta] \\ &=\cos ^{-1}[\cos (\theta+\phi)] \end{aligned} \\ &\text { Using, } \cos (A+B)=\cos A \cos B-\sin A \sin B \end{aligned}$
$\mathrm{y}=\phi+\theta \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\cos ^{-1}(\cos \theta)=\theta \text { if } 0 \varepsilon[0, \pi]\right]$
$\begin{aligned} &\mathrm{y}=\cos ^{-1}\left(\frac{2}{\sqrt{13}}\right)+\cos ^{-1} \mathrm{x} \\ &\text { since, } \mathrm{x}=\cos \theta \\ &\cos \phi=\frac{2}{\sqrt{13}} \end{aligned}$
Differentiating its with Respect to x,
$\begin{aligned} &\frac{d y}{d x}=\frac{-1}{\sqrt{1-\left(\frac{2}{\sqrt{13}}\right)^{2}}} \frac{d}{d x}\left(\frac{2}{\sqrt{13}}\right)+\left(\frac{-1}{\sqrt{1-x^{2}}}\right) \\ &\frac{d y}{d x}=\frac{-1}{\sqrt{1-\left(\frac{2}{\sqrt{13}}\right)^{2}} \times 0+\left(\frac{-1}{\sqrt{1-x^{2}}}\right)} \\ &\frac{d}{d x}\left(\cos ^{-1} x\right)=\frac{-1}{\sqrt{1-x^{2}}} \\ &\frac{d y}{d x}=0-\frac{1}{\sqrt{1-x^{2}}} \end{aligned}$

Differentiation Exercise 10.3 Question 48

Answer: $\frac{dy}{dx}=\frac{6}{\sqrt{1-9x^{2}}}$
Hint:
$\begin{aligned} &\left.\frac{\mathrm{d}}{\mathrm{dx}} \text { (constant }\right)=0 \\ &\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}-1} \end{aligned}$
Given:
$\begin{aligned} &y=\sin ^{-1}\left(6 x \sqrt{1-9 x^{2}}\right) \\ &\frac{1}{3 \sqrt{2}}<x<\frac{1}{3 \sqrt{2}} \end{aligned}$
Solution:
Let
$\begin{aligned} &\mathrm{x}=\frac{1}{3} \sin \theta \\ &\mathrm{y}=\sin ^{-1}\left(6 \times \frac{1}{3} \sin \theta \sqrt{\left.1-9\left(\frac{1}{3}\right)^{2} \sin ^{2} \theta\right)}\right. \\ &\mathrm{y}=\sin ^{-1}\left(2 \sin \theta \sqrt{1-9 \times \frac{1}{9} \sin ^{2} \theta}\right) \end{aligned}$
$\begin{aligned} &\mathrm{y}=\sin ^{-1}\left(2 \sin \theta \sqrt{1-\sin ^{2} \theta}\right) \\ &\mathrm{U} \sin g \sin ^{2} \theta+\cos ^{2} \theta=1 \\ &\mathrm{y}=\sin ^{-1}\left(2 \sin \theta \sqrt{\cos ^{2} \theta}\right) \\ &\mathrm{y}=\sin ^{-1}(2 \sin \theta \cos \theta) \\ &\mathrm{y}=\sin ^{-1}(\sin 2 \theta)-(\mathrm{i}) \\ &\mathrm{U} \sin g \sin 2 \theta=2 \sin \theta \cos \theta \end{aligned}$
Considering limits here
$\begin{aligned} &\frac{1}{3 \sqrt{2}}<\mathrm{x}<\frac{1}{3 \sqrt{2}} \\ &\frac{1}{3} \times \frac{1}{3 \sqrt{2}}<\frac{1}{3} \sin \theta<\frac{1}{3 \sqrt{2}} \times \frac{1}{3} \\ &\frac{1}{9 \sqrt{2}}<\frac{1}{3} \sin \theta<\frac{1}{9 \sqrt{2}} \end{aligned}$

From Equation (i)

$y=2\theta$ $\left\{\begin{array}{l} \sin ^{-1}(\sin \theta)=\theta \\ \text { if } \theta \varepsilon\left[\frac{-\pi}{2}, \frac{\pi}{2}\right] \end{array}\right\}$

$y=2\sin ^{-1}\left ( 3x \right )$

Differentiating it with respect to$x$

$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(2 \sin ^{-1} 3 x\right) \\ &\text { As we thow, } \frac{d}{d x}\left(\sin ^{-1} x\right)=\frac{1}{\sqrt{1-x^{2}}} \\ &\frac{d y}{d x}=2 \times \frac{1}{\sqrt{1-(3 x)^{2}}} \frac{d}{d x}(3 x) \\ &\frac{d y}{d x}=\frac{2}{\sqrt{1-9 x^{2}}} \times 3 \\ &\frac{d y}{d x}=\frac{6}{\sqrt{1-9 x^{2}}} \end{aligned}$