#### Answer

\[ = 2{e^x}\cos x\]

#### Work Step by Step

\[\begin{gathered}
y = {e^x}\sin x \hfill \\
\hfill \\
differentiate\,\,to\,\,find\,\,y{\,^,} \hfill \\
use\,\,the\,\,product\,\,rule \hfill \\
\hfill \\
y' = \,{\left( {{e^x}} \right)^\prime }\sin x + {e^x}\,{\left( {\sin x} \right)^\prime } \hfill \\
\hfill \\
= {e^x}\sin x + {e^x}\cos x \hfill \\
\hfill \\
differentiate\,\,to\,\,find\,\,{y^,}^, \hfill \\
use\,\,the\,\,product\,\,rule \hfill \\
\hfill \\
y'' = \,{\left( {{e^x}} \right)^\prime }\sin x + {e^x}\,{\left( {\sin x} \right)^\prime } + \,{\left( {{e^x}} \right)^\prime }\cos x + {e^x}\,{\left( {\cos x} \right)^\prime } \hfill \\
\hfill \\
= {e^x}\sin x + {e^x}\cos x + {e^x}\cos x - {e^x}\sin x \hfill \\
\hfill \\
Simplify \hfill \\
\hfill \\
= 2{e^x}\cos x \hfill \\
\hfill \\
\end{gathered} \]