Past exam of the mathematics course of the University of Cambridge / 2008 / ia / Paper 1 / 11D (Analysis I)

By differentiability implies continuity, $f(x+h)\to f(x)$. Add and subtract $f(x+h)g(x)$ in the difference quotient:The two quotient limits exist by the given differentiability. Taking the limit and using continuity yields the product rule:Thus the product is differentiable at the specified point; differentiability elsewhere is not needed.

At zero, $g(0)=0$ andby continuity of $f$ at zero. Thus $g$ is always differentiable there, with $g^{\prime }(0)=0$, regardless of whether $f^{\prime }(0)$ exists.

At $x\ne 0$, if $f$ is differentiable, the product rule applied to $t^{2}f(t)$ proves that $g$ is differentiable. Conversely, if $g$ is differentiable at $x$, the function $t^{-2}$ is differentiable on a neighborhood of $x$. Apply the product rule to $f(t)=t^{-2}g(t)$ on that neighborhood. This proves the differentiability restored by a quadratic zero equivalence:At a nonzero differentiability point, $g^{\prime }(x)=2xf(x)+x^2{f}^{\prime }(x)$.

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.