X^2 - 2xy + y^3 = c (implicit differentiation)?

What am I doing wrong? (please help me understand)





x^2 - 2xy + y^3 = c %26lt;---(c is a constant)


find the derivative of both sides of equation:


x - (2xy' + 2y) + 3y^2y' = 0


x - 2xy' - 2y + 3y^2y' = 0


solve for y':


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

X^2 - 2xy + y^3 = c (implicit differentiation)?
first step:


d(x^2)/dx=2x


therefore


2x-(2xy'+2y)+3y^2y'=0


y'(3y^2-2x)=2y-2x


y'=(2y-2x)/(3y^2-2x)
Reply:2x-2(y+xy')+3y^2*y'=0


(3y^2-2x)y'=2y-2x so y'= (2y-2x)/(3y^2-2x)


The derivative ofx^2 =2x
Reply:x^2 - 2xy + y^3 = c





Differentiating above implicitly on both sides with respect to x, we get:





2x - 2(x*y' + y) + 3y^2*y' = 0...................(1)





[I think you are missing that the derivative of x^2 is 2x and NOT x]





Expanding (1) and solving for y':





2x - 2xy' - 2y + 3y^2*y'=0 or





y'(-2x + 3y^2) = - 2x + 2y or





y' = 2( y - x)/(3y^2 - 2x) = 2(x - y)/(2x - 3y^2)

peony