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Math Help - Implicit Differentiation

  1. #1
    Member FalconPUNCH!'s Avatar
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    Implicit Differentiation

    I don't really understand how to do this but I tried to do a problem. Here's my work.

    The problem is  2x^3 + x^2 y - x y^3 = 2

    sorry I put up the work to another problem here's the real work:

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

    I moved everything except y' to one side and I got this:

     y' = \frac {-2x^3 - 2xy + y^3}{x^2 - 1}

    I have a feeling that I messed up somewhere but I don't know where since I do not fully understand Implicity Differentiation.
    Last edited by FalconPUNCH!; October 23rd 2007 at 06:49 PM. Reason: I just need them to check my work
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by FalconPUNCH! View Post
    The problem is  2x^3 + x^2 y - x y^3 = 2
    6x^2+(x^2y)'-(xy^3)'=0

    Product Rule.
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  3. #3
    Member FalconPUNCH!'s Avatar
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    Quote Originally Posted by Krizalid View Post
    6x^2+(x^2y)'-(xy^3)'=0

    Product Rule.
    Not sure if it's right

     y' = \frac {-6x^2 - 2xy + y^3}{x^2 - xy}
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  4. #4
    Member FalconPUNCH!'s Avatar
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    sorry I was looking at another problem's work. I put up the work I did for this problem.
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  5. #5
    Member FalconPUNCH!'s Avatar
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    Here's what I got:

     y' = \frac {-2x^3 - 2xy + y^3}{x^2 - 1}
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