# How do you implicitly differentiate  y^3+x^2-y/x^2-3y^2?

Jun 24, 2018

The answer is $= \frac{2 \left({x}^{4} + y\right)}{x \left(3 {x}^{2} {y}^{2} - 1 - 6 {x}^{2} y\right)}$

#### Explanation:

Let

$f \left(x , y\right) = {y}^{3} + {x}^{2} - \frac{y}{x} ^ 2 - 3 {y}^{2}$

Then the partial derivatives are

$\frac{\partial f}{\partial x} = 2 x + 2 \frac{y}{x} ^ 3 = 2 \left(x + \frac{y}{x} ^ 3\right) = 2 \frac{{x}^{4} + y}{x} ^ 3$

$\frac{\partial f}{\partial y} = 3 {y}^{2} - \frac{1}{x} ^ 2 - 6 y = \frac{3 {x}^{2} {y}^{2} - 1 - 6 {x}^{2} y}{x} ^ 2$

Therefore,

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}$

$= \frac{2 \frac{{x}^{4} + y}{x} ^ 3}{\frac{3 {x}^{2} {y}^{2} - 1 - 6 {x}^{2} y}{x} ^ 2}$

$= \frac{2 \left({x}^{4} + y\right)}{x \left(3 {x}^{2} {y}^{2} - 1 - 6 {x}^{2} y\right)}$