If f(x)=|x|+|sin x| for x in left(-frac{pi}{2 | vedclass

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(C) The given function is $f(x) = |x| + |\sin x|$.To find the left hand derivative $(LHD)$ at $x=0$, we use the definition:$LHD = \lim_{h 	o 0^+} \fr

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$-2$

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(C) The given function is $f(x) = |x| + |\sin x|$.To find the left hand derivative $(LHD)$ at $x=0$, we use the definition:$LHD = \lim_{h 	o 0^+} \frac{f(0-h) - f(0)}{-h}$Since $f(0) = |0| + |\sin 0| = 0$, we have:$LHD = \lim_{h 	o 0^+} \frac{|-h| + |\sin(-h)|}{-h}$For small $h > 0$, $|-h| = h$ and $|\sin(-h)| = |-\sin h| = \sin h$ (since $\sin h > 0$ for $h \in (0, \pi/2)$).$LHD = \lim_{h 	o 0^+} \frac{h + \sin h}{-h}$$LHD = \lim_{h 	o 0^+} -\left( \frac{h}{h} + \frac{\sin h}{h} ight)$$LHD = -(1 + 1) = -2$.

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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