Solve this following

Question:

Mark $(\sqrt{)}$ against the correct answer in the following:

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

A. $\frac{1}{(1+\mathrm{x})}$

B. $\frac{1}{\sqrt{x}(1+x)}$

C. $\frac{2}{\sqrt{x}(1+x)}$

D. $\frac{1}{2 \sqrt{x}(1+x)}$

Solution:

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