Question

The differentiation of ${\tan }^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)w.r.t.{\tan }^{-1}x$ is

• A. $1$
• B. $\frac{1}{2}$
• C. $\frac{3}{4}$
• D. $-\frac{1}{2}$

Answer 1

The correct option is B $\frac{1}{2}$

$\mathrm{Let}u={\tan }^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\mathrm{and}v={\tan }^{-1}x\mathrm{Putting}x=\tan \theta ,\theta \in \left(-\frac{\pi }{2},\frac{\pi }{2}\right),\mathrm{we}\mathrm{get}u={\tan }^{-1}\left(\frac{\sqrt{1+{\tan }^2\theta }-1}{\tan \theta }\right)={\tan }^{-1}\left(\frac{\sec \theta -1}{\tan \theta }\right)\Rightarrow u={\tan }^{-1}\left(\frac{1-\cos \theta }{\sin \theta }\right)={\tan }^{-1}\left(\frac{2{\sin }^2\frac{\theta }{2}}{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}\right)\Rightarrow u={\tan }^{-1}\left(\tan \frac{\theta }{2}\right)=\frac{\theta }{2}=\frac{1}{2}{\tan }^{-1}x\mathrm{Thus},\mathrm{we}\mathrm{have}u=\frac{1}{2}{\tan }^{-1}x\mathrm{and}v={\tan }^{-1}x.\therefore \frac{\mathrm{du}}{\mathrm{dx}}=\frac{1}{2}\times \frac{1}{1+x^2}\mathrm{and}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{1}{1+x^2}\therefore \frac{\mathrm{du}}{\mathrm{dv}}=\frac{\frac{\mathrm{du}}{\mathrm{dx}}}{\frac{\mathrm{dv}}{\mathrm{dx}}}=\frac{1}{2}$