$\begin{aligned} & y=\tan ^{-1}(\sec x-\tan x)=\tan ^{-1}\left(\frac{1-\sin x}{\cos x}\right)=\tan ^{-1}\left(\frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}\right) \\ & \Rightarrow y=\tan ^{-1}\left(\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}\right)=\tan ^{-1}\left(\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right)=\frac{\pi}{4}-\frac{x}{2} \\ & \Rightarrow \frac{d y}{d x}=\frac{-1}{2}\end{aligned}$