The derivative of  Sin^-1$\left(\frac{{\mathrm{x}}^2}{\sqrt{{\mathrm{x}}^4+{\mathrm{a}}^4}}\right)$ at x=1 is:

$$\begin{gathered} \mathrm{y}={\sin }^{-1}\frac{{\mathrm{x}}^2}{\sqrt{{\mathrm{x}}^4+{\mathrm{a}}^4}} \\ {\mathrm{x}}^2={\mathrm{a}}^2 \mathrm{tan\theta } \Rightarrow \mathrm{\theta }={\tan }^{-1}\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2} \\ \mathrm{y}={\sin }^{-1}\left(\frac{{\mathrm{a}}^2\mathrm{tan\theta }}{{\mathrm{a}}^2\mathrm{sec\theta }}\right) \\ ={\sin }^{-1}\left(\sin \mathrm{\theta }\right) \\ =\mathrm{\theta } \\ \mathrm{y}={\tan }^{-1}\left(\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}\right) \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{1+{\left({\displaystyle \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}}\right)}^2}.\frac{2\mathrm{x}}{{\mathrm{a}}^2} \\ =\frac{2{\mathrm{a}}^2\mathrm{x}}{{\mathrm{a}}^4+{\mathrm{x}}^4} \\ {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\mathrm{x}=1}=\frac{2{\mathrm{a}}^2}{{\mathrm{a}}^4+1} \end{gathered}$$