If \[y=\log {{\left( \frac{1+x}{1-x} \right)}^{1/4}}-\frac{1}{2}{{\tan }^{-1}}x,\]then \[\frac{dy}{dx}=\]

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

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

\[\frac{{{x}^{2}}}{2\,\,(1-{{x}^{4}})}\]

If \[y=\log {{\left( \frac{1+x}{1-x} \right)}^{1/4}}-\frac{1}{2}{{\tan

1.	If \[y=\log {{\left( \frac{1+x}{1-x} \right)}^{1/4}}-\frac{1}{2}{{\tan }^{-1}}x,\]then \[\frac{dy}{dx}=\]
A.	\[\frac{{{x}^{2}}}{1-{{x}^{4}}}\]
B.	\[\frac{2{{x}^{2}}}{1-{{x}^{4}}}\]
C.	\[\frac{{{x}^{2}}}{2\,\,(1-{{x}^{4}})}\]
D.	None of these
Answer» B. \[\frac{2{{x}^{2}}}{1-{{x}^{4}}}\]

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If \[y=\log {{\left( \frac{1+x}{1-x} \right)}^{1/4}}-\frac{1}{2}{{\tan }^{-1}}x,\]then \[\frac{dy}{dx}=\]

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