Explanation

(d)
$\because\ \ f(x) = \sin x - \cos x$
On differentiating w.r.t. $x$, we get
$f^{'}(x) = \cos x + \sin x$
$= \sqrt{2}\left( \frac{1}{\sqrt{2}}\cos x + \frac{1}{\sqrt{2}}\sin x \right)$
$= \sqrt{2}\left( \cos\frac{\pi}{4}\cos x + \sin\frac{\pi}{4}\sin x \right)$
$= \sqrt{2}\left\lbrack \cos\left( x - \frac{\pi}{4} \right) \right\rbrack$
For decreasing, $f^{'}(x) < 0$
$\frac{\pi}{2} < \left( x - \frac{\pi}{4} \right) < \frac{3\pi}{2}$(within $0 \leq x \leq 2\pi$)
$\Rightarrow \ \frac{\pi}{2} + \frac{\pi}{4} < \left( x - \frac{\pi}{4} + \frac{\pi}{4} \right) < \frac{3\pi}{2} + \frac{\pi}{4}$
$\Rightarrow \ \frac{3\pi}{4} < x < \frac{7\pi}{4}$