Question

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

Answer 1

Let, f(x) = |sin x| 

`|sin x| = {(-sin x, ,(2m-1),pi<2mpi \text { where m }∈ Z),(sin x, ,2mpi< x<(2m +1),pi\text { where m} ∈  Z),(-sin x, ,(2m +1)pi<x<2(m+1),pi \text { where m } ∈ Z):}`

\[\left( \text { LHD at x } = 2m\pi \right) = \lim_{x \to 2m \pi^-} \frac{f\left( x \right) - f\left( 2m\pi \right)}{x - 2m\pi}\]
\[ = \lim_{x \to 2m \pi^-} \frac{- \sin\left( x \right) - 0}{x - 2m\pi}\]
\[ = \lim_{h \to 0} \frac{- \sin\left( 2m\pi - h \right)}{2m\pi - h - 2m\pi}\]
\[ = \lim_{h \to 0} \frac{\sin\left( h \right)}{- h} = - 1\]

\[\left(\text {  RHD at x } = 2m\pi \right) = \lim_{x \to 2m \pi^+} \frac{f\left( x \right) - f\left( 2m\pi \right)}{x - 2m\pi}\]
\[ = \lim_{x \to 2m \pi^+} \frac{\sin\left( x \right) - 0}{x - 2m\pi}\]
\[ = \lim_{h \to 0} \frac{\sin\left( 2m\pi + h \right)}{2m\pi + h - 2m\pi}\]
\[ = \lim_{h \to 0} \frac{\sin\left( h \right)}{h} = 1\]

\[\text { Here, LHD } \neq\text {  RHD } \text{So, function is not differentiable at x} = 2m\pi, where, m \in Z . . . . . \left( 1 \right)\]
\[\]

\[\left[ \text { LHD at x } = \left( 2m + 1 \right)\pi \right] = \lim_{x \to \left( 2m + 1 \right) \pi^-} \frac{f\left( x \right) - f\left[ \left( 2m + 1 \right)\pi \right]}{x - \left( 2m + 1 \right)\pi}\]
\[ = \lim_{x \to \left( 2m + 1 \right) \pi^-} \frac{\sin \left( x \right) - 0}{x - \left( 2m + 1 \right)\pi}\]
\[ = \lim_{h \to 0} \frac{\sin \left[ \left( 2m + 1 \right)\pi - h \right]}{\left( 2m + 1 \right)\pi - h - \left( 2m + 1 \right)\pi}\]
\[ = \lim_{h \to 0} \frac{\sin \left( h \right)}{- h} = - 1\]

\[\left[ \text { RHD at x } = \left( 2m + 1 \right)\pi \right] = \lim_{x \to \left( 2m + 1 \right) \pi^+} \frac{f\left( x \right) - f\left( \left( 2m + 1 \right)\pi \right)}{x - \left( 2m + 1 \right)\pi}\]
\[ = \lim_{x \to \left( 2m + 1 \right) \pi^+} \frac{- \sin \left( x \right) - 0}{x - \left( 2m + 1 \right)\pi}\]
\[ = \lim_{h \to 0} \frac{- \sin \left[ \left( 2m + 1 \right)\pi + h \right]}{\left( 2m + 1 \right)\pi + h - \left( 2m + 1 \right)\pi}\]
\[ = \lim_{h \to 0} \frac{\sin \left( h \right)}{h} = 1\]

\[\text { Here, LHD } \neq \text { RHD . So, function is not differentiable at x }= \left( 2m + 1 \right)\pi, \text { where, m } \in Z . . . . . \left( 2 \right)\]
\[\text { From, } \left( 1 \right)\text {  and } \left( 2 \right), \text { we get }\]
\[f\left( x \right) = \left| \sin x \right| \text{is  not differentiable at x }= n\pi\]

We know that, 
\[\cos \left| x \right| = \cos x\text {  For all } x \in R\]
\[\text{Also we know that} \cos x\text {  is differentiable at all real points} . \]
\[\text{Therefore,} \cos \left| x \right| \text { is differentiable everywhere} .\]