Question
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 :
$$\log \,2\left( {\frac{{ - 1}}{{10}}} \right)$$
Solution :
Given expression can be written as
$$\eqalign{
& y = {\tan ^{ - 1}}\left[ {\frac{{{2^x}\left( {2 - 1} \right)}}{{1 + {2^x}{{.2}^{2x + 1}}}}} \right] \cr
& \,\,\,\,\, = {\tan ^{ - 1}}\left[ {\frac{{{2^{x + 1}} - {2^x}}}{{1 + {2^x}{{.2}^{2x + 1}}}}} \right] \cr
& \,\,\,\,\, = {\tan ^{ - 1}}\left( {{2^{x + 1}}} \right) - {\tan ^{ - 1}}\left( {{2^x}} \right) \cr
& \Rightarrow \,\frac{{dy}}{{dx}} = \frac{{{2^{x + 1}}\log \,2}}{{1 + {2^{2\left( {x + 1} \right)}}}} - \frac{{{2^x}\log \,2}}{{1 + {2^{2x}}}} \cr
& \therefore \,{\left( {\frac{{dy}}{{dx}}} \right)_{x = 0}} = \left( {\log \,2} \right)\left( {\frac{2}{5} - \frac{1}{2}} \right) = \log \,2\left( { - \frac{1}{{10}}} \right) \cr} $$