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

1 Answer
Oct 13, 2016

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#