If \[y={{\tan }^{-1}}\left( \frac{x}{1+\sqrt{1-{{x}^{2}}}} \right)\], then \[\frac{dy}{dx}=\]

\[\frac{1}{2\sqrt{1-{{x}^{2}}}}\]

\[\frac{1}{\sqrt{1-{{x}^{2}}}}\]

If \[y={{\tan }^{-1}}\left( \frac{x}{1+\sqrt{1-{{x}^{2}}}} \right)\],

1.	If \[y={{\tan }^{-1}}\left( \frac{x}{1+\sqrt{1-{{x}^{2}}}} \right)\], then \[\frac{dy}{dx}=\]
A.	\[\frac{1}{2\sqrt{1-{{x}^{2}}}}\]
B.	\[1-\sqrt{1-{{x}^{2}}}\]
C.	\[\frac{1}{2}\]
D.	\[\frac{1}{\sqrt{1-{{x}^{2}}}}\]
Answer» B. \[1-\sqrt{1-{{x}^{2}}}\]

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If \[y={{\tan }^{-1}}\left( \frac{x}{1+\sqrt{1-{{x}^{2}}}} \right)\], then \[\frac{dy}{dx}=\]

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