#### Answer

\[ - 2\tan x\]

#### Work Step by Step

\[\begin{gathered}
\frac{d}{{dx}}\,\,\left[ {\ln \,\left( {{{\cos }^2}x} \right)} \right] \hfill \\
\hfill \\
Differentiate\,\,use\,\,\,the\,\,formula\,\,\frac{d}{{dx}}\,\,\left[ {\ln u} \right] = \frac{{{u^,}}}{u} \hfill \\
\hfill \\
let\,\,u = {\cos ^2}x\,\, \hfill \\
\hfill \\
then\,\,\,{u^,} = 2\cos x\,\left( { - \sin x} \right) \hfill \\
{u^,} = - 2\sin x\cos x \hfill \\
\hfill \\
substitute\,\,u{\text{ and }}{u^,} \hfill \\
\hfill \\
\frac{d}{{dx}}\,\,\left[ {\ln \,\left( {{{\cos }^2}x} \right)} \right] = \frac{{ - 2\sin x\cos x}}{{{{\cos }^2}x}} \hfill \\
\hfill \\
Simplify \hfill \\
\hfill \\
= \frac{{2\sin x}}{{\cos x}} = - 2\tan x \hfill \\
\hfill \\
\hfill \\
\end{gathered} \]