Question

The derivative of $ta{n}^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ with respect to $ta{n}^{-1}\left(\frac{2x\sqrt{1-x^2}}{1-2x^2}\right)$ at x = 0, is

• A. $\frac{1}{8}$
• B. $\frac{1}{4}$
• C. $\frac{1}{2}$
• D. 1

Answer 1

Answer: (B)

$\frac{1}{4}$

Let $t={\tan }^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$

Put $x=\tan \theta$

$\Rightarrow t={\tan }^{-1}\left(\frac{\sqrt{1+{\tan }^2\theta }-1}{\tan \theta }\right)\Rightarrow t={\tan }^{-1}\left(\frac{\sec \theta -1}{\tan \theta }\right)\Rightarrow t={\tan }^{-1}\left(\frac{1-\cos \theta }{sin\theta }\right)$

$\Rightarrow t={\tan }^{-1}\left(\frac{1-\frac{1-{\tan }^2\frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}}{\frac{2\tan \frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}}\right)$

$\Rightarrow t={\tan }^{-1}\left(\tan \frac{\theta }{2}\right)=\frac{\theta }{2}\Rightarrow t=\frac{{\tan }^{-1}x}{2}$ .....(i)

$u={\tan }^{-1}\left(\frac{2x\sqrt{1-x^2}}{1-2x^2}\right)$

Put $x=\sin \theta$

$u={\tan }^{-1}\left(\frac{2\sin \theta \sqrt{1-{\sin }^2\theta }}{1-2{\sin }^2\theta }\right)u={\tan }^{-1}\left(\frac{2\sin \theta \cos \theta }{\cos 2\theta }\right)u={\tan }^{-1}\left(\tan 2\theta \right)u=2\theta \Rightarrow u=2{\tan }^{-1}x\frac{du}{dx}=\frac{2}{(1+x^2)}$ .......(ii)

Using $\left(i\right)$ and $\left(ii\right)$

$\frac{dt}{du}=\frac{\frac{1}{2(1+x^2)}}{\frac{2}{(1+x^2)}}=\frac{1}{4}$

So option B is correct.