Solution
= Solution
If $f$ is <differentiable> at $a$ and $g$ at $f(a)$, then $g\circ f$ is <differentiable> at $a$ and
$$
(g\circ f)'(a)=g'(f(a))f'(a).
$$
Write $g(f(a)+u)-g(f(a))=u\{g'(f(a))+\varepsilon(u)\}$, where $\varepsilon(u)\to0$, and put $u=f(a+h)-f(a)$. Division by $h$ and passage to the <limit> proves the formula.