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

Sep 26, 2016

Differentiate both sides of the equation with respect to $x$. Then isolate $\frac{\mathrm{dy}}{\mathrm{dx}}$.

#### Explanation:

Differentiate both sides of the equation with respect to $x$. By doing this, a $\frac{\mathrm{dy}}{\mathrm{dx}}$ should "appear" from the $y$ variable in the equation. Isolating the $\frac{\mathrm{dy}}{\mathrm{dx}}$ would give you the derivative implicitly.

Differentiating both sides with respect to $x$:

${D}_{x} \left[- {x}^{2} {y}^{2} - 3 {y}^{3} + 2\right] = {D}_{x} \left[5 {x}^{3}\right]$

This will give you:

$\left[\left(- 2 x\right) \left({y}^{2}\right) + \left(2 y \cdot \frac{\mathrm{dy}}{\mathrm{dx}}\right) \left(- {x}^{2}\right)\right] - 9 {y}^{2} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} + 0 = 15 {x}^{2}$

$- 2 x {y}^{2} - 2 y {x}^{2} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} - 9 {y}^{2} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 15 {x}^{2}$

Put all terms with $\frac{\mathrm{dy}}{\mathrm{dx}}$ to the left and shove the other terms to the right.

$- 2 y {x}^{2} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} - 9 {y}^{2} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 15 {x}^{2} + 2 x {y}^{2}$

Factor $\frac{\mathrm{dy}}{\mathrm{dx}}$ out.

$\frac{\mathrm{dy}}{\mathrm{dx}} \cdot \left(- 2 y {x}^{2} - 9 {y}^{2}\right) = 15 {x}^{2} + 2 x {y}^{2}$

Divide everything by $\left(- 2 y {x}^{2} - 9 {y}^{2}\right)$. You will get the derivative.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{15 {x}^{2} + 2 x {y}^{2}}{- 2 y {x}^{2} - 9 {y}^{2}}$