# How do you implicitly differentiate -y= 4x^3y^2+2x^2y^3-2xy^4 ?

Oct 6, 2016

$y ' = \frac{- 12 {x}^{2} {y}^{3} - 4 x {y}^{3} + 2 {y}^{4}}{1 + 8 {x}^{3} y + 6 {x}^{2} {y}^{2} - 8 x {y}^{3}}$

#### Explanation:

Implicit Differentiation is a special case of the chain rule. When you differentiate the y variable in a term or factor you differentiating with respect to x. Because of this you have include the factor of $\frac{\mathrm{dy}}{\mathrm{dx}}$ or $y '$.

We will have to use the product rule, power rule and chain rule.

Divide all the terms by -1 to remove the negative

$y = - 4 {x}^{3} {y}^{2} - 2 {x}^{2} {y}^{3} + 2 x {y}^{4}$

Differentiate Implicitly

$y ' = - 4 {x}^{3} 2 y y ' - 12 {x}^{2} {y}^{3} - \left(2 {x}^{2} 3 {y}^{2} y ' + 4 x {y}^{3}\right) + 2 x 4 {y}^{3} y ' + 2 {y}^{4}$

Distribute the negative

$y ' = - 4 {x}^{3} 2 y y ' - 12 {x}^{2} {y}^{3} - 2 {x}^{2} 3 {y}^{2} y ' - 4 x {y}^{3} + 2 x 4 {y}^{3} y ' + 2 {y}^{4}$

Gather the terms with the $y '$ factors

$y ' + 4 {x}^{3} 2 y y ' + 2 {x}^{2} 3 {y}^{2} y ' - 2 x 4 {y}^{3} y ' = - 12 {x}^{2} {y}^{3} - 4 x {y}^{3} + 2 {y}^{4}$

Factor out $y '$

$y ' \left(1 + 4 {x}^{3} 2 y + 2 {x}^{2} 3 {y}^{2} - 2 x 4 {y}^{3}\right) = - 12 {x}^{2} {y}^{3} - 4 x {y}^{3} + 2 {y}^{4}$

Divide to isolate $y '$

$\frac{y ' \cancel{\left(1 + 4 {x}^{3} 2 y + 2 {x}^{2} 3 {y}^{2} - 2 x 4 {y}^{3}\right)}}{\cancel{\left(1 + 4 {x}^{3} 2 y + 2 {x}^{2} 3 {y}^{2} - 2 x 4 {y}^{3}\right)}} = \frac{- 12 {x}^{2} {y}^{3} - 4 x {y}^{3} + 2 {y}^{4}}{1 + 4 {x}^{3} 2 y + 2 {x}^{2} 3 {y}^{2} - 2 x 4 {y}^{3}}$

$y ' = \frac{- 12 {x}^{2} {y}^{3} - 4 x {y}^{3} + 2 {y}^{4}}{1 + 4 {x}^{3} 2 y + 2 {x}^{2} 3 {y}^{2} - 2 x 4 {y}^{3}}$

Multiple numeric factors to simplify

$y ' = \frac{- 12 {x}^{2} {y}^{3} - 4 x {y}^{3} + 2 {y}^{4}}{1 + 8 {x}^{3} y + 6 {x}^{2} {y}^{2} - 8 x {y}^{3}}$

Below are a couple of tutorials include implicit differentiation