If \[y={{\tan }^{-1}}\sqrt{\frac{x+1}{x-1}}\]. Then \[\frac{dy}{dx}\]is

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

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

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

If \[y={{\tan }^{-1}}\sqrt{\frac{x+1}{x-1}}\]. Then \[\frac{dy}{dx}\]i

1.	If \[y={{\tan }^{-1}}\sqrt{\frac{x+1}{x-1}}\]. Then \[\frac{dy}{dx}\]is
A.	\[\frac{-1}{2\left\| x \right\|\sqrt{{{x}^{2}}-1}}\]
B.	\[\frac{-1}{2x\sqrt{{{x}^{2}}-1}}\]
C.	\[\frac{1}{2x\sqrt{{{x}^{2}}-1}}\]
D.	none of these
Answer» B. \[\frac{-1}{2x\sqrt{{{x}^{2}}-1}}\]

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If \[y={{\tan }^{-1}}\sqrt{\frac{x+1}{x-1}}\]. Then \[\frac{dy}{dx}\]is

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