# How do you find dy/dx by implicit differentiation given xy^2-3x^2y+x=1?

Sep 17, 2017

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

#### Explanation:

We need

$\left(u v\right) ' = u ' v + u v '$

The function is

$x {y}^{2} - 3 {x}^{2} y + x - 1 = 0$

We differentiate with respect to x

$\frac{d}{\mathrm{dx}} \left(x {y}^{2}\right) - \frac{d}{\mathrm{dx}} \left(3 {x}^{2} y\right) + \frac{d}{\mathrm{dx}} \left(x\right) - \frac{d}{\mathrm{dx}} \left(1\right) = 0$

${y}^{2} + 2 y x \frac{\mathrm{dy}}{\mathrm{dx}} - 6 x y - 3 {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} + 1 = 0$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(2 y x - 3 {x}^{2}\right) = 6 x y - {y}^{2} - 1$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(6 x y - {y}^{2} - 1\right)}{\left(2 y x - 3 {x}^{2}\right)}$