# How do you find the derivative of y^3 = x^2 -1 at P(2,1)?

Oct 13, 2016

The point $\left(2 , 1\right)$ is not on the curve. However, the derivative at any point is:
dy/dx = 2/3x/(y^2); x ne +-1 because x equal to plus or minus one will cause y to become zero and that is not allowed.

#### Explanation:

Let's check whether the point $\left(2 , 1\right)$ is on the curve by substituting 2 for x in the equation:

${y}^{3} = {2}^{2} - 1$

${y}^{3} = 4 - 1$

${y}^{3} = 3$

$y = \sqrt[3]{3}$

Let's find the derivative at any point:

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

dy/dx = 2/3x/(y^2); x ne +-1