If $f(x)=|\cos x-\sin x|$, then $f'(\pi / 4)$ is equal to

none of these

We have,

$f(x) =|\cos x-\sin x|$

$\Rightarrow f(x) = \begin{cases}\cos x-\sin x, & \text { for } 0<x \leq \frac{\pi}{4} \\
\sin x-\cos x, & \text { for } \frac{\pi}{4}<x<\frac{\pi}{2}\end{cases}$

Clearly,

(LHD at $x=\pi / 4$) = $\left\{\frac{d}{d x}(\cos x-\sin x)\right\}$

$=(-\sin x-\cos x)_{x=\pi / 4}=-\sqrt{2}$

and,

(RHD at $x=\pi / 4$) = $\left\{\frac{d}{d x}(\sin x-\cos x)\right\}$ at $x=\pi / 4$

$=(\cos x+\sin x)_{x=\pi / 4}=\sqrt{2}$

∴  (LHD at $x=\pi / 4$) ≠ (RHD at $x=\pi / 4$)

Thus, $f'(\pi / 4)$ does not exist.