Solution
= Solution
Differentiability lets us write
$$
f(a+h)=f(a)+f'(a)h+r(h),
\qquad \frac{r(h)}h\to0,
$$
and, with $k\to0$,
$$
g(f(a)+k)=g(f(a))+g'(f(a))k+s(k),
\qquad \frac{s(k)}k\to0.
$$
Substitute $k=f(a+h)-f(a)=f'(a)h+r(h)$. Since $k=O(h)$, the remainder $s(k)$ is $o(h)$. Thus
$$
g(f(a+h))-g(f(a))
=g'(f(a))f'(a)h+o(h),
$$
and the <chain rule> is
$$
\boxed{(g\circ f)'(a)=g'(f(a))f'(a)}.
$$