A) \[\sqrt{1-{{x}^{2}}}\]
B) \[\frac{1}{\sqrt{1-{{x}^{2}}}}\]
C) \[\frac{1}{2\sqrt{1-{{x}^{2}}}}\]
D) x
Correct Answer: C
Solution :
Let \[y={{\tan }^{-1}}\left( \frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} \right)\] Put \[x=\cos 2\theta \Rightarrow \theta =\frac{1}{2}{{\cos }^{-1}}x\] \[y={{\tan }^{-1}}\left( \frac{\sqrt{1+\cos 2\theta }-\sqrt{1-\cos 2\theta }}{\sqrt{1+\cos 2\theta }+\sqrt{1-\cos 2\theta }} \right)\] \[y={{\tan }^{-1}}\left( \frac{\sqrt{2{{\cos }^{2}}\theta }-\sqrt{2{{\sin }^{2}}\theta }}{\sqrt{2{{\cos }^{2}}\theta }+\sqrt{2{{\sin }^{2}}\theta }} \right)\] \[y={{\tan }^{-1}}\left( \frac{\cos \theta -\sin \theta }{\cos \theta +\sin \theta } \right)={{\tan }^{-1}}\left( \frac{1-\tan \theta }{1+\tan \theta } \right)\] \[y={{\tan }^{-1}}(\tan (\pi /4-\theta ))\] Þ \[y=\frac{\pi }{4}-\theta \] \[y=\frac{\pi }{4}-\frac{1}{2}{{\cos }^{-1}}x\] Þ \[\frac{dy}{dx}=\frac{1}{2}\left( \frac{1}{\sqrt{1-{{x}^{2}}}} \right)\].
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