Implicit Differentiation Problem

[James2]

1. Given that y^{2}-2xy+x^{3}=0, find \frac{dy}{dx}



2. (no relevant equations other than the problem statement)



3. So, I solved it like this,
\frac{dy}{dx}y^{2}-2xy+x^{3}=0

2y\frac{dy}{dx}-2+3x^{2}=0

Solving for dy/dx I got...

\frac{dy}{dx}=\frac{-3x^{2}+2}{2y}

However, wolfram alpha on my phone when I checked it said that the derivative of -2xy was -2y...? What I am asking is, if there are two variables in one term, the derivative is different?

 

[bossman27]

I think your problem here was with the middle term:

[D(2xy)] = [2xD(y) + yD(2x)] = [2xy' + 2y]

This gave me a solution of:

y' = \frac{2y-3x^{2}}{2y-2x}

 

[James2]

Yeah, I figured that out shortly after posting. I asked a friend and he said that the problem was that I didn't use the product rule on -2xy. We got the same solution once we worked it out, thanks! Is there a way to mark this thread as solved or something?