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

Apr 11, 2016

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

#### Explanation:

$2 x y y ' + {y}^{2} - 2 {x}^{3} y ' - 6 {x}^{2} y = 1 + y '$

$2 x y y ' - 2 {x}^{3} y ' - y ' = 1 - {y}^{2} + 6 {x}^{2} y$

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

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