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

$$\begin{gathered} \mathrm{u}={\tan }^{-1}\left(\frac{\sqrt{1+{\mathrm{x}}^2}-1}{\mathrm{x}}\right) \\ \mathrm{x}=\mathrm{tan\theta }\Rightarrow \mathrm{\theta }={\tan }^{-1}\mathrm{x} \\ \mathrm{u}={\tan }^{-1}\left(\frac{1-\mathrm{cos\theta }}{\mathrm{sin\theta }}\right) \\ ={\tan }^{-1}\left(\tan \raisebox{1ex}{$\mathrm{\theta }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right) \\ =\raisebox{1ex}{$\mathrm{\theta }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.=\frac{1}{2}{\tan }^{-1}\mathrm{x} \\ \mathrm{v}={\tan }^{-1}\left(\frac{2\mathrm{x}\sqrt{1-{\mathrm{x}}^2}}{1+2{\mathrm{x}}^2}\right) \\ \mathrm{x}=\sin \mathrm{\alpha } \Rightarrow \mathrm{\alpha } = {\sin }^{-1}\mathrm{x} \\ \mathrm{v}={\tan }^{-1}\left(\frac{\sin 2\mathrm{\alpha }}{\cos 2\mathrm{\alpha }}\right)=2\mathrm{\alpha } \end{gathered}$$

$$\begin{gathered} \mathrm{v}=2{\sin }^{-1}\mathrm{x} \\ \frac{\mathrm{du}}{\mathrm{dx}}=\frac{1}{1+{\mathrm{x}}^2}·\frac{1}{2} \\ \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{2}{\sqrt{1-{\mathrm{x}}^2}} \end{gathered}$$

derivative of u w.r.t v

$$\begin{gathered} \frac{\mathrm{du}}{\mathrm{dv}}=\frac{{\displaystyle \frac{1}{2(1+{\mathrm{x}}^2)}}}{{\displaystyle \frac{2}{\sqrt{1-{\mathrm{x}}^2}}}}=\frac{\sqrt{1-{\mathrm{x}}^2}}{4(1+{\mathrm{x}}^2)} \\ {\left(\frac{\mathrm{du}}{\mathrm{dv}}\right)}_{\mathrm{x}=0}=\frac{1}{4} \end{gathered}$$