Answer
\[{f^,}\,\left( x \right) = \frac{{4{e^{4x}}}}{{1 + {e^{8x}}}}\]
Work Step by Step
\[\begin{gathered}
f\,\left( x \right) = {\tan ^{ - 1}}\,\left( {{e^{4x}}} \right) \hfill \\
\hfill \\
Use\,\,\frac{d}{{dy}}\,\,\,\left[ {{{\tan }^{ - 1}}u} \right] = \frac{{{u^,}}}{{1 + {u^2}}} \hfill \\
then \hfill \\
\hfill \\
{f^,}\,\left( x \right) = \frac{{\,\left( {{e^{4x}}} \right)}}{{1 + \,{{\left( {{e^{4x}}} \right)}^2}}} \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
{f^,}\,\left( x \right) = \frac{{4{e^{4x}}}}{{1 + {e^{8x}}}} \hfill \\
\end{gathered} \]
