#### Answer

\[ = {\sec ^2}\,\left( {x{e^x}} \right) \cdot \,\left( {{e^x} + x{e^x}} \right)\]

#### Work Step by Step

\[\begin{gathered}
y = \,\tan \,\left( {x{e^x}} \right) \hfill \\
\hfill \\
differentiate\, \hfill \\
\hfill \\
{y^,} = \frac{d}{{dx}}\,\left( {\tan \,\left( {x{e^x}} \right)} \right) \hfill \\
\hfill \\
use\,\,the\,\,\,Chain\,\,rule \hfill \\
\hfill \\
= {\sec ^2}\,\left( {x{e^x}} \right) \cdot \,{\left( {x{e^x}} \right)^,} \hfill \\
\hfill \\
use\,\,the\,Product\,\,rule \hfill \\
\hfill \\
= {\sec ^2}\,\left( {x{e^x}} \right) \cdot \,\left( {{x^,} \cdot {e^x} + x\, \cdot {{\left( {{e^x}} \right)}^,}} \right) \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
= {\sec ^2}\,\left( {x{e^x}} \right) \cdot \,\left( {{e^x} + x{e^x}} \right) \hfill \\
\end{gathered} \]