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