Let f(x) = begin{cases} sin x, & text{for } x | vedclass

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(D) To find $(g \circ f)'(0)$,we must check the left-hand derivative and right-hand derivative at $x = 0$.For the right-hand derivative $(g \circ f)'(

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None of these

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(D) To find $(g \circ f)'(0)$,we must check the left-hand derivative and right-hand derivative at $x = 0$.For the right-hand derivative $(g \circ f)'(0^+)$:$(g \circ f)(x) = g(f(x)) = g(\sin x) = e^{\sin x}$ for $x \ge 0$.$(g \circ f)'(x) = e^{\sin x} \cdot \cos x$.At $x = 0$,$(g \circ f)'(0^+) = e^{\sin 0} \cdot \cos 0 = e^0 \cdot 1 = 1$.For the left-hand derivative $(g \circ f)'(0^-)$:$(g \circ f)(x) = g(f(x)) = g(1 - \cos x) = e^{1 - \cos x}$ for $x \le 0$.$(g \circ f)'(x) = e^{1 - \cos x} \cdot \sin x$.At $x = 0$,$(g \circ f)'(0^-) = e^{1 - \cos 0} \cdot \sin 0 = e^0 \cdot 0 = 0$.Since the left-hand derivative $(0)$ is not equal to the right-hand derivative $(1)$,the derivative $(g \circ f)'(0)$ does not exist.Therefore,the correct option is $D$.

Let $f(x) = \begin{cases} \sin x, & \text{for } x \ge 0 \\ 1 - \cos x, & \text{for } x \le 0 \end{cases}$ and $g(x) = e^x$. Then $(g \circ f)'(0)$ is

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