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

Question:

Is |sinĀ 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.