If $y=\tan^{-1}\left(\dfrac{\sqrt{a+\sqrt{x}}}{1-\sqrt{ax}}\right)$ then $\dfrac{dy}{dx}=?$

$\begin{array}{l} y=\tan ^{-1}\left(\frac{\sqrt{a}+\sqrt{x}}{1-\sqrt{a x}}\right) \\ y=\tan ^{-1}(\sqrt{a})+\tan ^{-1}(\sqrt{x}) \\ \frac{d y}{d x}=0+\frac{d}{d x}\left(\tan ^{-1} \sqrt{x}\right) \\ =\frac{1}{2 \sqrt{x}(1+x)} \end{array}$

Diff both sides with respect to x,

$\begin{array}{l} \frac{d y}{d x}=\frac{d\left\{\tan ^{-1}(\sqrt{a})\right\}}{d(\sqrt{a})} \times \frac{d(\sqrt{a})}{d x}+ \\ \frac{d\left\{\tan ^{-1}(\sqrt{x})\right\}}{d(\sqrt{x})} \times \frac{d(\sqrt{x})}{d x} \\ \frac{d y}{d x}=0+\frac{1}{2 \sqrt{x}(1+x)} \\ \frac{d y}{d x}=\frac{1}{2 \sqrt{x}(1+x)} \end{array}$