## Calculus: Early Transcendentals (2nd Edition)

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