Question

If x^3+y^3=26, find the value of d^2y/dx^2 at the point (-1,3).

The value at d^2y/dx^2 at the point (−1,3) is

Answer 1

x^3+y^3=26
Take derivative both sides
3x^2+3y^2dy/dx = 0
dy/dx =-x^2/y^2
d^2y/dx^2 =(y^2(2x)-2xdy/dx)/y^4
=(2xy^2-2x(-x^2/y^2))/y^4
=(2xy^2+2x^4/y^2)/y^4
Substitute x=-1 and y=3

Source: http://www.mathskey.com/question2answer/

Answer 2

Thanks.