How do you implicitly differentiate -1=-y^3x-2xy+3x^2y ?

1 Answer
Feb 17, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{y}^{3} + 2 y - 6 x y}{6 {x}^{2} - 3 x {y}^{2} - 2 x}$

Explanation:

Differentiate both sides of the equation with respect to $x$:

$0 = - 3 x {y}^{2} y ' - {y}^{3} - 2 y - 2 x y ' + 6 x y + 6 {x}^{2} y '$

$0 = y ' \left(- 3 x {y}^{2} - 2 x + 6 {x}^{2}\right) - {y}^{3} - 2 y + 6 x y$

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