Q.Examine the differentiability of the following functions in $\mathbb{R}$ by drawing the diagrams.

Step 1. General principle. If $g$ is differentiable and has a simple zero at $x_0$ (i.e. $g(x_0)=0$ but $g'(x_0)\neq0$), then near $x_0$, $g$ changes sign, so $|g(x)|$ switches between $-g(x)$ and $g(x)$ exactly at $x_0$. Its one-sided derivatives there are $\pm g'(x_0)$ with opposite signs, so $|g|$ has a corner at $x_0$ and is not differentiable there. Away from any zero of $g$, $|g(x)|$ locally equals either $g(x)$ or $-g(x)$ with no sign change, so it is differentiable there with derivative $\pm g'(x)$.

Step 2. (i) $|\sin x|$. $\sin x=0$ exactly at $x=n\pi$, $n\in\mathbb{Z}$, and $\dfrac{d}{dx}\sin x=\cos x$, with $\cos(n\pi)=\pm1\neq0$ — every zero of $\sin x$ is simple. So by Step 1, $|\sin x|$ has a corner at each $x=n\pi$: e.g. at $x=0$, RHD $=\lim_{h\to0^+}\frac{\sin h}{h}=1$ while LHD $=\lim_{h\to0^-}\frac{-\sin h}{h}=-1$ (since $\sin h<0$ there so $|\sin h|=-\sin h$), confirming the corner. At every other $x$, $\sin x\neq0$ so $|\sin x|$ coincides locally with $\pm\sin x$ and is differentiable, with derivative $\cos x$ where $\sin x>0$ and $-\cos x$ where $\sin x<0$. Graphically, $|\sin x|$ is a series of identical humps of height $1$ touching the $x$-axis at each $x=n\pi$, each hump meeting the next in a sharp point. …

• Claiming $|\sin x|$ or $|\cos x|$ is differentiable everywhere because $\sin$/$\cos$ themselves are smooth — the absolute value still creates corners exactly at their zeros …