How do you find the derivative of y^3 = x^2 -1y3=x21 at P(2,1)?

1 Answer
Oct 13, 2016

The point (2,1)(2,1) is not on the curve. However, the derivative at any point is:
dy/dx = 2/3x/(y^2); x ne +-1dydx=23xy2;x±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)(2,1) is on the curve by substituting 2 for x in the equation:

y^3 = 2^2 - 1y3=221

y^3 = 4 - 1y3=41

y^3 = 3y3=3

y = root(3)3y=33

Let's find the derivative at any point:

3y^2(dy/dx) = 2x3y2(dydx)=2x

dy/dx = 2/3x/(y^2); x ne +-1dydx=23xy2;x±1