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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