Question

\[\frac{\sqrt{a} + \sqrt{x}}{\sqrt{a} - \sqrt{x}}\] 

Answer 1

\[\text{ Let } u = \sqrt{a} + \sqrt{x}; v = \sqrt{a} - \sqrt{x}\]
\[\text{ Then }, u' = \frac{1}{2\sqrt{x}}; v' = \frac{- 1}{2\sqrt{x}}\]
\[\text{ Using thequotient rule }:\]
\[\frac{d}{dx}\left( \frac{u}{v} \right) = \frac{vu' - uv'}{v^2}\]
\[\frac{d}{dx}\left( \frac{\sqrt{a} + \sqrt{x}}{\sqrt{a} - \sqrt{x}} \right) = \frac{\left( \sqrt{a} - \sqrt{x} \right)\frac{1}{2\sqrt{x}} - \left( \sqrt{a} + \sqrt{x} \right)\left( \frac{- 1}{2\sqrt{x}} \right)}{\left( \sqrt{a} - \sqrt{x} \right)^2}\]
\[ = \frac{\sqrt{a} - \sqrt{x} + \sqrt{a} + \sqrt{x}}{2\sqrt{x} \left( \sqrt{a} - \sqrt{x} \right)^2}\]
\[ = \frac{2\sqrt{a}}{2\sqrt{x} \left( \sqrt{a} - \sqrt{x} \right)^2}\]
\[ = \frac{\sqrt{a}}{\sqrt{x} \left( \sqrt{a} - \sqrt{x} \right)^2}\]