Explanation

(b)
(Slope) $f^{'}(x) = e^{x}\cos x + \sin xe^{x}$
$= e^{x}\sqrt{2}\sin{(x + \pi/4)}$
$f^{''}(x) = \sqrt{2}\ e^{x}\left\{ \sin{x + \pi/4} \right\} + \cos{\{ x + \pi/4\}}$
$= 2\ e^{x}.\sin{(x + \pi/2)}$
For maximum slope put, $f^{''}(x) = 0$
$\Rightarrow \sin{(x + \pi/2)} = 0$
$\Rightarrow \ \cos x = 0$
$\therefore\ \ x = \pi/2,\ 3\pi/2$
$f^{'''}(x) = 2e^{x}\cos{(x + \pi/2)}$
$f^{'''}(\pi/2) = 2e^{x}.\cos\pi = - ve$
Maximum slope is at $x = \pi/2$