# How do you use implicit differentiation to find (dy)/(dx) given 3x^2y^2=4x^2-4xy?

Aug 15, 2017

$\frac{\mathrm{dx}}{\mathrm{dy}} = \frac{8 x - 4 y - 6 x {y}^{2}}{6 {x}^{2} y + 4 x}$

#### Explanation:

Given -

$3 {x}^{2} {y}^{2} = 4 {x}^{2} - 4 x y$
$6 x {y}^{2} + 6 {x}^{2} y . \frac{\mathrm{dx}}{\mathrm{dy}} = 8 x - 4 y + \left(- 4 x . \frac{\mathrm{dy}}{\mathrm{dx}}\right)$
$6 x {y}^{2} + 6 {x}^{2} y . \frac{\mathrm{dx}}{\mathrm{dy}} = 8 x - 4 y - 4 x . \frac{\mathrm{dy}}{\mathrm{dx}}$

$6 x {y}^{2} + 6 {x}^{2} y . \frac{\mathrm{dx}}{\mathrm{dy}} + 4 x . \frac{\mathrm{dy}}{\mathrm{dx}} = 8 x - 4 y$
$6 {x}^{2} y . \frac{\mathrm{dx}}{\mathrm{dy}} + 4 x . \frac{\mathrm{dy}}{\mathrm{dx}} = 8 x - 4 y - 6 x {y}^{2}$
$\left(6 {x}^{2} y + 4 x\right) \frac{\mathrm{dx}}{\mathrm{dy}} = 8 x - 4 y - 6 x {y}^{2}$
$\frac{\mathrm{dx}}{\mathrm{dy}} = \frac{8 x - 4 y - 6 x {y}^{2}}{6 {x}^{2} y + 4 x}$

Aug 15, 2017

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

#### Explanation:

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

Product and power rule:
$3 \left(2 x\right) \left({y}^{2}\right) + 3 \left({x}^{2}\right) \left(2 y\right) \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 8 x - 4 \left(y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right)$

$6 x {y}^{2} + 6 {x}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} = 8 x - 4 y - 4 x \frac{\mathrm{dy}}{\mathrm{dx}}$

Move all terms that include $\frac{\mathrm{dy}}{\mathrm{dx}}$ to one side:
$4 x \frac{\mathrm{dy}}{\mathrm{dx}} + 6 {x}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} = 8 x - 4 y - 6 x {y}^{2}$

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

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