Find y" by implicit differentiation
x^3 + y^3 = 1
Here's my work:
3x^2 + 3y^2y' = 0
y' = (-3x^2)/(3y^2) = (-x^2)/(y^2)
y" = (-2xy^2 + 2x^2y(dy/dx)) / y^4
= 2(-xy - x^4y^-2) / y^3
The back of the book says that the answer is -2x / y^5. I don't understand how they're simplifying it to that. Thanks for any help.