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