Given that \(y = {\left( {\cos x} \right)^x}\).
Taking log both side to get.
\(\ln y = x\ln \cos x\)
Now differentiating with respect to \(x\) we get,
\(\begin{aligned}{c}\frac{1}{y}\frac{{dy}}{{dx}}&= \frac{d}{{dx}}\left( {x\ln \cos x} \right)\\\frac{{dy}}{{dx}}&= y\left( {x\frac{d}{{dx}}\left( {\ln \cos x} \right) + \ln \cos x} \right)\\&= {\left( {\cos x} \right)^x}\left( {\frac{x}{{\cos x}}\left( { - \sin x} \right) + \ln \cos x} \right)\\&= {\left( {\cos x} \right)^x}\left( {\ln \cos x - x\tan x} \right)\end{aligned}\)
Hence \(\frac{{dy}}{{dx}} = {\left( {\cos x} \right)^x}\left( {\ln \cos x - x\tan x} \right)\).