KC Sinha Mathematics Solution Class 12 Chapter 11 अवकलन (Differentiation) Exercise 11.2 (Q4-Q6)

Exercise 11.2

Question 4

निम्नलिखित फलनों को x के सापेक्ष अवकलित करें।

[Differentiate the following with respect to x]

(i)

$\cos ^{-1}(\sin x)$
Sol:
Let

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

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

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

Differentiating with respect to x

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

(ii)

$\tan ^{-1}(\cot x)$
Sol :
Let y=

$\tan ^{-1}(\cot x)$

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

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

Differentiating with respect to x

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

(iii)

$\sin ^{-1} \sqrt{\frac{1+\cos 2 x}{2}}$
Sol :
Let y=

$\sin ^{-1} \sqrt{\frac{1+\cos 2 x}{2}}$

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

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

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

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

Differentiating with respect to x

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

(iv)

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

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

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

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

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

Differentiating with respect to x

$\frac{d y}{dx}=\frac{1}{2}$

(v)

$\tan ^{-1} \frac{\sin x}{1+\cos x}$
Sol :
Let y=

$\tan ^{-1} \frac{\sin x}{1+\cos x}$

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

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

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

Differentiating with respect to x

$\frac{d y}{dx}=\frac{1}{2}$

(vi)

$\tan ^{-1} \frac{\cos x}{1+\sin x}$
Sol :
Let y=

$\tan ^{-1} \frac{\cos x}{1+\sin x}$

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

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

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

$y=\tan ^{-1}\left(\frac{\tan \frac{\pi}{4}-\tan \frac{x}{2}}{\left.1+\tan \frac{\pi}{4} \cdot \tan \frac{x}{2}\right.}\right)$

$y=\tan ^{-1} \tan \left(\frac{\pi}{4}-\frac{x}{2}\right)$

$y=\frac{\pi}{4}-\frac{x}{2}$

Differentiating with respect to x

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

(vii)

$\tan ^{-1} \sqrt{\frac{1+\sin x}{1-\sin x}}$
Sol :
Let y=

$\tan ^{-1} \sqrt{\frac{1+\sin x}{1-\sin x}}$

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

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

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

$y=\tan ^{-1}\left(\frac{\tan \frac{\pi}{4}+\tan \frac{x}{2}}{1-\tan \frac{\pi}{4} \cdot \tan \frac{x}{2}}\right)$

$y=\tan ^{-1} \tan \left(\frac{\pi}{4}+\frac{x}{2}\right)$

$y=\frac{\pi}{4}+\frac{x}{2}$

Differentiating with respect to x

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

(viii)

$\tan ^{-1} \sqrt{\frac{1+\cos x}{1-\cos x}}$
Sol :
Let y=

$\tan ^{-1} \sqrt{\frac{1+\cos x}{1-\cos x}}$

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

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

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

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

Differentiating with respect to x

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

(ix)

$\tan ^{-1} \sqrt{\frac{1-\cos x}{1+\cos x}}$
Sol :
Let y=

$\tan ^{-1} \sqrt{\frac{1-\cos x}{1+\cos x}}$

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

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

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

Differentiating with respect to x

$\frac{d y}{dx}=\frac{1}{2}$

Question 5

निम्नलिखित फलनों को x के सापेक्ष अवकलित करें।

[Differentiating the following functions with respect to x]

(i)

$\cot ^{-1} \frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}, 0<x<\frac{\pi}{2}$
Sol :

$\cot ^{-1} \frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}$

$=\operatorname{cot}^{-1} \frac{\sqrt{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^{2}}+\sqrt{\left(\cos \frac{x}{2}-\sin\frac{x}{2}\right)^{2}}}{\sqrt{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^{2}}-\sqrt{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)^{2}}}$

$=\cot^{-1} \frac{\cos \frac{x}{2}+\sin \frac{x}{2}+\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}-\cos \frac{x}{2}+\sin \frac{x}{2}}$

$=\cot^{-1} \frac{\cos \frac{x}{2}+\sin \frac{x}{2}+\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin\frac{x}{2}-\frac{\cos x}{2}+\sin \frac{x}{2}}$

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

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

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

Differentiating with respect to x

$\frac{d y}{dx}=\frac{1}{2}$

(ii)

$\tan ^{-1} \frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}, 0<x<\frac{\pi}{2}$
Sol :

(iii)

$\sin ^{-1}\left(\frac{\sin x+\cos x}{\sqrt{2}}\right)$
Sol :
Let y=

$\sin ^{-1}\left(\frac{\sin x+\cos x}{\sqrt{2}}\right)$

$y=\sin^{-1}\left(\sin x.\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} \cdot \cos x\right)$

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

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

$y=\frac{\pi}{4}+2$

Differentiating with respect to x

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

(iv)

$\cos ^{-1}\left(\frac{\cos x+\sin x}{\sqrt{2}}\right)$
Sol :
Let y=

$\cos ^{-1}\left(\frac{\cos x+\sin x}{\sqrt{2}}\right)$

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

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

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

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

Differentiating with respect to x

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

Question 6

निम्नलिखित फलनों को x के सापेक्ष अवकलित करें।

[Differentiating the following functions with respect to x]

(i)

$\tan ^{-1}(\sqrt{1+x^{2}}+x)$

Sol :

Let y=

$\tan ^{-1}(\sqrt{1+x^{2}}+x)$

[put x=cot θ

⇒θ=cot^-1x]

$y=\tan ^{-1}(\sqrt{1+\cot^{2} \theta}+\cot \theta)$

$y=\tan ^{-1}(\operatorname{cosec} \theta+\cot \theta)$

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

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

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

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

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

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

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

Differentiating with respect to x

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

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

(ii)

$\cot ^{-1}(\sqrt{1+x^{2}}+x)$

Sol :

Let y=

$\cot ^{-1}(\sqrt{1+x^{2}}+x)$

[put x=cot θ

⇒θ=cot^-1x]

$y=\cot ^{-1}(\sqrt{1+\cot^{2} \theta}+\cot \theta)$

$y=\cot^{-1}(\operatorname{cosec} \theta+\cot \theta)$

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

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

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

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

y=\frac{\theta}{2}

y=\frac{1}{2} \cot ^{-1} x

Differentiating with respect to x

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

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

(iii)

$\tan ^{-1}(\sqrt{1+x^{2}}-x)$
Sol :
Let

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

[put x=cot θ

⇒θ=cot^-1x]

$y=\tan ^{-1}(\sqrt{1+\cot ^{2} \theta}-\cot \theta]$

$y=\tan ^{-1}(\operatorname{cosec} \theta-\cot \theta)$

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

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

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

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

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

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

Differentiating with respect to x

$\frac{d y}{dx}=\frac{1}{2}\left[-\frac{1}{1+x^{2}}\right]$

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

(iv)

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

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

[put x=tan θ

⇒θ=tan^-1x]

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

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

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

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

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

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

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

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

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

Differentiating with respect to x

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

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

(v)

$\cot ^{-1} \frac{x}{\sqrt{1-x^{2}}}$
Sol :
Let y=

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

[put x=cos θ

⇒θ=cos^-1x]

$y=\cot^{-1}\frac{\cos \theta}{\sqrt{1-\cos ^{2} \theta}}$

$y=\cot^{-1} \frac{\cos \theta}{\sin \theta}$

$y=\cot ^{-1}(\cot \theta)$

y=θ

$y=\cos ^{-1} x$

Differentiating with respect to x

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

(vi)

$\tan ^{-1} \frac{x}{\sqrt{a^{2}-x^{2}}}$
Sol :
Let y=

$\tan ^{-1} \frac{x}{\sqrt{a^{2}-x^{2}}}$

[put x=a sin θ

⇒

$\frac{x}{a}=\sin \theta$ ]

$\Rightarrow \theta=\sin^{-1}{\frac{x}{a}}$

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

$y=\tan ^{-1} \frac{a \sin \theta}{\sqrt{a^{2}-a^{2}\sin^{2} \theta}}$

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

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

$y=\tan ^{-1} \frac{a \sin \theta}{a \cos \theta}$

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

y=θ

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

Differentiating with respect to x

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

$=\frac{1}{\sqrt{1-\left(\frac{x}{a}\right)^{2}}} \times \frac{1}{a}$

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

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

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