Solutions

a) Let $g(x)=|x|$. We know that $g$ is differentiable for all $x\ne0$. In addition, $f$ is differentiable at $a$. Hence, by the chain rule,

$$h'(a)=g(f(a))'=g'(f(a))f'(a),$$

given $f(a)\ne0$.

b) Consider $f(x)=x-a$. Then $f(a)=a-a=0$ and $|f|$ is not differentiable at $a$.

c) Since both $f$ and $g$ are differentiable at $a$, it follows that they also are continuous at $a$. Hence, there exists an interval centered at $a$ where $\max(f,g)=f$ or $\max(f,g)=g$, given $f(a)\ne g(a)$. Without loss of generality assume $\max(f,g)=f$ in this interval. Therefore $(\max(f,g))'=f'(a)$. An analogous argument can be made to prove that $\min(f,g)$ is differentiable at $a$.

d) Consider $f(x)=x$ and $g(x)=-x$. Both are clearly differentiable at $x=0$ and $f(0)=g(0)=0$, but neither $\max(f,g)=|x|$ nor $\min(f,g)=-|x|$ is differentiable at $x=0$.