Differentiate the functions with respect to x.

Let $f(x)=\cos (\sin x), u(x)=\sin x$, and $v(t)=\cos t$

Then, $(v o u)(x)=v(u(x))=v(\sin x)=\cos (\sin x)=f(x)$

Thus, f is a composite function of two functions.

Put t = u (x) = sin x

$\therefore \frac{d v}{d t}=\frac{d}{d t}[\cos t]=-\sin t=-\sin (\sin x)$

$\frac{d t}{d x}=\frac{d}{d x}(\sin x)=\cos x$

By chain rule, $\frac{d f}{d x}=\frac{d v}{d t} \cdot \frac{d t}{d x}=-\sin (\sin x) \cdot \cos x=-\cos x \sin (\sin x)$

Alternate method

$\frac{d}{d x}[\cos (\sin x)]=-\sin (\sin x) \cdot \frac{d}{d x}(\sin x)=-\sin (\sin x) \cdot \cos x=-\cos x \sin (\sin x)$