Differentiate $\tan^{-1}\left(\dfrac{\cos x - \sin x}{\cos x + \sin x}\right)$ with respect to x
$
\begin{aligned}
\text{Let } y & = \tan^{-1} \left(\dfrac{\cos x - \sin x}{\cos x + \sin x}\right) \\
& = \tan^{-1} \left(\dfrac{1-\dfrac{\sin x}{\cos x}}{1 + \dfrac{\sin x}{\cos x}}\right) \\
& = \tan^{-1} \left(\dfrac{1-\tan x}{1 + \tan x}\right)
\end{aligned}
$
Now, $\tan \left(\dfrac{\pi}{4} - x\right) = \dfrac{\tan \dfrac{\pi}{4} - \tan x}{1 + \tan \dfrac{\pi}{4} \tan x} = \dfrac{1- \tan x}{1 + \tan x}$
$\therefore$ $y = \tan^{-1} \left[\tan \left(\dfrac{\pi}{4} - x\right)\right] = \dfrac{\pi}{4} - x$
$\therefore$ $\dfrac{dy}{dx} = -1$