# How do you find (dy)/(dx) given 2xy+y^2=x+y?

Jun 11, 2018

$\setminus \frac{1 - 2 y}{2 x + 2 y - 1}$

#### Explanation:

$2 x y + {y}^{2} = x + y$
$\setminus \frac{d}{\mathrm{dx}} \left(2 x y + {y}^{2}\right) = \setminus \frac{d}{\mathrm{dx}} \left(x + y\right)$
$2 \setminus \frac{d}{\mathrm{dx}} \left(x y\right) + 2 y \setminus \frac{d}{\mathrm{dx}} = 1 + \setminus \frac{d}{\mathrm{dx}}$

Use product rule and finish it off to get:

$2 \left(x \setminus \frac{d}{\mathrm{dx}} + y\right) + 2 y \setminus \frac{d}{\mathrm{dx}} = 1 + \setminus \frac{d}{\mathrm{dx}}$

to make it clearer, let $\setminus \frac{d}{\mathrm{dx}} = y '$

so you get:
$2 \left(x y ' + y\right) + 2 y y ' = 1 + y '$

Rearrange to make $y '$ the factor to get:

$y ' = \setminus \frac{1 - 2 y}{2 x + 2 y - 1}$

bring back $y ' = \setminus \frac{d}{\mathrm{dx}}$

so you get:
$\setminus \frac{d}{\mathrm{dx}} : \setminus \frac{1 - 2 y}{2 x + 2 y - 1}$

Jun 11, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - 2 y}{2 x + 2 y - 1}$

#### Explanation:

$\textcolor{b l u e}{\text{differentiate implicitly with respect to x}}$

$\text{noting that "d/dx(y)=dy/dx" and } \frac{d}{\mathrm{dx}} \left({y}^{2}\right) = 2 y \frac{\mathrm{dy}}{\mathrm{dx}}$

$\text{differentiate "xy" using the "color(blue)"product rule}$

$2 x \frac{\mathrm{dy}}{\mathrm{dx}} + 2 y + 2 y \frac{\mathrm{dy}}{\mathrm{dx}} = 1 + \frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(2 x + 2 y - 1\right) = 1 - 2 y$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - 2 y}{2 x + 2 y - 1}$