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:
