The derivative of

Question:

The derivative of $\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)$ with

respect to $\tan ^{-1}\left(\frac{2 x \sqrt{1-x^{2}}}{1-2 x^{2}}\right)$ at $x=\frac{1}{2}$ is :

1. $\frac{\sqrt{3}}{12}$

2. $\frac{\sqrt{3}}{10}$

3. $\frac{2 \sqrt{3}}{5}$

4. $\frac{2 \sqrt{3}}{3}$

Correct Option: , 2

Solution:

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

Put $x=\tan \theta \Rightarrow \theta=\tan ^{-1} x$

$f=\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right)$

$f=\tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right)=\frac{\theta}{2}$

$f=\frac{\tan ^{-1} x}{2} \Rightarrow \frac{d f}{d x}=\frac{1}{2\left(1+x^{2}\right)} \ldots$ (i)

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

Put $x=\sin \theta \Rightarrow \theta=\sin ^{-1} x$

$g=\tan ^{-1}\left(\frac{2 \sin \theta \cos \theta}{1-2 \sin ^{2} \theta}\right)$

$\mathrm{g}=\tan ^{-1}(\tan 2 \theta)=2 \theta$

$g=2 \sin ^{-1} x$

$\frac{\mathrm{dg}}{\mathrm{dx}}=\frac{2}{\sqrt{1-\mathrm{x}^{2}}}$....$\ldots$ (ii)

$\frac{d f}{d g}=\frac{1}{2\left(1+x^{2}\right)} \frac{\sqrt{1-x^{2}}}{2}$

at $x=\frac{1}{2}\left(\frac{d f}{d g}\right)_{x=\frac{1}{2}}=\frac{\sqrt{3}}{10}$