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

##### 2 Answers
Oct 21, 2016

Given a choice, I would solve for $y$ first.

#### Explanation:

$y = {x}^{2} + x y$

Making the function explicit
$y - x y = {x}^{2}$

$y = {x}^{2} / \left(1 - x\right)$

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

$= \frac{2 x - {x}^{2}}{1 - x} ^ 2$

Leaving the function implicit

$y = {x}^{2} + x y$

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

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

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

Oct 21, 2016

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

#### Explanation:

You have to use Implicit Differentiation, which is a special case of the chain rule, and the product rule.

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

Simplify

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

Solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$

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

Factor out $\frac{\mathrm{dy}}{\mathrm{dx}}$

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

Isolate $\frac{\mathrm{dy}}{\mathrm{dx}}$

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

Simplify

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

Here is an example of using the implicit differentiation.