If \[y={{\tan }^{-1}}\left( \frac{{{2}^{x}}}{1+{{2}^{2x+1}}} \right),\]then \[\frac{dy}{dx}at\,x=0\] is

\[\log \,2\left( \frac{-1}{10} \right)\]

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

1.	If \[y={{\tan }^{-1}}\left( \frac{{{2}^{x}}}{1+{{2}^{2x+1}}} \right),\]then \[\frac{dy}{dx}at\,x=0\] is
A.	\[\frac{3}{5}\log \,2\]
B.	\[\frac{2}{5}\log \,2\]
C.	\[-\frac{3}{2}\log \,2\]
D.	\[\log \,2\left( \frac{-1}{10} \right)\]
Answer» E.

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If \[y={{\tan }^{-1}}\left( \frac{{{2}^{x}}}{1+{{2}^{2x+1}}} \right),\]then \[\frac{dy}{dx}at\,x=0\] is

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