If \(u=\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)\) and \(v=\tan ^{-1}\left(\frac{2 x \sqrt{1-x^{2}}}{1-2 x^{2}}\right)\), then \(\frac{d u}{d v}\) at \(x=0\) is

Given \(u=\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)\)
Put \(x=\tan \theta\)
\(u=\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right) \quad=\tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right)=\tan ^{-1}\) \(\left(\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos }\right)\)
\(\therefore u=\frac{\tan ^{-1} x}{2} \Rightarrow \frac{d u}{d x}=\frac{1}{2\left(1+x^{2}\right)}\)
We have, \(v=\tan ^{-1}\left(\frac{2 x \sqrt{1-x^{2}}}{1-2 x^{2}}\right)\)
Put \(x=\sin \theta\)
\(\therefore v=\tan ^{-1}\left(\frac{2 \sin \theta \cdot \cos \theta}{\cos 2 \theta}\right)=\tan ^{-1}\left(\frac{\sin 2 \theta}{\cos 2 \theta}\right)=\tan ^{-1}\) \((\tan 2 \theta)=2 \theta\)
\(\therefore v=2 \tan ^{-1} x \Rightarrow \frac{d v}{d x}=\frac{2}{1+x^{2}}\)
\(\left.\frac{d u}{d v}=\frac{d u}{d x}\right)=\frac{1}{2\left(1+x^{2}\right)} \times \frac{\left(1+x^{2}\right)}{2}=\frac{1}{4}\)