question_answer

Find \[\frac{dy}{dx},\] if \[y={{\sin }^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right),\] \[0<x<1.\]

Answer:

We have, \[y={{\sin }^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right).\] Put \[x=\tan \theta \] \[\Rightarrow \] \[{{\tan }^{-1}}x=\theta \] \[\therefore \]      \[y={{\sin }^{-1}}\left( \frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta } \right)={{\sin }^{-1}}(cos2\theta )\] \[\left[ \because \,\,\cos 2\theta =\frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta } \right]\] \[\Rightarrow \]   \[y={{\sin }^{-1}}\left[ \sin \left( \frac{\pi }{2}-2\theta  \right) \right]\] \[\Rightarrow \]   \[y=\frac{\pi }{2}-2\theta \]                \[[\because si{{n}^{-1}}(sin\theta )=\theta ]\] \[\Rightarrow \]   \[y=\frac{\pi }{2}-2{{\tan }^{-1}}x\] On differentiating both sides w.r.t. x, we get \[\frac{dy}{dx}=\frac{d}{dx}\left( \frac{\pi }{2} \right)-2\frac{d}{dx}({{\tan }^{-1}}x)\] \[\Rightarrow \] \[\frac{dy}{dx}=0-\frac{2}{1+{{x}^{2}}}\] \[\therefore \] \[\frac{dy}{dx}=\frac{-\,2}{1+{{x}^{2}}}\]                     \[\left[ \because \,\,\frac{d}{dx}({{\tan }^{-1}}x)=\frac{1}{1+{{x}^{2}}} \right]\]