If \[y={{\tan }^{-1}}\sqrt{\frac{a-x}{a+x}}\], then \[\frac{dy}{dx}=\]

1.	If \[y={{\tan }^{-1}}\sqrt{\frac{a-x}{a+x}}\], then \[\frac{dy}{dx}=\]
A.	\[{{\cos }^{-1}}\frac{x}{a}\]
B.	\[-{{\cos }^{-1}}\frac{x}{a}\]
C.	\[\frac{1}{2}{{\cos }^{-1}}\frac{x}{a}\]
D.	None of these
Answer» E.

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If \[y={{\tan }^{-1}}\sqrt{\frac{a-x}{a+x}}\], then \[\frac{dy}{dx}=\]

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