= Absolute value at a simple zero
{title2=$(|g|)'_\pm(a)=\pm|g'(a)|$}
Suppose a real function $g$ is differentiable at $a$, with $g(a)=0$ and $g'(a)\ne0$. The <derivative> definition gives $g(a+h)=g'(a)h+o(|h|)$. The <reverse triangle inequality> then implies
$$
|g(a+h)|=|g'(a)|\,|h|+o(|h|).
$$
Thus $|g|$ has unequal <one-sided derivatives>, $|g'(a)|$ on the right and $-|g'(a)|$ on the left. More generally, multiplication by a function $w$ continuous at $a$ with $w(a)\ne0$ gives right and left slopes $w(a)|g'(a)|$ and $-w(a)|g'(a)|$. This elementary expansion explains how a smooth oscillation can acquire infinitely many corners under the <absolute value function>, even after multiplication by a factor that makes it differentiable at an accumulation point.
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