Contents
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
$\frac{d y}{dx}=0$
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
$\frac{d y}{d x}=\frac{d\left(\tan ^{-1} \frac{x}{a}\right)}{d\left(\frac{x}{a}\right)} \times \frac{d\left(\frac{x}{a}\right)}{d x}$
$=\frac{1}{\sqrt{1-\left(\frac{x}{a}\right)^{2}}} \times \frac{1}{a}$
$=\frac{1}{\sqrt{\frac{a^{2}-x^{2}}{a^{2}}}} \times \frac{1}{a}$
$=\frac{a}{\sqrt{a^{2}-x^{2}}} \times \frac{1}{a}$
$=\frac{1}{\sqrt{a^{2}-x^{2}}}$
Question 7
निम्नलिखित फलनों को x के सापेक्ष अवकलित करें।
[Differentiating the following with respect to x]
(i) $\sin ^{-1}\left(1-2 x^{2}\right)$
Sol :
Let y=$\sin ^{-1}\left(1-2 x^{2}\right)$
[put x=a sin θ
⇒θ=sin-1x]
$y=\sin ^{-1}\left(1-2 \sin ^{2} \theta\right)$
$y=\operatorname{sen}^{-1}(\cos 2 \theta)$
$y=\sin ^{-1}\left[\sin \left(\frac{\pi}{2}-2 \theta\right)\right]$
$y=\frac{\pi}{2}-2 \theta$
$y=\frac{\pi}{2}-2 \sin ^{-1} x$
Differentiating with respect to x
$\frac{d y}{d x}=-2 \times \frac{1}{\sqrt{1-x^{2}}}$
$=\frac{-2}{\sqrt{1-x^{2}}}$
(ii) $\sin ^{-1} \frac{1}{\sqrt{1+x^{2}}}$
Sol :
Let y=$\sin ^{-1} \frac{1}{\sqrt{1+x^{2}}}$
[put x=tan θ
⇒θ=tan-1x$]
$y=\sin ^{-1} \frac{1}{\sqrt{1+\tan ^{2} \theta}}$
$y=\sin ^{-1} \frac{1}{\sec \theta}$
$y=\sin ^{-1}(\cos \theta)$
$y=\sin ^{-1}\left[\sin \left(\frac{\pi}{2}-\theta\right)\right]$
$y=\frac{\pi}{2}-\theta$
$y=\frac{\pi}{2}-\tan ^{-1} x$
Differentiating with respect to x
$\frac{d y}{dx}=-\frac{1}{1+x^{2}}$
(iii) $\tan ^{-1} \frac{x}{1+\sqrt{1-x^{2}}}$
Sol :
Let y=$\tan ^{-1} \frac{x}{1+\sqrt{1-x^{2}}}$
[put x=sin θ
$y=\tan ^{-1} \frac{\sin \theta}{1+\sqrt{1-\sin ^{2} \theta}}$
$y=\tan ^{-1} \frac{\sin \theta}{1+\cos \theta}$
$y=\tan ^{-1} \frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}}$
$y=\tan ^{-1}\left(\tan \frac{\theta}{2}\right)$
$y=\frac{\theta}{2}$
$y=\frac{1}{2} \sin ^{-1} x$
Differentiating with respect to x
$\frac{d y}{dx}=\frac{1}{2} \times \frac{1}{\sqrt{1-x^{2}}}$
$=\frac{1}{2 \sqrt{1-x^{2}}}$
(iv) $\sin ^{-1}\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right)$
Sol :
Let y=$\sin ^{-1}\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right)$
Putting x=cos θ
⇒2θ=cos-1x
$\theta=\frac{1}{2} \cos ^{-1} x$
$y=\sin ^{-1}\left(\frac{\sqrt{1+\cos 2 \theta}+\sqrt{1-\cos 2 \theta}}{2}\right)$
$y=\sin ^{-1}\left(\frac{\sqrt{2 \cos ^{2} \theta}+\sqrt{2 \sin ^{2} \theta}}{2}\right)$
$y=\sin ^{-1}\left(\frac{\sqrt{2} \cos \theta+\sqrt{2} \sin \theta}{2}\right)$
$y=\sin ^{-1}\left(\frac{\sqrt{2} \cos \theta}{2}+\frac{\sqrt{2}}{2} \sin \theta\right)$
$y=\sin ^{-1}\left(\sin \frac{\pi}{4} \cos \theta+\cos \frac{\pi}{4} \sin \theta\right)$
$y=\sin ^{-1}\left(\sin \left[\frac{\pi}{4}+\theta\right)\right]$
$y=\frac{\pi}{4}+\theta$
$y=\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} x$
Differentiating with respect to x
$\frac{d{y}}{d x}=0+\frac{1}{2} \times \frac{-1}{\sqrt{1-x^{2}}}$
$=\frac{-1}{2 \sqrt{1-x^{2}}}$
(v) $\cos ^{-1} \sqrt{\frac{1+x^{2}}{2}}$
Sol :
Let y=$\cos ^{-1} \sqrt{\frac{1+x^{2}}{2}}$
Put x2=cos2θ
cos-1(x2)=2θ
$\frac{1}{2} \cos ^{-1}\left(x^{2}\right)=\theta$
$y=\cos ^{-1} \sqrt{1+\frac{\cos 2\theta}{2}}$
$y=\cos ^{-1} \sqrt{\frac{2\cos ^{2} \theta}{2}}$
$y=\cos^{-1}(\cos \theta)$
y=θ
$y=\frac{1}{2} \cos ^{-1}\left(x^{2}\right)$
Differentiating with respect to x
$\frac{d y}{dx}=\frac{1}{2} \times \frac{-1}{\sqrt{1-(x)^{2}}} \times 2x$
$=\frac{-x}{\sqrt{1-x^{4}}}$
(vi) $\tan ^{-1} \sqrt{\frac{a-x}{a+x}}$
Sol :
Let y=$\tan ^{-1} \sqrt{\frac{a-x}{a+x}}$
Put x=a cosθ
$\frac{x}{a}=\cos \theta$
$\cos^{-1} \frac{x}{a}=\theta$
$y=\tan ^{-1} \sqrt{\frac{a-a \cos \theta}{a+a \cos \theta}}$
$y=\tan ^{-1} \sqrt{\frac{a(1-\cos \theta)}{a(1+\cos \theta)}}$
$y=\tan ^{-1} \sqrt{\frac{2\cos^{2} \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}}}$
$y=\tan ^{-1} \sqrt{\tan ^{2} \frac{\theta}{2}}$
$y=\tan ^{-1}\left(\tan \frac{\theta}{2}\right)$
$y=\frac{\theta}{2}$
$y=\frac{1}{2} \cos ^{-1} \frac{x}{a}$
Differentiating with respect to x
$\frac{d y}{dx}=\frac{1}{2} \times \frac{-1}{\sqrt{1-\left(\frac{x}{a}\right)^{2}}} \times \frac{1}{a}$
$=\frac{-1}{2 a \sqrt{1-\frac{x^{2}}{a^{2}}}}$
$=\frac{-1}{2 a \sqrt{\frac{a^{2}-x^{2}}{a^{2}}}}$
$\frac{d y}{d x}=\frac{-1}{2a \frac{\sqrt{a^{2}-x^{2}}}{a}}$
$=\frac{-1}{2 \sqrt{a^{2}-x^{2}}}$
Question 8
निम्नलिखित फलनों को x के सापेक्ष अवकलित करें।
[Differentiating the following function with respect to x]
(i) $\sin ^{-1} \frac{2 x}{1+x^{2}},|x|<1$
Sol :
Let y=$\sin ^{-1} \frac{2 x}{1+x^{2}},|x|<1$
[Put x=tanθ]
$y=2 \tan ^{-1} x$
Differentiating with respect to x
$\frac{dy}{dx}=2 \times \frac{1}{1+x^2}$
$=\frac{2}{1+x^{2}}$
(ii) $\cos ^{-1} \frac{1-x^{2}}{1+x^{2}},|x|<1$
Sol :
(iii) $\tan ^{-1} \frac{2 x}{1-x^{2}},|x|<1$
Sol :
$y=2 \tan ^{-1} x$
Differentiating with respect to x
(iv) $\sec ^{-1}\left(\frac{1}{4 x^{3}-3 x}\right)$
Sol :
Let y=$\sec ^{-1}\left(\frac{1}{4 x^{3}-3 x}\right)$
$y=\cos ^{-1}\left(4 x^{3}-3 x\right)$
$y=3 \cos ^{-1} x$
Differentiating with respect to x
$\frac{dy}{xx}=3 \times \frac{-1}{\sqrt{1-x^2}}$
$=\frac{-3}{\sqrt{1-x^{2}}}$
Question 9
निम्नलिखित फलनों को x के सापेक्ष अवकलित करें।
[Differentiating the following function with respect to x]
(i) $\tan ^{-1} \frac{4 \sqrt{x}}{1-4 x}$
Sol :
Let y=$\tan ^{-1} \frac{4 \sqrt{x}}{1-4 x}$
$y=\tan ^{-1} \frac{2 \cdot(2 \sqrt{x})}{1-(2 \sqrt{x})^{2}}$
$y=\tan ^{-1} 2 \sqrt{x}$
Differentiating with respect to x
$\frac{d y}{d x}=2\times \frac{1}{1+(2 \sqrt{x})^{2}} \times 2 \times \frac{1}{2 \sqrt{x}}$
$=\frac{2}{\sqrt{x}(1+4 x)}$
(ii) $\tan ^{-1} \frac{\sqrt{x}+\sqrt{a}}{1-\sqrt{xa} }$
Sol :
Let y=$\tan ^{-1} \frac{\sqrt{x}+\sqrt{a}}{1-\sqrt{xa} }$
[∵ $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1} \frac{x+y}{1-xy}$]
$y=\tan ^{-1} \sqrt{x}+\tan ^{-1} \sqrt{a}$
Differentiating with respect to x
$\frac{dy}{dx}=\frac{1}{1+(\sqrt{x})^{2}} \times \frac{1}{2 \sqrt{x}}+0$
$=\frac{1}{2 \sqrt{x}(1+x)}$
(iii) $\tan ^{-1} \frac{5 x}{1-6 x^{2}}$
Sol :
Let y=$\tan ^{-1} \frac{5 x}{1-6 x^{2}}$
$y=\tan^{-1} \frac{3 x+2 x}{1-3 x.2x}$
[$\because \tan ^{-1} x+\tan^{-1}y=\tan ^{-1} \frac{x+y}{1-xy}$]
$y=\tan ^{-1} 3 x+\tan ^{-1} 2 x$
Differentiating with respect to x
$\frac{d{y}}{dx}=\frac{1}{1+(3 x)^{2}} \times 3+\frac{1}{1+(2 x)^{2}} \times 2$
$=\frac{3}{1+9 x^{2}}+\frac{2}{1+4 x^{2}}$
(iv) $\tan ^{-1} \frac{a+x}{1-a x}$
Sol :
Let y=$\tan ^{-1} \frac{a+x}{1-a x}$
[$\because \tan ^{-1} x+\tan^{-1}y=\tan ^{-1} \frac{x+y}{1-xy}$]
$y=\tan ^{-1} a+\tan ^{-1} x$
Differentiating with respect to x
$\frac{d y}{d x}=\frac{1}{1+x^{2}}$
(v) $\sin ^{-1} \frac{2^{x+1}}{1+4^{x}}$
Sol :
Let y=$\sin ^{-1} \frac{2^{x+1}}{1+4^{x}}$
$y=\sin ^{-1} \frac{2^{x} \cdot 2}{1+\left(2^{2}\right)^{x}}$
$y=\sin^{-1} \frac{2 \cdot\left(2^{x}\right)}{1+\left(2^{x}\right)^{2}}$
[$\because 2 \tan ^{-1} x=\sin ^{-1} \frac{2x}{1+x^{2}}$]
$y=2 \tan ^{-1}\left(2^{x}\right)$
Differentiating with respect to x
$\frac{d y}{d x}=2 \times \frac{1}{1+\left(2^{x}\right)^{2}} \times 2^{x} \cdot \log 2$
$=\frac{2^{x+1} \cdot \log 2}{1+4^{x}}$
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
$\frac{d y}{d x}=\frac{-1}{\sqrt{1-x^{2}}}-\frac{1}{\sqrt{1-\left(x^{\frac{1}{3}}\right)^{2}}} \times \frac{1}{3} \cdot x^{-\frac{2}{3}}$
$=\frac{-1}{\sqrt{1-x^{2}}}-\frac{1}{3 x^{2 / 3} \sqrt{1-x^{2 / 3}}}$