# What is the implicit derivative of y=(x-y)^2+4xy+5y^2 ?

May 29, 2017

The implicit derivative is $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x + 2 y}{- 12 y - 2 x + 1}$.

#### Explanation:

First, you would derive this by turning $y$ into $\frac{\mathrm{dy}}{\mathrm{dx}}$ when needed since it is a function while treating $x$ as the variable. For the ${\left(x - y\right)}^{2}$, the chain rule is applied. This will result in:

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 \left(x - y\right) \cdot \left(1 - \frac{\mathrm{dy}}{\mathrm{dx}}\right) + 4 y + 4 x \frac{\mathrm{dy}}{\mathrm{dx}} + 10 y \frac{\mathrm{dy}}{\mathrm{dx}}$

For simplicity, we can distribute the values in the parentheses and then implement quadratic multiplication.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(2 x - 2 y\right) \cdot \left(1 - \frac{\mathrm{dy}}{\mathrm{dx}}\right) + 4 y + 4 x \frac{\mathrm{dy}}{\mathrm{dx}} + 10 y \frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x - 2 x \frac{\mathrm{dy}}{\mathrm{dx}} - 2 y + 2 y \frac{\mathrm{dy}}{\mathrm{dx}} + 4 y + 4 x \frac{\mathrm{dy}}{\mathrm{dx}} + 10 y \frac{\mathrm{dy}}{\mathrm{dx}}$

We now simplify like terms to get:

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x + 12 y \frac{\mathrm{dy}}{\mathrm{dx}} + 2 y + 2 x \frac{\mathrm{dy}}{\mathrm{dx}}$

Move over all of the terms with $\frac{\mathrm{dy}}{\mathrm{dx}}$ to the other side:

$- 12 y \frac{\mathrm{dy}}{\mathrm{dx}} - 2 x \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x + 2 y$

Factor the $\frac{\mathrm{dy}}{\mathrm{dx}}$ from the left side to get:

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(- 12 y - 2 x + 1\right) = 2 x + 2 y$

Divide the left and right sides by $\left(- 12 y - 2 x + 1\right)$ to achieve:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x + 2 y}{- 12 y - 2 x + 1}$