The derivative of \(\tan ^{-1}\left[\frac{\sin x}{1+\cos x}\right]\) with respect to \(\tan ^{-1}\left[\frac{\cos x}{1+\sin x}\right]\) is

Let
\[
u=\tan ^{-1}\left(\frac{\sin x}{1+\cos x}\right), v=\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)
\]
\(\begin{aligned} u &=\tan ^{-1}\left(\frac{2 \sin x / 2 \cdot \cos x / 2}{1+2 \cos ^{2} x / 2-1}\right) \\ &=\tan ^{-1}(\tan x / 2)=x / 2 \end{aligned}\)
\(v=\tan ^{-1}\left(\frac{\cos x}{\cos ^{2} \frac{x}{2}+\sin ^{2} \frac{x}{2}+2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}}\right)\)
\(=\tan ^{-1}\left(\frac{\left(\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}\right)}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^{2}}\right)\)
\(=\tan ^{-1}\left(\frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}\right)=\tan ^{-1}\left(\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}\right)\)
\(\begin{aligned} u &=\tan ^{-1}\left(\frac{2 \sin x / 2 \cdot \cos x / 2}{1+2 \cos ^{2} x / 2-1}\right) \\ &=\tan ^{-1}(\tan x / 2)=x / 2 \end{aligned}\)
\(v=\tan ^{-1}\left(\frac{\cos x}{\cos ^{2} \frac{x}{2}+\sin ^{2} \frac{x}{2}+2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}}\right)\)
\(=\tan ^{-1}\left(\frac{\left(\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}\right)}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^{2}}\right)\)
\(=\tan ^{-1}\left(\frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}\right)=\tan ^{-1}\left(\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}\right)\)
\(=\left[\tan ^{-1}\left(\tan ^{-1}\left(\frac{\pi}{4}-\frac{\pi}{2}\right)\right]=\left(\frac{\pi}{4}-\frac{x}{2}\right)\right.\)
Now, \(\quad \frac{d u}{d x}=\frac{1}{2}\) and \(\frac{d v}{d x}=\frac{1}{2}\)
\[
\Rightarrow \quad \frac{d u}{d v}=\frac{d u}{d x} \times \frac{d x}{d v}=\left(\frac{1}{2}\right) \times(-2)=-1
\]