Answer
\[{f^,}\,\left( x \right) = \frac{{x{e^{3x}} + {e^{3x}}}}{{x{e^x} + 1}} + 2{e^{2x}}\ln \,\left( {x{e^x} + 1} \right)\]
Work Step by Step
\[\begin{gathered}
f\,\left( x \right) = {e^{2x}}\ln \,\left( {x{e^x} + 1} \right) \hfill \\
Find\,\,the\,\,derivative\,\,of\,\,the\,\,\,function \hfill \\
{f^,}\,\left( x \right) = \,\,{\left[ {{e^{2x}}\ln \,\left( {x{e^x} + 1} \right)} \right]^,} \hfill \\
Use\,\,the\,\,product\,\,rule \hfill \\
{f^,}\,\left( x \right) = {e^{2x}}\,\,{\left[ {\ln \,\left( {x{e^x} + 1} \right)} \right]^,} + \ln \,\left( {x{e^x} + 1} \right)\,{\left( {{e^{2x}}} \right)^,} \hfill \\
Where\,\,\frac{d}{{dx}}\,\,\left[ {\ln g\,\left( x \right)} \right] = \frac{{{g^,}\,\left( x \right)}}{{g\,\left( x \right)}} \hfill \\
{f^,}\,\left( x \right) = {e^{2x}}\,\left( {\frac{{x{e^x} + {e^x}}}{{x{e^x} + 1}}} \right) + 2{e^{2x}}\ln \,\left( {x{e^x} + 1} \right) \hfill \\
Multiplying \hfill \\
{f^,}\,\left( x \right) = \frac{{x{e^{3x}} + {e^{3x}}}}{{x{e^x} + 1}} + 2{e^{2x}}\ln \,\left( {x{e^x} + 1} \right) \hfill \\
\end{gathered} \]