KC Sinha Mathematics Solution Class 12 Chapter 11 अवकलन (Differentiation) Exercise 11.2 (Q10-Q11)

Exercise 11.2

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Question 10

निम्नलिखित फलनों को x के सापेक्ष अवकलित करें।

[Differentiating the following function with respect to x]

(i) $\tan ^{-1}\left(\frac{a+b x}{b-a x}\right)$
Sol :
Let y=$\tan ^{-1}\left(\frac{a+b x}{b-a x}\right)$

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

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

$y=\tan ^{-1} \frac{a}{b}+\tan ^{-1} x$

Differentiating with respect to x

$\frac{d{y}}{dx}=\frac{1}{1+x^{2}}$


(ii) $\tan ^{-1}\left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)$
Sol :
Let y=$\tan ^{-1}\left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)$

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

$y=\tan ^{-1}\left(\frac{\tan \frac{\pi}{4}+\tan x}{1-\tan\frac{\pi}{4}.\tan x}\right)$

$y=\tan ^{-1}\left(\tan \left(\frac{\pi}{4}+x\right)\right)$

$y=\frac{\pi}{4}+x$

Differentiating with respect to x

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


(iii) $\tan ^{-1}\left(\frac{a+b \tan x}{b-a \tan x}\right)$
Sol :
Let y=$\tan ^{-1}\left(\frac{a+b \tan x}{b-a \tan x}\right)$

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

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

$y=\tan ^{-1} \frac{a}{b}+\tan ^{-1}(\tan x)$

$y=\tan ^{-1} \frac{a}{b}+x$

Differentiating with respect to x

$\frac{d y}{dx}=1$


(iv) $\tan ^{-1}\left(\frac{3 a^{2} x-x^{3}}{a^{3}-3 a x^{2}}\right)$
Sol :
Let y=$\tan ^{-1}\left(\frac{3 a^{2} x-x^{3}}{a^{3}-3 a x^{2}}\right)$

$y=\tan ^{-1}\left[\frac{\frac{3 a^{2} x}{a^{3}}-\frac{x^{3}}{a^{3}}}{\frac{a^{3}}{a^{3}}-\frac{3 a x^{2}}{a^{3}}}\right]$

$y=\tan ^{-1}\left[\frac{3\left(\frac{x}{a}\right)-\left(\frac{x}{a}\right)^{3}}{1-3\left(\frac{x}{a}\right)^{2}}\right]$

$\left[\because \tan ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right)=3 \tan ^{-1} x \right]$

$y=3 \tan ^{-1}\left(\frac{x}{a}\right)$

Differentiating with respect to x

$\frac{d y}{d x}=3 \times \frac{1}{1+\left(\frac{x}{a}\right)^{2}} \times \frac{1}{a}$

$=\frac{3}{a\left(1+\frac{x^{2}}{a^{2}}\right)}$

$=\frac{3}{a\left(\frac{a^{2}+x^{2}}{a^{2}}\right)}$

$=\frac{3 a}{a^{2}+x^{2}}$


(v) $\tan ^{-1}\left(\frac{a \cos x-b \sin x}{b \cos x+a \sin x}\right)$
Sol :
Let y=$\tan ^{-1}\left(\frac{a \cos x-b \sin x}{b \cos x+a \sin x}\right)$

$y=\tan ^{-1}\left[\frac{\frac{a \cos x}{b \cos x}-\frac{b \sin x}{b \cos x}}{\frac{b \cos x}{b \cos x}+\frac{a \sin x}{b \cos x}}\right]$

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

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

$y=\tan ^{-1} \frac{a}{b}-\tan ^{-1}(\tan x)$

$y=\tan ^{-1} \frac{a}{b}-x$

Differentiating with respect to x

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


(vi) $\tan ^{-1}\left(\frac{2 a^{x}}{1-a^{2 x}}\right)$
Sol :
Let y=$\tan ^{-1}\left(\frac{2 a^{x}}{1-a^{2 x}}\right)$

$y=\tan ^{-1}\left(\frac{2 \cdot a^{x}}{1-\left(a^{2 x}\right)^{2}}\right]$

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

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

Differentiating with respect to x

$\frac{d y}{d x}=2 \times \frac{1}{1+\left(a^{x}\right)^{2}} \times a^{x} \cdot \log a$

$=\frac{2 a^{x} \log a}{1+a^{2 x}}$

Question 11

(i) यदि (If) $y=\sec ^{-1}\left(\frac{x+1}{x-1}\right)+\sin ^{-1}\left(\frac{x-1}{x+1}\right)$ दिखाएँ कि (show that) $\frac{d y}{d x}=0$
Sol :
$y=\sec ^{-1}\left(\frac{x+1}{x-1}\right)+\sin ^{-1}\left(\frac{x-1}{x+1}\right)$

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

$y=\frac{\pi}{2}$

Differentiating with respect to x

$\frac{d y}{d x}=0$


(ii) यदि (If) $y=\tan ^{-1}(\cot x)+\cot ^{-1}(\tan x)$ दिखाएँ कि ( show that )$\frac{d y}{d x}=-2$
Sol :
$y=\tan ^{-1}(\cot x)+\cot ^{-1}(\tan x)$

$y=\tan ^{-1}\left[\tan \left(\frac{\pi}{2}-x\right)\right]+\cot^{-1}\left[\cot \left(\frac{\pi}{2}-x\right)\right]$

$y=\frac{\pi}{2}-x+\frac{\pi}{2}-x$

y=𝜋-2x

Differentiating with respect to x

$\frac{d y}{d x}=-2$


(iii) यदि (If) $y=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)+\sec ^{-1}\left(\frac{1+x^{2}}{1-x^{2}}\right)$ दिखाएँ कि (show that) $\frac{d y}{d x}=\frac{4}{1+x^{2}}$
Sol :
Let y=$y=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)+\sec ^{-1}\left(\frac{1+x^{2}}{1-x^{2}}\right)$

$y=2 \tan ^{-1} x+\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$

$y=2 \tan ^{-1} x+2 \tan ^{-1} x$

Differentiating with respect to x

$\frac{d y}{dx}=4 \times \frac{1}{1+x^{2}}=\frac{4}{1+x^{2}}$


(iv) यदि (If) $y=\sin \left\{2 \tan ^{-1} \sqrt{\left(\frac{1-x}{1+x}\right)}\right\}$ show that $\frac{d y}{d x}=\frac{-x}{\sqrt{1-x^{2}}}$
Sol :
$y=\sin \left\{2 \tan ^{-1} \sqrt{\left(\frac{1-x}{1+x}\right)}\right\}$

Put x=cosθ
⇒θ=cos^-1x

$y=\sin \left\{2 \tan ^{-1} \sqrt{\frac{1-\cos \theta}{1+\cos \theta}}\right\}$

$y=\sin \left[2 \tan ^{-1} \sqrt{\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}}}\right]$

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

$y=\sin\left\{2 \times \frac{\theta}{2}\right\}$

y=sin θ

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

Differentiating with respect to x

$\frac{d y}{dx}=\cos \left(\cos^{-1} x\right) \times \frac{-1}{\sqrt{1-x^{2}}}$

$=\frac{-x}{\sqrt{1-x^{2}}}$


(v) यदि (If) $y=\cos ^{-1}(2 x)+2 \cos ^{-1} \sqrt{1-4 x^{2}}$ दिखाएँ कि (show that) $\frac{d y}{d x}=\frac{2}{\sqrt{1-4 x^{2}}}$
Sol :
$y=\cos ^{-1}(2 x)+2 \cos ^{-1} \sqrt{1-4 x^{2}}$

$y=\cos ^{-1}(2x)+2 \cos ^{-1} \sqrt{1-(2 x)^{2}}$

$y=\cos^{-1}(2 x)+2 \cdot \sin ^{-1} 2 x$

$\left[\because\cos^{-1} \sqrt{1-x^{2}}=\sin ^{-1} x\right]$

$y=\cos ^{-1}(2 x)+\sin ^{-1}(2 x)+\sin^{-1}(2 x)$

$y=\frac{\pi}{2}+\sin ^{-1}(2 x)$

Differentiating with respect to x

$\frac{dy}{dx}=\frac{1}{\sqrt{1-(2 x)^{2}}} \times 2$

$=\frac{2}{\sqrt{1-4 x^{2}}}$


(vi) यदि (If) $y=\sin ^{2}\left(\cot ^{-1} \sqrt{\frac{1+x}{1-x}}\right)$ दिखाएँ कि (prove that ) $\frac{d y}{d x}=-\frac{1}{2}$
Sol :
$y=\sin^{2}\left(\cot ^{-1} \sqrt{\frac{1+x}{1-x}}\right)$

Put x=cos2θ

$y=\sin ^{2}\left[\cot ^{-1} \sqrt{\frac{1+\cos 2\theta}{1-\cos 2 \theta}}\right]$

$y=\sin ^{2}\left[\cot^{-1} \sqrt{\frac{2\cos^2 \theta}{2 \sin^{2}\theta}}\right]$

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

$y=\sin ^{2} \theta$

Differentiating with respect to θ

$\frac{d x}{d \theta}=-\sin 2 \theta \times 2$

$\frac{d \theta}{d x}=-\frac{1}{2 \sin 2 \theta}$


Differentiating with respect to x

$\frac{d y}{d x}=\frac{d\left(\sin ^{2} \theta\right)}{d(\sin \theta)} \times \frac{d(\sin \theta)}{d \theta} \times \frac{d \theta}{d x}$

$=2 \sin \theta \cdot \cos \theta \cdot\left(-\frac{1}{2 \sin 2 \theta}\right)$

$=\sin 2 \theta \cdot\left(\frac{-1}{2 \sin 2\theta}\right)$

$\frac{d y}{d x}=-\frac{1}{2}$


(vii) यदि(If) $y=\sin ^{-1}\left(\frac{1}{\sqrt{1+x^{2}}}\right)+\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)$ निकाले (find) $\frac{d y}{d x}$
Sol :
$y=\sin ^{-1}\left(\frac{1}{\sqrt{1+x^{2}}}\right)+\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)$

Put x=tan θ
θ=tan^-1x

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

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

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

$y=\sin ^{-1}\left[\sin \left(\frac{\pi}{2}-\theta\right)\right]+\tan ^{-1}\left(\frac{\frac{1-\cos \theta}{\cos \theta}}{\frac{\sin \theta}{\cos \theta}}\right)$

$y=\frac{x}{2}-\sigma+\tan ^{-1}\left(\frac{2\sin^2 \frac{\theta}{2}}{2 \sin \frac{\theta}{2}\cos \frac{\theta}{2}}\right)$

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

$y=\frac{\pi}{2}-\theta+\frac{\theta}{2}$

$y=\frac{\pi}{2}-\frac{\theta}{2}$

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

Differentiating with respect to x

$\frac{d y}{d x}=0-\frac{1}{2} \times \frac{1}{1+x^{2}}$

$=\frac{-1}{2\left(1+x^{2}\right)}$


(viii) यदि(If) $y=\cos ^{-1}\left[x^{4 / 3}-\sqrt{\left(1-x^{2}\right)\left(1-x^{2 / 3}\right)}\right], 0 \leq x \leq 1$ दिखाएँ कि(show that) $\frac{d y}{d x}=-\frac{1}{\sqrt{1+x^{2}}}-\frac{1}{3 x^{2 / 3} \sqrt{1-x^{2 / 3}}}$
Sol :
$y=\cos ^{-1}\left[x^{4 / 3}-\sqrt{\left(1-x^{2}\right)\left(1-x^{2 / 3}\right)}\right]$

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

$y=\cos ^{-1} x+\cos ^{-1}\left(x^{\frac{1}{3}}\right)$

Differentiating with respect to x

$\frac{d y}{d x}=\frac{-1}{\sqrt{1-x^{2}}}-\frac{1}{\sqrt{1-\left(x^{\frac{1}{3}}\right)^{2}}} \times \frac{1}{3} \cdot x^{-\frac{2}{3}}$

$=\frac{-1}{\sqrt{1-x^{2}}}-\frac{1}{3 x^{2 / 3} \sqrt{1-x^{2 / 3}}}$