# How do you find dy/dx by implicit differentiation of x^3-3x^2y+2xy^2=12?

Jan 10, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6 x y - 3 {x}^{2} - 2 {y}^{2}}{4 x y - 3 {x}^{2}}$

#### Explanation:

To differentiate $- 3 {x}^{2} y \text{ and } 2 x {y}^{2}$ we use the $\textcolor{b l u e}{\text{product rule}}$

$\Rightarrow 3 {x}^{2} - \left(3 {x}^{2.} \frac{\mathrm{dy}}{\mathrm{dx}} + 6 x y\right) + \left(2 x .2 y \frac{\mathrm{dy}}{\mathrm{dx}} + 2 {y}^{2}\right) = 0$

$\Rightarrow 3 {x}^{2} - 3 {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} - 6 x y + 4 x y \frac{\mathrm{dy}}{\mathrm{dx}} + 2 {y}^{2} = 0$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} \left(4 x y - 3 {x}^{2}\right) = 6 x y - 3 {x}^{2} - 2 {y}^{2}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6 x y - 3 {x}^{2} - 2 {y}^{2}}{4 x y - 3 {x}^{2}}$