\[\frac{d}{dx}\left( {{\tan }^{-1}}\frac{\sqrt{1+{{x}^{2}}}-1}{x} \right)\] is equal to[MP PET 2004]

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

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

1.	\[\frac{d}{dx}\left( {{\tan }^{-1}}\frac{\sqrt{1+{{x}^{2}}}-1}{x} \right)\] is equal to[MP PET 2004]
A.	\[\frac{1}{1+{{x}^{2}}}\]
B.	\[\frac{1}{2(1+{{x}^{2}})}\]
C.	\[\frac{{{x}^{2}}}{2\sqrt{1+{{x}^{2}}}(\sqrt{1+{{x}^{2}}}-1)}\]
D.	\[\frac{2}{1+{{x}^{2}}}\]
Answer» C. \[\frac{{{x}^{2}}}{2\sqrt{1+{{x}^{2}}}(\sqrt{1+{{x}^{2}}}-1)}\]