Differentiate the function with respect to x.

Let $y=(\log x)^{\cos x}$

Taking logarithm on both the sides, we obtain

$\log y=\cos x \cdot \log (\log x)$

Differentiating both sides with respect to x, we obtain

$\frac{1}{y} \cdot \frac{d y}{d x}=\frac{d}{d x}(\cos x) \times \log (\log x)+\cos x \times \frac{d}{d x}[\log (\log x)]$

$\Rightarrow \frac{1}{y} \cdot \frac{d y}{d x}=-\sin x \log (\log x)+\cos x \times \frac{1}{\log x} \cdot \frac{d}{d x}(\log x)$

$\Rightarrow \frac{d y}{d x}=y\left[-\sin x \log (\log x)+\frac{\cos x}{\log x} \times \frac{1}{x}\right]$

$\therefore \frac{d y}{d x}=(\log x)^{\cos x}\left[\frac{\cos x}{x \log x}-\sin x \log (\log x)\right]$