 Solution

If $f$ is differentiable at $a$ and $g$ at $f(a)$, then $g\circ f$ is differentiable at $a$ andWrite $g(f(a)+u)-g(f(a))=u\{g^{\prime }(f(a))+\epsilon (u)\}$, where $\epsilon (u)\to 0$, and put $u=f(a+h)-f(a)$. Division by $h$ and passage to the limit proves the formula.