 Differentiability restored by a quadratic zero

For a continuous function $f:\mathbb{R}\to \mathbb{R}$, $g(t)=t^{2}f(t)$ is differentiable at zero with $g^{\prime }(0)=0$, because $[g(h)-g(0)]/h=hf(h)\to 0$. At any nonzero point $x$, $g$ is differentiable exactly when $f$ is: one direction uses the product rule, and the converse applies it to $f(t)=t^{-2}g(t)$ near $x$.