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Implicit Differentiation Problem

  1. Oct 21, 2012 #1
    1. Given that [itex]y^{2}-2xy+x^{3}=0[/itex], find [itex]\frac{dy}{dx}[/itex]



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



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


    [itex]2y\frac{dy}{dx}-2+3x^{2}=0[/itex]

    Solving for dy/dx I got...

    [itex]\frac{dy}{dx}=\frac{-3x^{2}+2}{2y}[/itex]

    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?
     
  2. jcsd
  3. Oct 21, 2012 #2
    I think your problem here was with the middle term:

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

    This gave me a solution of:

    [itex] y' = \frac{2y-3x^{2}}{2y-2x} [/itex]
     
  4. Oct 21, 2012 #3
    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?
     
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