Let $f(x) = \left\{ \begin{array}{r} \sin{x,\ \ for\ \ x \geq 0} \ 1 - \cos{x,\ \ \ for\ \ x \leq 0} \end{array} ight.\$ and$g(x) = e^{x}$. Then, $(gof)^{'}(0)$ is

Question

Let $f(x) = \left\{ \begin{array}{r} \sin{x,\ \ for\ \ x \geq 0} \ 1 - \cos{x,\ \ \ for\ \ x \leq 0} \end{array} ight.\$ and$g(x) = e^{x}$. Then, $(gof)^{'}(0)$ is

Dr. Abubacker · Accepted Answer

(c)
Here, $gof = \left\{ \begin{array}{r}
e^{\sin x},\ \ \ \ x \geq 0 \
e^{1 - \cos x},\ \ \ \ \ x \leq 0
\end{array} ight.\$
$	herefore$ LHD$= \lim_{h ightarrow 0}\frac{gof(0 - h) - gof(h)}{- h}$
$= \lim_{h ightarrow 0}\frac{e^{1 - \cos h} - e^{1 - \cos h}}{- h} = 0$
RHD$= \lim_{h ightarrow 0}\frac{gof(0 + h) - gof(h)}{h}$
$= \lim_{h ightarrow 0}\frac{e^{\sin h} - e^{\sin h}}{h} = 0$
Since, RHD=LHD=0
$	herefore\ \ (gof)^{'}\ (0) = 0$

Q. Let $f(x) = \left\{ \begin{array}{r} \sin{x,\ \ for\ \ x \geq 0} \\ 1 - \cos{x,\ \ \ for\ \ x \leq 0} \end{array} \right.\$ and$g(x) = e^{x}$. Then, $(gof)^{'}(0)$ is

Explanation

Reviewed by Dr. Abubacker

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