If $y=|\cos x|+|\sin x|$, then $\frac{d y}{d x}$ at $x=\frac{2 \pi}{3}$ is

$\frac{\sqrt{3}-1}{2}$

In the neighbourhood of $x=\frac{2 \pi}{3}$, we have

$\cos x<0$ and $\sin x>0$

$\Rightarrow |\cos x|=-\cos x$ and $|\sin x|=\sin x$

∴  $y=|\cos x|+|\sin x|$

$\Rightarrow y=-\cos x+\sin x$

$\Rightarrow \frac{d y}{d x}=\sin x+\cos x$

$\Rightarrow\left(\frac{d y}{d x}\right)_{x=\frac{2 \pi}{3}}=\sin \frac{2 \pi}{3}+\cos \frac{2 \pi}{3}=\frac{\sqrt{3}-1}{2}$