Exercise 11.2
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]
$y=\operatorname{sen}^{-1}(\cos 2 \theta)$
$y=\sin ^{-1}\left[\sin \left(\frac{\pi}{2}-2 \theta\right)\right]$
$y=\frac{\pi}{2}-2 \theta$
$y=\frac{\pi}{2}-2 \sin ^{-1} x$
$\frac{d y}{d x}=-2 \times \frac{1}{\sqrt{1-x^{2}}}$
$=\frac{-2}{\sqrt{1-x^{2}}}$
(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$]
$y=\sin ^{-1} \frac{1}{\sec \theta}$
$y=\sin ^{-1}(\cos \theta)$
$y=\sin ^{-1}\left[\sin \left(\frac{\pi}{2}-\theta\right)\right]$
$y=\frac{\pi}{2}-\theta$
$y=\frac{\pi}{2}-\tan ^{-1} x$
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
⇒θ=sin-1x$]
$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]
Differentiating with respect to x
$\frac{dy}{dx}=2 \times \frac{1}{1+x^2}$
$=\frac{2}{1+x^{2}}$
(ii) $\cos ^{-1} \frac{1-x^{2}}{1+x^{2}},|x|<1$
(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=$\tan ^{-1} \frac{4 \sqrt{x}}{1-4 x}$
$y=\tan ^{-1} \frac{2 \cdot(2 \sqrt{x})}{1-(2 \sqrt{x})^{2}}$
$\frac{d y}{d x}=2\times \frac{1}{1+(2 \sqrt{x})^{2}} \times 2 \times \frac{1}{2 \sqrt{x}}$
$=\frac{2}{\sqrt{x}(1+4 x)}$
(ii) $\tan ^{-1} \frac{\sqrt{x}+\sqrt{a}}{1-\sqrt{xa} }$
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
$\frac{d y}{d x}=2 \times \frac{1}{1+\left(2^{x}\right)^{2}} \times 2^{x} \cdot \log 2$
$=\frac{2^{x+1} \cdot \log 2}{1+4^{x}}$
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