Differentiate $\tan^{-1}\left[\dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right]$ w.r.t x
Let $y=\tan^{-1}\left[\dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right]$
$
\begin{aligned}
\text{Now, } \dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} & = \dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} \times \dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}} \\
& \\
& = \dfrac{\left(\sqrt{1+x} - \sqrt{1-x}\right)^2}{\left(\sqrt{1+x}\right)^2 - \left(\sqrt{1-x}\right)^2} \\
& \\
& = \dfrac{1+x+1-x-2\sqrt{1-x^2}}{1+x-\left(1-x\right)} \\
& \\
& = \dfrac{2-2\sqrt{1-x^2}}{1+x-1+x} \\
& \\
& = \dfrac{2-2\sqrt{1-x^2}}{2x} \\
& \\
& = \dfrac{1-\sqrt{1-x^2}}{x}
\end{aligned}
$
Let $x = \sin \theta$
Then, $\sqrt{1-x^2} = \sqrt{1-\sin^2 \theta} = \cos \theta$
$
\begin{aligned}
\therefore \dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} & = \dfrac{1-\sqrt{1-x^2}}{x} \\
& \\
& = \dfrac{1-\cos \theta}{\sin \theta} \\
& \\
& = \dfrac{2 \sin^2 \left(\dfrac{\theta}{2}\right)}{2 \sin \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{\theta}{2}\right)} \\
& \\
& = \dfrac{\sin \left(\dfrac{\theta}{2}\right)}{\cos \left(\dfrac{\theta}{2}\right)} = \tan \left(\dfrac{\theta}{2}\right)
\end{aligned}
$
$\therefore$ $y=\tan^{-1}\left[\dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right] = \tan^{-1}\left[\tan\left(\dfrac{\theta}{2}\right)\right] = \dfrac{\theta}{2}$
$\therefore$ $\dfrac{dy}{dx} = \dfrac{1}{2} \dfrac{d\theta}{dx}$
Now, $x = \sin \theta$ $\implies$ $\theta = \sin^{-1}x$
$\therefore$ $\dfrac{d\theta}{dx} = \dfrac{1}{\sqrt{1-x^2}}$
$\therefore$ $\dfrac{dy}{dx}= \dfrac{1}{2}\dfrac{d\theta}{dx} = \dfrac{1}{2\sqrt{1-x^2}}$