= Differentiability of the square of a continuous function
{title2=$f^2\text{ differentiable at }x\iff f\text{ differentiable at }x\text{ or }f(x)=0,\ f(x+h)=o(\sqrt{|h|})$}
At a point with $f(x)\ne0$, factor the difference of squares and divide by $f(x+h)+f(x)\to2f(x)$ to show that <differentiability> of $f^2$ is equivalent to that of $f$. At a zero of $f$, its square has a local minimum; if <differentiable>, its derivative must vanish. The difference quotient then vanishes exactly when $f(x+h)^2/|h|\to0$, equivalently $f(x+h)/\sqrt{|h|}\to0$. For example, $f(t)=|t|^\alpha$ with $1/2<\alpha<1$ is not <differentiable> at zero, but its square is.
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