If $f(x)=|x|+|\sin x|$ for $x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, then its left hand derivative at $x=0$ is

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Aicharya · Accepted Answer

$\begin{aligned} & f(x)=|x|+|\sin x| \ & 	ext { LHD }=\lim _{h ightarrow 0} \frac{f(0-h)-f(0)}{0-h} \ & =\lim _{h ightarrow 0} \frac{|0-h|+|\sin (0-h)|-(0+0)}{0-h} \ & =\lim _{h ightarrow 0} \frac{h+\sin h}{-h}=-\lim _{h ightarrow 0}\left(1+\frac{\sin h}{h}ight) \ & =-(1+1)=-2\end{aligned}$

If $f(x)=|x|+|\sin x|$ for $x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, then its left hand derivative…

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