# What is the implicit derivative of 1= 3y-ysqrt(x-y) ?

Mar 26, 2018

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

#### Explanation:

You differentiate both sides of the expression

$1 = 3 y - y \sqrt{x - y}$

with respect to $x$ to get

$0 = 3 \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{\mathrm{dy}}{\mathrm{dx}} \sqrt{x - y} - y \frac{1}{2 \sqrt{x - y}} \left(1 - \frac{\mathrm{dy}}{\mathrm{dx}}\right)$
$\quad = \left(3 - \sqrt{x - y} + \frac{y}{2 \sqrt{x - y}}\right) \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{y}{2 \sqrt{x - y}} \implies$

$\left(6 \sqrt{x - y} - 2 \left(x - y\right) + y\right) \frac{\mathrm{dy}}{\mathrm{dx}} - y = 0 \implies$

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

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