Question

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

• A. $\frac{\sqrt{3}}{12}$
• B. $\frac{\sqrt{3}}{10}$
• C. $\frac{2\sqrt{3}}{5}$
• D. $\frac{2\sqrt{3}}{3}$

Answer 1

Answer: (B)

Let $f={\tan }^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$
Put $x=\tan \theta$
$\Rightarrow \theta ={\tan }^{-1}x$
$f={\tan }^{-1}\left(\frac{\sec \theta -1}{\tan \theta }\right)$
$f={\tan }^{-1}\left(\frac{1-\cos \theta }{\sin \theta }\right)=\frac{\theta }{2}$
$f=\frac{{\tan }^{-1}x}{2}$
$\Rightarrow \frac{df}{dx}=\frac{1}{2\left(1+x^2\right)}\dots$(i)
Let $g={\tan }^{-1}\left(\frac{2x\sqrt{1-x^2}}{1-2x^2}\right)$
Put $x=\sin \theta$
$\Rightarrow \theta ={\sin }^{-1}x$
$g={\tan }^{-1}\left(\frac{2\sin \theta \cos \theta }{1-2{\sin }^2\theta }\right)$
$g={\tan }^{-1}(\tan 2\theta )=2\theta$
$g=2{\sin }^{-1}x$
$\frac{dg}{dx}=\frac{2}{\sqrt{1-x^2}}\dots$(ii)
$\frac{df}{dg}=\frac{1}{2\left(1+x^2\right)}\frac{\sqrt{1-x^2}}{2}$
at $x=\frac{1}{2}{\left(\frac{df}{dg}\right)}_{x=\frac{1}{2}}=\frac{\sqrt{3}}{10}$