Is |sin x| differentiable? What about cos |x|?

Solution:

Let, $f(x)=|\sin x|$

$\sin x=0$, for $x=n \pi$

$|\sin x|=\left\{\begin{array}{cc}-\sin x & (2 m-1) \pi

$(\mathrm{LHD}$ at $x=2 m \pi)=\lim _{x \rightarrow 2 m \pi^{-}} \frac{f(x)-f(2 m \pi)}{x-2 m \pi}$

$=\lim _{x \rightarrow 2 m \pi^{-}} \frac{-\sin (x)-0}{x-2 m \pi}$

$=\lim _{h \rightarrow 0} \frac{-\sin (2 m \pi-h)}{2 m \pi-h-2 m \pi}$

$=\lim _{h \rightarrow 0} \frac{\sin (h)}{-h}=-1$

$(\mathrm{RHD}$ at $x=2 m \pi)=\lim _{x \rightarrow 2 m \pi^{+}} \frac{f(x)-f(2 m \pi)}{x-2 m \pi}$

$=\lim _{x \rightarrow 2 m \pi^{+}} \frac{\sin (x)-0}{x-2 m \pi}$

$=\lim _{h \rightarrow 0} \frac{\sin (2 m \pi+h)}{2 m \pi+h-2 m \pi}$

$=\lim _{h \rightarrow 0} \frac{\sin (h)}{h}=1$

Here, $\mathrm{LHD} \neq \mathrm{RHD}$ So, function is not differentiable at $x=2 m \pi$, where, $m \in \mathrm{Z}$

$[\operatorname{LHD}$ at $x=(2 m+1) \pi]=\lim _{x \rightarrow(2 m+1) \pi^{-}} \frac{f(x)-f[(2 m+1) \pi]}{x-(2 m+1) \pi}$

$=\lim _{x \rightarrow(2 m+1) \pi^{-}} \frac{\sin (x)-0}{x-(2 m+1) \pi}$

$=\lim _{h \rightarrow 0} \frac{\sin [(2 m+1) \pi-h]}{(2 m+1) \pi-h-(2 m+1) \pi}$

$=\lim _{h \rightarrow 0} \frac{\sin (h)}{-h}=-1$

$[\mathrm{RHD}$ at $x=(2 m+1) \pi]=\lim _{x \rightarrow(2 m+1) \pi^{+}} \frac{f(x)-f((2 m+1) \pi)}{x-(2 m+1) \pi}$

Here, LHD $\neq$ RHD. So, function is not differentiable at $x=(2 m+1) \pi$, where, $m \in \mathrm{Z} \quad \ldots \ldots$ (2)

From, $(1)$ and $(2)$, we get

$f(x)=|\sin x|$ is not differentiable at $x=n \pi$

We know that,

$\cos |x|=\cos x \quad$ For all $x \in R$

Also we know that $\cos x$ is differentiable at all real points.

Therefore, $\cos |x|$ is differentiable everywhere.