# √ a + √ X √ a − √ X - Mathematics

$\frac{\sqrt{a} + \sqrt{x}}{\sqrt{a} - \sqrt{x}}$

#### Solution

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

Concept: The Concept of Derivative - Algebra of Derivative of Functions
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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 30 Derivatives
Exercise 30.5 | Q 15 | Page 44