 Solution

For the necessity in the differentiability of the square of a continuous function, suppose $g=f^2$ is differentiable at $x$. If $f(x)\ne 0$, continuity makes $f(x+h)+f(x)$ nonzero for small $h$, andHence $f$ is differentiable at $x$, giving alternative (i).

If $f(x)=0$, $g(x)=0$ is a minimum because $g\ge 0$. The right difference quotients of $g$ are nonnegative and the left ones are nonpositive; since their common limit exists, it must be zero. Thus $g^{\prime }(x)=0$, andTaking square roots of this nonnegative expression shows $f(x+h)/\sqrt{|h|}\to 0$, giving alternative (ii). The two labelled parts below establish sufficiency of the respective alternatives. Together they prove the full stated equivalence, including the zero of $f$ where squaring can remove a failure of differentiability.