What is the derivative of ^ -1 ( 1+x^2 -1 x ) with respect to ^ -1...

Let $y=\tan ^{-1} \left[\frac{\sqrt{1+x^2}-1}{x}\right] and$u=\tan ^{-1} x$Put$x=\tan $\theta \Rightarrow \theta=\tan ^{-1}$x$Then,$y=\tan ^{-1}\left[\frac{\sqrt{1+\tan ^2 \theta}-1}{\tan \theta}\right]=\tan ^{-1}\left[\frac{\sqrt{\sec ^2 \theta}-1}{\tan \theta}\right]$$\begin{aligned}& =\tan ^{-1}\left[\frac{\sec \theta-1}{\tan \theta}\right]=\tan ^{-1} \left[\frac{\frac{1}{\cos \theta}-1}{\frac{\sin \theta}{\cos \theta}}\right]\\ & =\tan ^{-1}\left[\frac{1-\cos \theta}{\sin \theta}\right]=\tan ^{-1} \left[\frac{2 \sin ^2 \frac{\theta}{2}}{2 \sin \frac{\theta}{2}, \cos \frac{\theta}{2}}\right]\\ & \binom{\because 1-\cos \theta=2 \sin ^2 \frac{\theta}{2} \text{ and }}{\sin x=2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}} \end{aligned}$$=\tan ^{-1}\left[\tan \frac{\theta}{2}\right]\Rightarrow \mathrm{y}=\frac{\theta}{2} \Rightarrow \mathrm{y}=\frac{\tan ^{-1} \mathrm{x}}{2} \quad\left[\because \theta=\tan ^{-1} \mathrm{x}\right]$$\begin{aligned} \Rightarrow \quad \mathrm{y}& =\frac{\mathrm{u}}{2}\\ \frac{\mathrm{dy}}{\mathrm{du}}& =\frac{1}{2} \end{aligned}$$\therefore \quad$Option (B) is correct.