Results 1 to 4 of 4

Thread: Q about using the product rule with implicit differentiation

  1. #1
    Newbie
    Joined
    Mar 2009
    Posts
    5

    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; Mar 19th 2009 at 05:22 PM. Reason: correction
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    76
    Awards
    1

    Smile

    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).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2009
    Posts
    5
    Ohh, I see now. Thanks a bunch!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Mar 2007
    Posts
    1,240

    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!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: Aug 17th 2011, 07:48 AM
  2. Simple Power Rule in Implicit Differentiation
    Posted in the Calculus Forum
    Replies: 7
    Last Post: Jul 5th 2011, 02:28 PM
  3. Replies: 3
    Last Post: Jun 8th 2010, 12:00 AM
  4. Replies: 1
    Last Post: Oct 5th 2009, 08:22 AM
  5. Replies: 5
    Last Post: Oct 14th 2008, 02:47 PM

Search Tags


/mathhelpforum @mathhelpforum