Solution:
Let’s take $y=(\log x)^{\cos x}$
On both the sides taking log, we get
$\log y=\log (\log x)^{\cos x}=\cos x \log (\log x)$
$\frac{d}{d x} \log y=\frac{d}{d x}[\cos x \log (\log x)]$
$\frac{1}{y} \frac{d y}{d x}=\cos x \frac{d}{d x} \log (\log x)+\log (\log x) \frac{d}{d x} \cos x$
[Using the Product rule]
$\frac{1}{y} \frac{d y}{d x}=\cos x \frac{1}{\log x} \frac{d}{d x}(\log x)+\log (\log x)(-\sin x)$
$\frac{1}{y} \frac{d y}{d x}=\frac{\cos x}{\log x} \cdot \frac{1}{x}-\sin x \log (\log x)$
$\frac{d y}{d x}=y\left[\frac{\cos x}{x \log x}-\sin x \log (\log x)\right.$
$=(\log x)^{\cos x}\left[\frac{\cos x}{x \log x}-\sin x \log (\log x)\right]$