A) \[\frac{2}{1+{{x}^{2}}}\]
B) \[\frac{2x}{1+{{x}^{2}}}\]
C) \[\frac{-2}{1+{{x}^{2}}}\]
D) \[\frac{-x}{1+{{x}^{2}}}\]
Correct Answer: A
Solution :
\[y={{\sin }^{-1}}\frac{2x}{1+{{x}^{2}}}\] Put \[x=\tan \theta \] \[x\] \[y=2\theta =2{{\tan }^{-1}}x\] \[(\because \theta ={{\tan }^{-1}}x)\] Differentiating with respect to x, we get \[\frac{dy}{dx}=\frac{2}{1+{{x}^{2}}}\] \[\left( \text{Since }\,\,0<x<1\text{ and }0<y<\frac{\pi }{2} \right)\].
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