#### Answer

\[\frac{{dy}}{{dx}} = {\cos ^2}x - {\sin ^2}x\]

#### Work Step by Step

\[\begin{gathered}
y = \sin x\cos x \hfill \\
\hfill \\
\,\,use\,\,the\,\,product\,\,rule. \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = \,\left( {\sin x} \right)\,{\left( {\cos x} \right)^,} + \,\left( {\cos x} \right)\,{\left( {\sin x} \right)^,} \hfill \\
\hfill \\
then \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = \,\left( {\sin x} \right)\,\left( { - \sin x} \right) + \,\left( {\cos x} \right)\,\left( {\cos x} \right) \hfill \\
\hfill \\
multiply \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = - {\sin ^2}x + {\cos ^2}x \hfill \\
\hfill \\
or \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = {\cos ^2}x - {\sin ^2}x \hfill \\
\end{gathered} \]