Answer
$$y' =\left(\cos x\right)^x\left[ -x\tan x+ \ln \left(\cos x\right )\right]$$
Work Step by Step
Given $$y=\left(\cos x\right)^x $$
Then
\begin{align*}
\ln y &=\ln \left(\cos x\right)^x\\
&=x\ln \left(\cos x\right )
\end{align*}
Hence
\begin{align*}
\frac{ y'}{y} &=x\frac{-\sin x}{\cos x}+ \ln \left(\cos x\right ) \\
y'&=\left(\cos x\right)^x\left[ -x\tan x+ \ln \left(\cos x\right )\right]
\end{align*}
