Past exam of the mathematics course of the University of Cambridge / 2008 / ia / Paper 1 / 11D / c

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.

If $f$ is differentiable at $x$, the product rule applied to $f\cdot f$ givesThis proves sufficiency of the first alternative. No additional growth condition is required; if $f(x)=0$, the derivative of the square is zero.

Suppose $f(x)=0$ and $f(x+h)/\sqrt{|h|}\to 0$. ThenThereforeThis proves sufficiency of the second alternative even if $f$ itself is not differentiable at $x$. For example, $f(t)=|t{|}^{3/4}$ is continuous and not differentiable at zero, while $g(t)=|t{|}^{3/2}$ is differentiable there with derivative zero.