Chain rule with sine and cosine composition

Write $f(x) = \sin(\cos(x))$ as a composition of an outer function $\sin(u)$ and inner function $u = \cos(x)$. By the chain rule, $f'(x) = \cos(\cos(x)) \cdot \frac{d}{dx}[\cos(x)] = \cos(\cos(x)) \cdot (-\sin(x))$. So $f'(x) = -\sin(x)\cos(\cos(x))$.

Now substitute $x = \pi/2$: we have $\sin(\pi/2) = 1$ and $\cos(\pi/2) = 0$, so $\cos(\cos(\pi/2)) = \cos(0) = 1$. Therefore $f'(\pi/2) = -1 \cdot 1 = -1$.