The derivative of $\cos ^{-1}\left(\frac{\sin x+\cos x}{\sqrt{2}}\right), -\frac{\pi}{4}<x<\frac{\pi}{4}$ with respect to x is :

-1

$\cos ^{-1}\left[\frac{1}{\sqrt{2}} \sin x+\frac{1}{\sqrt{2}} \cos x\right]$

$\cos ^{-1}\left[\cos \frac{\pi}{4} \cos x+\sin \frac{\pi}{4} \sin x\right]$

(as $cos(\frac{\pi}{4}) = \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$)

$y = \cos ^{-1}\left(\cos (-x + \frac{\pi}{4})\right]$   as  [cos(x - y) = cos x cos y + sin x + sin y]

$y=\frac{\pi}{4}-x$

as  $-\frac{\pi}{4}<x<\pi / 4$

so  $-\frac{\pi}{4}<-x<+\frac{\pi}{4}$

so  $0 < \frac{\pi}{4}-x < \frac{\pi}{2}$

domain lies in (0, π)

so y' = -1