# How do you find the slope of the curve y^3-xy^2=4 at the point where y=2?

Jul 2, 2018

The slope of the curve is $\frac{1}{2}$

#### Explanation:

We need to start by finding the x-coordinate.

${2}^{3} - x {\left(2\right)}^{2} = 4$

$- 4 x = - 4$

$x = 1$

Now we differentiate.

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

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

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

The slope is simply the value of $\frac{\mathrm{dy}}{\mathrm{dx}}$ at $\left(1 , 2\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{6 - 2} = \frac{2}{4} = \frac{1}{2}$

Hopefully this helps!