#### Answer

\[{y^,} = - \sin x\,\left( {2+\ln \,\left( {{{\cos }^2}x} \right)} \right)\]

#### Work Step by Step

\[\begin{gathered}
y = \cos x\ln \,\left( {{{\cos }^2}x} \right) \hfill \\
\hfill \\
Differentiate,\,{\text{ using}}\,\,{\text{the product rule}} \hfill \\
\hfill \\
{\text{and}}{\left( {{\text{cos}}\left( u \right)} \right)^,} = - \sin u \cdot {u^,} \hfill \\
\hfill \\
then \hfill \\
\hfill \\
{y^,} = \cos x\,\left( {\frac{{2\cos x\,\left( { - \sin x} \right)}}{{{{\cos }^2}x}}} \right) + \ln \,\left( {{{\cos }^2}x} \right)\,\left( { - \sin x} \right) \hfill \\
\hfill \\
multiply \hfill \\
\hfill \\
{y^,} = - 2\sin x - \sin x\ln \,\left( {{{\cos }^2}x} \right) \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
{y^,} = - \sin x\,\left( {2+\ln \,\left( {{{\cos }^2}x} \right)} \right) \hfill \\
\end{gathered} \]