# How do you find the derivative of y^2-3xy+x^2=7?

##### 1 Answer
Dec 14, 2017

Differentiate each term.
When a term is a function of y, treat $\frac{\mathrm{dy}}{\mathrm{dx}}$ as a dependent variable.
After differentiating, solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$.

#### Explanation:

Given:

${y}^{2} - 3 x y + {x}^{2} = 7$

Differentiate each term:

$\frac{d \left({y}^{2}\right)}{\mathrm{dx}} - \frac{d \left(3 x y\right)}{\mathrm{dx}} + \frac{d \left({x}^{2}\right)}{\mathrm{dx}} = \frac{d \left(7\right)}{\mathrm{dx}}$

The derivative of the constant term is 0:

$\frac{d \left({y}^{2}\right)}{\mathrm{dx}} - \frac{d \left(3 x y\right)}{\mathrm{dx}} + \frac{d \left({x}^{2}\right)}{\mathrm{dx}} = 0$

For the first term, I shall use the chain rule, $\frac{d \left(f \left(y\right)\right)}{\mathrm{dx}} = \frac{d \left(f \left(y\right)\right)}{\mathrm{dy}} \frac{\mathrm{dy}}{\mathrm{dx}}$:

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

Returning to the equation:

$2 y \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{d \left(3 x y\right)}{\mathrm{dx}} + \frac{d \left({x}^{2}\right)}{\mathrm{dx}} = 0$

For the second term, I shall use the linear property of the derivative and the product rule:

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

Returning to the equation:

$2 y \frac{\mathrm{dy}}{\mathrm{dx}} - 3 y - 3 x \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{d \left({x}^{2}\right)}{\mathrm{dx}} = 0$

For the third term, I shall use the power rule, $\frac{d \left({x}^{n}\right)}{\mathrm{dx}} = n {x}^{n - 1}$:

$2 y \frac{\mathrm{dy}}{\mathrm{dx}} - 3 y - 3 x \frac{\mathrm{dy}}{\mathrm{dx}} + 2 x = 0$

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

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

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

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