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

$\begin{array}{l}\mathrm{Let} \mathrm{y}={\tan }^{-1}\left(\frac{\sqrt{1+{\mathrm{x}}^2}-1}{\mathrm{x}}\right) \mathrm{and} \mathrm{z}={\tan }^{-1}\left(\frac{2\mathrm{x}\sqrt{1-{\mathrm{x}}^2}}{1-2{\mathrm{x}}^2}\right)\\ \mathrm{Putting} \mathrm{x}=\mathrm{tan\theta } \mathrm{in} \mathrm{y}, \mathrm{we} \mathrm{get}\\ \mathrm{y}={\tan }^{-1}\left(\frac{\mathrm{sec\theta }-1}{\mathrm{tan\theta }}\right)={\tan }^{-1}\left(\tan \frac{\mathrm{\theta }}{2}\right)=\frac{1}{2}{\tan }^{-1}\mathrm{x}\\ \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2\left(1+{\mathrm{x}}^2\right)}\\ \mathrm{Putting} \mathrm{x}=\mathrm{sin\theta } \mathrm{in} \mathrm{z}, \mathrm{we} \mathrm{get}\\ \mathrm{z}={\tan }^{-1}\left(\frac{2\mathrm{sin\theta cos\theta }}{\cos 2\mathrm{\theta }}\right)={\tan }^{-1}\left(\tan 2\mathrm{\theta }\right)=2\mathrm{\theta }=2{\sin }^{-1}\mathrm{x}\\ \therefore \frac{\mathrm{dz}}{\mathrm{dx}}=\frac{2}{\sqrt{1-{\mathrm{x}}^2}}\\ \mathrm{Thus}, \frac{\mathrm{dy}}{\mathrm{dz}}=\frac{\frac{\mathrm{dy}}{\mathrm{dx}}}{\frac{\mathrm{dz}}{\mathrm{dx}}}=\frac{1}{4\left(1+{\mathrm{x}}^2\right)} \mathrm{or} {\left(\frac{\mathrm{dy}}{\mathrm{dz}}\right)}_{\mathrm{x}=0}=\frac{1}{4}\end{array}$