The derivate of $y=\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ is equal to

Question

Aicharya · Accepted Answer

Given, $y=\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ Let, $x=\tan \theta$ So, $y=\tan ^{-1}\left(\frac{\sqrt{1+\tan ^2 \theta}-1}{\tan \theta}\right)$ $\begin{aligned} & =\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right)=\tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right) \ & =\tan ^{-1}(\tan \theta / 2)=\theta / 2 \ \Rightarrow y & =\frac{1}{2} \tan ^{-1} x \Rightarrow \frac{d y}{d x}=\frac{1}{2}\left(\frac{1}{1+x^2}\right) \end{aligned}$ Hence, option (b) is correct.

The derivate of \(y=\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\) is equal to

Solution