# The derivative

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. (1) $\frac{2 \sqrt{3}}{5}$

2. (2) $\frac{\sqrt{3}}{12}$

3. (3) $\frac{2 \sqrt{3}}{3}$

4. (4) $\frac{\sqrt{3}}{10}$

Correct Option: , 4

Solution:

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

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

$\therefore u=\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right)=\tan ^{-1}\left(\tan \frac{\theta}{2}\right)$

$=\frac{\theta}{2}=\frac{1}{2} \tan ^{-1} x$

$\therefore \frac{d u}{d x}=\frac{1}{2} \times \frac{1}{\left(1+x^{2}\right)}$

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

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

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

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

$\frac{d v}{d x}=2 \frac{1}{\sqrt{1-x^{2}}}$

$\frac{d u}{d v}=\frac{d u / d x}{d v / d x}=\frac{\sqrt{1-x^{2}}}{4\left(1+x^{2}\right)}$

$\therefore\left(\frac{d u}{d v}\right)_{\left(x=\frac{1}{2}\right)}=\frac{\sqrt{3}}{10}$

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