Differentiate the following functions with respect to x :

$y=\tan ^{-1}\left(\frac{\sqrt{x}+\sqrt{a}}{1-\sqrt{x a}}\right)$

Using, $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$

$y=\tan ^{-1} \sqrt{x}+\tan ^{-1} \sqrt{a}$

Differentiating w.r.t $x$ we get

$\frac{d y}{d x}=\frac{d}{d x}\left(\tan ^{-1} \sqrt{x}+\tan ^{-1} \sqrt{a}\right)$

$\frac{d y}{d x}=\frac{1}{1+(\sqrt{x})^{2}} \frac{d}{d x}(\sqrt{x})$

$\frac{d y}{d x}=\frac{1}{2 \sqrt{x}\left(1+x^{2}\right)}$