Differentiation of $y = \tan^{-1} \Bigl\{ \frac{\sqrt{1+x^{2}} - \sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+ \sqrt{1-x^{2}}}\Bigr\}$ How do i differentiate the following: $$y = \tan^{-1} \biggl\{ \frac{\sqrt{1+x^{2}} - \sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+ \sqrt{1-x^{2}}}\biggr\}$$
I know that $\text{derivative}$ of $\tan^{-1}{x}$ is $\frac{1}{1+x^{2}}$ but not sure as to how to do this.

Here are all the steps in excruciating detail.
Using the tangent of a sum formula gives
$$
\frac{1-\tan{\phi}}{1+\tan{\phi}}=\tan\left(\frac\pi4-\phi\right)\tag{1}
$$
and letting $\tan(\phi)=\sqrt{\frac{1-x^2}{1+x^2}}$ yields
$$
\begin{align}
\tan^{-1}\left(\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right)
&=\tan^{-1}\left(\frac{1-\sqrt{\frac{1-x^2}{1+x^2}}}{1+\sqrt{\frac{1-x^2}{1+x^2}}}\right)\\
&=\frac\pi4-\tan^{-1}\left(\sqrt{\frac{1-x^2}{1+x^2}}\right)\tag{2}
\end{align}
$$
Furthermore, we have
$$
\begin{align}
\sqrt{\frac{1-\cos(\psi)}{1+\cos(\psi)}}
&=\frac{\sin(\psi/2)}{\cos(\psi/2)}\\
&=\tan(\psi/2)\tag{3}
\end{align}
$$
so letting $x^2=\cos(\psi)$ yields
$$
\begin{align}
\tan^{-1}\left(\sqrt{\frac{1-x^2}{1-x^2}}\right)
&=\psi/2\\
&=\frac12\cos^{-1}(x^2)\tag{4}
\end{align}
$$
Combining $(2)$ and $(4)$ shows
$$
\begin{align}
\tan^{-1}\left(\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right)
&=\frac\pi4-\frac12\cos^{-1}(x^2)\\
&=\frac12\sin^{-1}(x^2)\tag{5}
\end{align}
$$
Therefore,
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}\tan^{-1}\left(\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right)
&=\frac{\mathrm{d}}{\mathrm{d}x}\frac12\sin^{-1}(x^2)\\
&=\frac{x}{\sqrt{1-x^4}}\tag{6}
\end{align}
$$
