# How do you find (dy)/(dx) given -x^3y^2+4=5x^2+3y^3?

Nov 22, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{10 x + 3 {x}^{2} {y}^{2}}{2 y {x}^{3} + 9 {y}^{2}}$

#### Explanation:

$- {x}^{3} {y}^{2} + 4 = 5 {x}^{2} + 3 {y}^{3}$
$\text{ }$
$\Rightarrow - 3 {x}^{2} {y}^{2} + 2 y \frac{\mathrm{dy}}{\mathrm{dx}} \times \left(- {x}^{3}\right) = 10 x + 9 {y}^{2} \times \frac{\mathrm{dy}}{\mathrm{dx}}$
$\text{ }$
$\Rightarrow - 3 {x}^{2} {y}^{2} - 2 y {x}^{3} \frac{\mathrm{dy}}{\mathrm{dx}} = 10 x + 9 {y}^{2} \times \frac{\mathrm{dy}}{\mathrm{dx}}$
$\text{ }$
$\Rightarrow - 2 y {x}^{3} \frac{\mathrm{dy}}{\mathrm{dx}} - 9 {y}^{2} \times \frac{\mathrm{dy}}{\mathrm{dx}} = 10 x + 3 {x}^{2} {y}^{2}$
$\text{ }$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} \left(- 2 y {x}^{3} - 9 {y}^{2}\right) = 10 x + 3 {x}^{2} {y}^{2}$
$\text{ }$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{10 x + 3 {x}^{2} {y}^{2}}{- 2 y {x}^{3} - 9 {y}^{2}}$
$\text{ }$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{10 x + 3 {x}^{2} {y}^{2}}{2 y {x}^{3} + 9 {y}^{2}}$