Let f(x) = begin{cases} 1, & forall x

Question

(D) To check the differentiability of $f(x)$ at $x = 0$,we calculate the Left-Hand Derivative $(LHD)$ and the Right-Hand Derivative $(RHD)$.For $x < 0

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does not exist

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(D) To check the differentiability of $f(x)$ at $x = 0$,we calculate the Left-Hand Derivative $(LHD)$ and the Right-Hand Derivative $(RHD)$.For $x $LHD = \lim_{h 	o 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h 	o 0^-} \frac{1 - 1}{h} = 0$.For $x \ge 0$,$f(x) = 1 + \sin x$. Therefore,the $RHD$ at $x = 0$ is:$RHD = \lim_{h 	o 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h 	o 0^+} \frac{(1 + \sin h) - 1}{h} = \lim_{h 	o 0^+} \frac{\sin h}{h} = 1$.Since $LHD 
eq RHD$ (i.e.,$0 
eq 1$),the derivative $f'(0)$ does not exist.

Let $f(x) = \begin{cases} 1, & \forall x < 0 \\ 1 + \sin x, & \forall 0 \le x \le \pi/2 \end{cases}$,then what is the value of $f'(x)$ at $x = 0$?

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