Is | | differentible? What about |x|?

Question

Is $|\sin\text{x}|$ differentible? What about $\cos|\text{x}|?$

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Answer

Let, d(x) = |sin x|
$\sin\text{x}=0,$ for $\text{x}=\text{n}\pi,$
$|\sin\text{x}|=\begin{cases}-\sin\text{x}\ (2\text{m}-1)\pi<\text{x}<2\text{mx},&\text{where m}\in\text{Z}\\\sin\text{x}\ 2\text{mx}<\text{x}<(2\text{m}+1)\pi,&\text{where m}\in\text{Z}\\-\sin\text{x}\ (2\text{m}+1)\pi<\text{x}<2(\text{m}+1)\pi,&\text{where m}\in\text{Z}\end{cases}$
$(\text{LHL at x}=2\text{mx})=\lim_\limits{\text{x}\rightarrow2\text{mx}^{-}}\frac{\text{f(x)}-\text{f}(2\text{mx})}{\text{x}-2\text{mx}}$
$=\lim_\limits{\text{x}\rightarrow2\text{mx}^{-}}\frac{-\sin(\text{x}-0)}{\text{x}-2\text{mx}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{-\sin(2\text{mx}-\text{h})}{2\text{mx}-\text{h}-2\text{mx}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{h})}{\text{h}}=-1$
$(\text{RHL at x}=2\text{mx})=\lim_\limits{\text{x}\rightarrow2\text{mx}^{+}}\frac{\text{f(x)}-\text{f}(2\text{mx})}{\text{x}-2\text{mx}}$
$=\lim_\limits{\text{x}\rightarrow2\text{mx}^{+}}\frac{\sin(\text{x)}-0}{\text{x}-2\text{mx}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(2\text{mx}+\text{h})}{2\text{mx}+\text{h}-2\text{mx}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{h})}{\text{h}}=1$
Here, $\text{LHL}\neq\text{RHL}$ So, function is not differentiable at $\text{x}=2\text{m}\pi,$ where, $\text{m}\in\text{Z}\ \dots(1)$
$[\text{LHL at x}=(2\text{m}+1)\pi]=\lim_\limits{\text{x}\rightarrow(2\text{m}+1)\pi^{-}}\frac{\text{f(x)}-\text{f}[(2\text{m}+1)\pi]}{\text{x}-(2\text{m}+1)\pi}$
$=\lim_\limits{\text{x}\rightarrow(2\text{m}+1)\pi^{-}}\frac{\sin(\text{x})-0}{\text{x}-(2\text{m}+1)\pi}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{\sin[(2\text{m}+1)\pi-\text{h}]}{(2\text{m}+1)\pi-\text{h}-(2\text{m}+1)\pi}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{\sin(\text{h})}{\text{h}}=-1$
$[\text{RHL at x}=(2\text{m}+1)\pi]=\lim_\limits{\text{x}\rightarrow(2\text{m}+1)\pi^{+}}\frac{\text{f(x)}-\text{f}[(2\text{m}+1)\pi]}{\text{x}-(2\text{m}+1)\pi}$
$=\lim_\limits{\text{x}\rightarrow(2\text{m}+1)\pi^{+}}\frac{-\sin(\text{x})-0}{\text{x}-(2\text{m}+1)\pi}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{-\sin[(2\text{m}+1)\pi+\text{h}]}{(2\text{m}+1)\pi+\text{h}-(2\text{m}+1)\pi}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{\sin(\text{h})}{\text{h}}=1$
Here, $\text{LHL}\neq\text{RHL}.$
So, function is not differentiable at $\text{x}=(2\text{m}+1)\pi,$ where, $\text{m}\in\text{Z}\ \dots(2)$
From, (1) and (2), we get
$\text{f(x)}=|\sin\text{x}|$ is not differentiable at $\text{x}=\text{n}\pi$
We know that,
$\cos|\text{x}|=\cos\text{x}$ For all $\text{x}\in\text{R}$
Also we know that cos x is differentiable at all real points.
Therefore, cos |x| is differentiable everywhere.