Explanation

(b)
$f(x) = e^{x}\sin x$
$f^{'}(x) = e^{x}cosx + sin\ xe^{x}$
$f^{''}(x) = - e^{x}sin\ x + \cos{xe^{x} + e^{x}\cos x + e^{x}\sin x}$
$2e^{x}cosx$
For maximum slope
$f^{''}(x) = 0$
$\Rightarrow \ \ 2e^{x}\cos x = 0$
$\Rightarrow \ \ \ \ \ \ \ \cos{x = 0}$
$\ \Rightarrow \ \ x = \frac{\pi}{2}\ ,\frac{3\pi}{2},\forall x \in \lbrack 0,2\pi\rbrack$
${\ \ \ \ \ \ f}^{'''}(x) = 2\ \ \ \ \ \ \ \ \left\lbrack - e^{x}\sin{x + e^{x}\cos x} \right\rbrack$
$\ \ at\ \ x = \frac{\pi}{2}\ \ ,f^{'''}(x) < 0$
$and\ \ at\ \ \ \ x = \frac{3\pi}{2}f^{'''}(x) > 0$
$\therefore slop\ is\ maximum\ at\ x = \frac{\pi}{2}\ \ \ \ \ \ \$