If $y=|\cos x-\sin x|$, then $\frac{dy}{dx}$ at $x=\pi /4$ is:

$f(x)=|\cos x-\sin x|$.At $x=\pi /4$, $\cos x-\sin x=0$. We need to check LHD and RHD.$f(x)=\sqrt{2}|\frac{1}{\sqrt{2}}\cos x-\frac{1}{\sqrt{2}}\sin x|=\sqrt{2}|\sin (\frac{\pi }{4}-x)|$.Near $x=\pi /4$:If $x<\pi /4$, $\sin x<\cos x$, so $f(x)=\cos x-\sin x⟹f^{\prime }(x)=-\sin x-\cos x$.LHD $=-\sin (\pi /4)-\cos (\pi /4)=-1/\sqrt{2}-1/\sqrt{2}=-\sqrt{2}$.If $x>\pi /4$, $\sin x>\cos x$, so $f(x)=\sin x-\cos x⟹f^{\prime }(x)=\cos x+\sin x$.RHD $=\cos (\pi /4)+\sin (\pi /4)=1/\sqrt{2}+1/\sqrt{2}=\sqrt{2}$.Since LHD $\ne$ RHD, the derivative is not defined at $x=\pi /4$.