\[\frac{d}{dx}\left[ {{\tan }^{-1}}\frac{\sqrt{1+{{x}^{2}}}+\sqrt{1-{{x}^{2}}}}{\sqrt{1+{{x}^{2}}}-\sqrt{1-{{x}^{2}}}} \right]=\] [Roorkee 1980; Karnataka CET 2005]

\[\frac{x}{\sqrt{1-{{x}^{4}}}}\]

\[\frac{-x}{\sqrt{1-{{x}^{4}}}}\]

\[\frac{-1}{2\sqrt{1-{{x}^{4}}}}\]

\[\frac{1}{2\sqrt{1-{{x}^{4}}}}\]

\[\frac{d}{dx}\left[ {{\tan }^{-1}}\frac{\sqrt{1+{{x}^{2}}}+\sqrt{1-{{

1.	\[\frac{d}{dx}\left[ {{\tan }^{-1}}\frac{\sqrt{1+{{x}^{2}}}+\sqrt{1-{{x}^{2}}}}{\sqrt{1+{{x}^{2}}}-\sqrt{1-{{x}^{2}}}} \right]=\] [Roorkee 1980; Karnataka CET 2005]
A.	\[\frac{-x}{\sqrt{1-{{x}^{4}}}}\]
B.	\[\frac{x}{\sqrt{1-{{x}^{4}}}}\]
C.	\[\frac{-1}{2\sqrt{1-{{x}^{4}}}}\]
D.	\[\frac{1}{2\sqrt{1-{{x}^{4}}}}\]
Answer» B. \[\frac{x}{\sqrt{1-{{x}^{4}}}}\]

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\[\frac{d}{dx}\left[ {{\tan }^{-1}}\frac{\sqrt{1+{{x}^{2}}}+\sqrt{1-{{x}^{2}}}}{\sqrt{1+{{x}^{2}}}-\sqrt{1-{{x}^{2}}}} \right]=\] [Roorkee 1980; Karnataka CET 2005]

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