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

Mar 7, 2016

You can do it like this:

#### Explanation:

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

Differentiate both sides implicitly:

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

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

$\therefore 4 y . y ' - 3 {x}^{2.} y ' - 6 x y + 1 = 4 y '$

$\therefore 4 y . y ' - 4 y ' - 3 {x}^{2.} y ' = 6 x y - 1$

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

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