The derivative of ${\mathrm{Tan}}^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\text{ w.r.t }{\mathrm{Tan}}^{-1}\left(\frac{2x\sqrt{1-x^2}}{1-2x^2}\right)$ at x = 0 is

$\text{ Let }y={\tan }^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\text{ and }z={\tan }^{-1}\left(\frac{2x\sqrt{1-x^2}}{1-2x^2}\right)$

$\text{ Putting }x=\tan \theta \text{ in }y\text{, we get }$

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

$\therefore \frac{dy}{dx}=\frac{1}{2\left(1+x^2\right)}$

$\text{ Putting }x=\sin \theta \text{ in }z\text{, we get }$

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

$\therefore \frac{dz}{dx}=\frac{2}{\sqrt{1-x^2}}$

$\text{ Thus, }\frac{dy}{dz}=\frac{\frac{dy}{dx}}{\frac{dz}{dx}}=\frac{1}{4\left(1+x^2\right)}\sqrt{1-x^2}\text{ or }{\left(\frac{dy}{dz}\right)}_{x=0}=\frac{1}{4}$