Question

I have the answer can somone please check me ....
Find (dy/dx)
x^3+y^3=3x

Answer 1

Hi

When differentiating a function of y with respect to x, take the derivative with respect to y and multiply it by y'. Also, remember to use the product rule.

x^3 + y^3 = 6xy
d/dx[x^3 + y^3] = d/dx[6xy]
3x^2 + (3y^2)y' = 6[x' * y + y' * x]
3x^2 + (3y^2)y' = 6y + 6xy'

Solve for y' now.

3x^2 + (3y^2)y' = 6y + 6xy'
3x^2 + (3y^2)y' - 6xy' = 6y
(3y^2)y' - 6xy' = 6y - 3x^2
(y')(3x^2 - 6) = 6y - 3x^2
y' = (6y - 3x^2)/(3x^2 - 6)
y' = (2y - x^2)/(x^2 - 2)

http://ca.answers.yahoo.com/question/index?qid=20101008020900AAMdNZM

Answer 2

so then what would the answer be I;m confused cause that is not what I got.