KC Sinha Mathematics Solution Class 12 Chapter 11 अवकलन (Differentiation) Exercise 11.2 (Q1-Q3)

Exercise 11.2

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Question 1

निम्नलिखित फलनों को x के सापेक्ष अवकलित करें।

[Differentiate the following w.r.t x]
(i) sin^-13x
Sol :
Let y=sin^-13x

Differentiating with respect to x

$\frac{d y}{dx}=\frac{d\left(\sin^{-1} 3x\right)}{d(3 x)} \times \frac{d(3 x)}{dx}$

$=\frac{1}{\sqrt{1-(3x)^{2}}} \times 3$

$=\frac{3}{\sqrt{1-9 x^{2}}}$


(ii) $\cos ^{-1} \sqrt{\cos x}$
Sol :
Let y=$\cos ^{-1} \sqrt{\cos x}$

Differentiating with respect to x

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

$=\frac{-1}{\sqrt{1-\sqrt{\cos x}}}+\frac{1}{2 \sqrt{\cos x}} \times(-\sin x)$

$=\frac{\sin x}{2 \sqrt{\cos x} \cdot \sqrt{1-\cos x}} \times \frac{\sqrt{1+\cos x}}{\sqrt{1+\cos x}}$

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

$=\frac{\sin x \sqrt{1+\cos x}}{2 \sqrt{\cos x}.\sin x}$

$=\frac{1}{2} \sqrt{\frac{1+\cos x}{\cos x}}$

$=\frac{1}{2} \sqrt{\frac{1}{\cos x}+\frac{\cos x}{\cos x}}$

$=\frac{1}{2} \sqrt{\sec x+1}$


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

Differentiating with respect to x

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

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

$=\frac{-1}{2 \sqrt{x}(1+x)}$


(iv) $\sin ^{-1} x \sqrt{x}, 0 \leq x \leq 1$
Sol :
Let y=$\sin ^{-1} x \sqrt{x}, 0 \leq x \leq 1$

Differentiating with respect to x

$\frac{d y}{d x}=\frac{d\left(\sin ^{-1} x\right.\sqrt x)}{d(x \sqrt{x})} \times \frac{d(x\sqrt{x})}{d x}$

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

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

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

Differentiating with respect to x

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

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

$=\frac{1}{4 \sqrt{x}(1+x) \cdot \sqrt{\tan ^{-1} \sqrt{x}}}$

Question 2

निम्नलिखित फलनों को x के सापेक्ष अवकलित करें।

[Differentiate the following with respect to x]

(i) (tan^-1x)²
Sol :
Let y=(tan^-1x)²

Differentiating with respect to x

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

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

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


(ii) cos^-1x⁴
Sol :
Let y=cos^-1x⁴

Differentiating with respect to x

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

$=\frac{-1}{\sqrt{1-\left(x^{4}\right)^{2}}}\times {4 x^{3}}$

$=\frac{-4 x^{3}}{\sqrt{1-x^{8}}}$


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

Differentiating with respect to x

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

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

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


(iv) $\sin ^{-1} \sqrt{1-x^{2}}$
Sol :
Let y=$\sin ^{-1} \sqrt{1-x^{2}}$

Differentiating with respect to x

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

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

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

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

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

Question 3

निम्नलिखित फलनों को x के सापेक्ष अवकलित करें।

[Differentiate the following with respect to x]

(i) $x \sec ^{-1} x$

Sol :

Let y=$x \sec ^{-1} x$

Differentiating with respect to x

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

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

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

(ii) $\sqrt{1-x^{2}} \sin ^{-1} x-x$

Sol :

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

Differentiating with respect to x

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

$=\frac{1}{2 \sqrt{1-x^{2}}} \times (-2 x) \cdot \sin ^{-1} x+\sqrt{1-x^{2}} \cdot \frac{1}{\sqrt{1-x^{2}}}-1$

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

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


(iii) $\frac{\cos ^{-1} \frac{x}{2}}{\sqrt{2 x+7}},-2<x<2$
Sol :
Let y=$\frac{\cos ^{-1} \frac{x}{2}}{\sqrt{2 x+7}}$

Differentiating with respect to x

$\frac{d{y}}{d{x}}=\frac{\frac{d\left(\cos^{-1} \frac{x}{2}\right)}{d\left(\frac{x}{2}\right)}\times \frac{d\left(\frac{x}{2}\right)}{dx} \cdot \sqrt{2 x+7}-\cos^{-1} \frac{x}{2} \cdot \frac{d(\sqrt{2 x+7})}{d(2 x+7)}\times\frac{d(2x+7)}{dx}}{(\sqrt{2 x+7})^{2}}$

$=\frac{\frac{-1}{\sqrt{1-\left(\frac{x}{2}\right)^{2}}} \times \frac{1}{2} \sqrt{2 t+7}-\cos^{\frac{x}{2}}\cdot \frac{1}{2 \sqrt{2 x+7}}\times2x}{2 x+7}$

$=\frac{\frac{-1}{\sqrt{1-\frac{x^{2}}{4}}} \times \frac{1}{2} \sqrt{2 x+7}-\frac{cos \frac{-1 x}{2}}{\sqrt{2 x+7}}}{2 x+7}$

$\frac{\frac{-1}{\sqrt{\frac{4-x^{2}}{4}}} \times \frac{1}{2} \sqrt{2 x+7}-\frac{cos^{-1} \frac{x}{2}}{\sqrt{2 x+7}}}{2 x+7}$

$\frac{\frac{-2}{\sqrt{4-x^{2}}} \cdot \frac{1}{2} \sqrt{2 x+7}-\frac{\cos^{-1} \frac{x}{2}}{\sqrt{2 x+7}}}{2 x+7}$

$=\dfrac{\frac{-(2 x+7)-\sqrt{4-x^{2}}.\cos^{-1} \frac{x}{2}}{\sqrt{4-x^{2}} \sqrt{2x+7}}}{\sqrt{2x+7}}$

$=-\frac{2 x+7+\sqrt{4-x^{2}} \cdot \cos^{-1} \frac{x}{2}}{(2 x+7)^{3 / 2} \sqrt{4-x^{2}}}$


(iv) $\sin ^{-1} x+\sin ^{-1} \sqrt{1-x^{2}},-1 \leq x \leq 1$
Sol :
Let y = $\sin ^{-1} x+\sin ^{-1} \sqrt{1-x^{2}}$

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

 $y=\frac{\pi}{2}$

Differentiating with respect to x

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