# How do you find the derivative of sqrt(x-3)/sqrt(x+3)?

Apr 2, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6}{2 \sqrt{x - 3} {\left(x + 3\right)}^{\frac{3}{2}}}$

#### Explanation:

Given -

$y = \frac{\sqrt{x - 3}}{\sqrt{x + 3}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left[\sqrt{x + 3} \left(\frac{1}{2 \sqrt{x - 3}}\right)\right] - \left[\sqrt{x - 3} \left(\frac{1}{2 \sqrt{x + 3}}\right)\right]}{\sqrt{x + 3}} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left[\frac{\sqrt{x + 3}}{2 \sqrt{x - 3}}\right] - \left[\frac{\sqrt{x - 3}}{2 \sqrt{x + 3}}\right]}{x + 3}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\sqrt{x + 3}}{2 \sqrt{x - 3}} - \frac{\sqrt{x - 3}}{2 \sqrt{x + 3}}}{x + 3}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\left(\sqrt{x + 3} \sqrt{x + 3}\right) - \left(\sqrt{x - 3} \sqrt{x - 3}\right)}{2 \sqrt{x - 3} \sqrt{x + 3}}}{x + 3}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\left(x + 3\right) - \left(x - 3\right)}{2 \sqrt{x - 3} \sqrt{x + 3}}}{x + 3}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{x + 3 - x + 3}{2 \sqrt{x - 3} \sqrt{x + 3}}}{x + 3}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\cancel{x} + 3 \cancel{- x} + 3}{2 \sqrt{x - 3} \sqrt{x + 3}}}{x + 3}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{6}{2 \sqrt{x - 3} \sqrt{x + 3}}}{x + 3}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left\{\frac{6}{2 \sqrt{x - 3} \sqrt{x + 3}}\right\} \div \left(x + 3\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left\{\frac{6}{2 \sqrt{x - 3} \sqrt{x + 3} \left(x + 3\right)}\right\}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6}{2 \sqrt{x - 3} {\left(x + 3\right)}^{\frac{3}{2}}}$