Answer
\[ = - {e^x}\sin x\]
Work Step by Step
\[\begin{gathered}
y = \frac{1}{2}{e^x}\cos x \hfill \\
\hfill \\
differentiate\,\,to\,\,find\,\,y{\,^,} \hfill \\
\hfill \\
y' = \frac{1}{2}\,\,\left[ {\,{{\left( {{e^x}} \right)}^\prime }\cos x + {e^x}\,{{\left( {\cos x} \right)}^\prime }} \right] \hfill \\
\hfill \\
= \frac{1}{2}\,\,\left[ {{e^x}\cos x - {e^x}\sin x} \right] \hfill \\
\hfill \\
differentiate\,\,to\,\,find\,\,{y^,}^, \hfill \\
use\,\,the\,\,product\,\,rule \hfill \\
\hfill \\
y'' = \frac{1}{2}\,\,\left[ {\,{{\left( {{e^x}} \right)}^\prime }\cos x + {e^x}\,{{\left( {\cos x} \right)}^\prime } - \,{{\left( {{e^x}} \right)}^\prime }\sin x - {e^x}\,{{\left( {\sin x} \right)}^\prime }} \right] \hfill \\
\hfill \\
= \frac{1}{2}\,\,\left[ {{e^x}\cos x - {e^x}\sin x - {e^x}\sin x - {e^x}\cos x} \right] \hfill \\
\hfill \\
Simplify \hfill \\
\hfill \\
= \frac{1}{2}\,\,\left[ { - 2{e^x}\sin x} \right] \hfill \\
\hfill \\
= - {e^x}\sin x \hfill \\
\hfill \\
\end{gathered} \]
