# How do you differentiate 5xy + y^3 = 2x + 3y?

Aug 4, 2015

I found: $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 - 5 y}{5 x + 3 {y}^{2} - 3}$

#### Explanation:

You can use implicit differentiation rememberig that $y$ represents a function of $x$ and needs to be differentiated accordingly;
for example: if you have ${y}^{2}$ you differentiate it to get:
$2 y \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$ where you use $\frac{\mathrm{dy}}{\mathrm{dx}}$ to take into account the dependence with $x$.
$5 y + 5 x \frac{\mathrm{dy}}{\mathrm{dx}} + 3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} = 2 + 3 \frac{\mathrm{dy}}{\mathrm{dx}}$
collect $\frac{\mathrm{dy}}{\mathrm{dx}}$:
$\frac{\mathrm{dy}}{\mathrm{dx}} \left(5 x + 3 {y}^{2} - 3\right) = 2 - 5 y$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 - 5 y}{5 x + 3 {y}^{2} - 3}$