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Math Help - Q about using the product rule with implicit differentiation

  1. #1
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    Q about using the product rule with implicit differentiation

    [edit] ignore the bit about the product rule in the title [edit]


    Hi everyone, this is my first time posting here I hope you guys can help me.
    I am trying to get the derivative of x^2 - xy + y^2 = 3

    After the first step I got:
    2x - y - x(dy/dx) + 2y(dy/dx) = 0

    When I get to the step where I would move all like terms to one side, I could either move all non(dy/dx) terms to the right hand side and have the equation
    -x(dy/dx) + 2y(dy/dx) = y - 2x

    OR if I were to move the (dy/dx) terms to the right hand side I would get:
    2x - y = x(dy/dx) - 2y(dy/dx)

    The first route will give me the derivative (y - 2x)/(2y - x)
    The second route will give me derivative (2x - y)/(x - 2y)

    It seems like I attacked both possible routes correctly, but I'm pretty sure one of the final derivatives is incorrect because (y - 2x)/(2y - x) does not equal (2x - y)/(x - 2y).... or do they?

    feedback would be much appreciated!!
    Last edited by robo_robb; March 19th 2009 at 05:22 PM. Reason: correction
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  2. #2
    MHF Contributor Reckoner's Avatar
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    Your work is correct.

    Quote Originally Posted by robo_robb View Post
    ...but I'm pretty sure one of the final derivatives is incorrect because (y - 2x)/(2y - x) does not equal (2x - y)/(x - 2y).... or do they?
    They are equal. Factor a -1 out of the numerator and denominator:

    \frac{y-2x}{2y-x}

    =\frac{-(-y+2x)}{-(-2y+x)}

    =\frac{-y+2x}{-2y+x}

    =\frac{2x-y}{x-2y}.

    It is helpful to remember that, in general, -(a-b)=(b-a).
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  3. #3
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    Ohh, I see now. Thanks a bunch!
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    Talking

    Quote Originally Posted by Reckoner View Post
    It is helpful to remember that, in general, -(a-b)=(b-a).
    And it is also helpful to remember that, if this fact was necessary in the homework, it will almost certainly be necessary on the next test!
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