\(y = tan^{-1}\left(\frac{\sqrt{1+x^2}+1}{x}\right)
\)
Let \(x = tan \theta \)
⇒ \(\theta = tan^{-1}x\)
\(\therefore y= tan^{-1}\left(\frac{\sqrt{1+tan^2\theta}+1}{tan\theta}\right)\)
\(= tan^{-1}\left(\frac{sec\theta + 1}{tan\theta}\right)\)
\(= tan^{-1}\left(\frac{1+cos\theta}{sin\theta}\right)\)
\(= tan^{-1} \left(\frac{2cos^2\frac{\theta}{2}}{2sin\frac{\theta}2cos\frac{\theta}{2}}\right)\)
\(= tan^{-1}\left(cot\frac{\theta}{2}\right)\)
\(= tan^{-1}\left(tan\frac{\pi}{2}- \frac{\theta}{2}\right)\)
⇒ \(y = \frac{\pi}{2}- \frac{\theta}{2}\)
\(\therefore \frac{dy}{dx} = \frac{-1}{2} \frac{d\theta}{dx}\)
\(= \frac{-1}{2} \frac{d}{dx} tan^{-1}x\) \((\because \theta = tan^{-1}x)\)
\(= \frac{-1}{2}\frac{1 }{1+x^2}\)
\(\therefore \frac{d}{dx} tan^{-1} (\frac{\sqrt{1 +x^2} + 1}{x})= \frac{-1}{2(1 + x^2)}\)
