Question

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

Answer 1

The point `(2,1)` 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 `(2, 1)` 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 = root(3)3`

Let's find the derivative at any point:

`3y^2(dy/dx) = 2x`

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