A) \[\frac{1}{2\sqrt{1-{{x}^{2}}}}\]
B) \[1-\sqrt{1-{{x}^{2}}}\]
C) \[\frac{1}{2}\]
D) \[\frac{1}{\sqrt{1-{{x}^{2}}}}\]
Correct Answer: A
Solution :
\[y={{\tan }^{-1}}\left( \frac{x}{1+\sqrt{1-{{x}^{2}}}} \right)\] Put \[x=\sin \theta \] \[\therefore y={{\tan }^{-1}}\left( \frac{\sin \theta }{1+\sqrt{1-{{\sin }^{2}}\theta }} \right)={{\tan }^{-1}}\left( \frac{\sin \theta }{1+\cos \theta } \right)\] \[={{\tan }^{-1}}\frac{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{2{{\cos }^{2}}\frac{\theta }{2}}={{\tan }^{-1}}\tan \frac{\theta }{2}=\frac{\theta }{2}\] So, \[y=\frac{{{\sin }^{-1}}x}{2}\Rightarrow \frac{dy}{dx}=\frac{1}{2\sqrt{1-{{x}^{2}}}}\].
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