Results 1 to 5 of 5
Like Tree1Thanks
  • 1 Post By romsek

Math Help - Mysterous negative sign

  1. #1
    Member
    Joined
    May 2009
    Posts
    109

    Mysterious negative sign

    why is the differential of \frac{x}{p} - \frac{n-x}{1-p} = \frac{x}{p^2} - \frac{n-x}{(1-p)^2} and not \frac{x}{p^2} + \frac{n-x}{(1-p)^2}
    Last edited by alyosha2; January 31st 2014 at 05:20 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    2,484
    Thanks
    961

    Re: Mysterous negative sign

    Quote Originally Posted by alyosha2 View Post
    why is the differential of \frac{x}{p} - \frac{n-x}{1-p} = \frac{x}{p^2} - \frac{n-x}{(1-p)^2} and not \frac{x}{p^2} + \frac{n-x}{(1-p)^2}
    I don't think you have this correct. I assume that p is the variable you are differentiating with respect to.

    \begin{align*}&\frac{d}{dp} \left(\frac{x}{p}-\frac{n-x}{1-p}\right) =\\ \\&-\frac{x}{p^2} - \left(-\frac{(n-x)}{(1-p)^2}(-1)\right) = \\ \\ &-\frac{x}{p^2} - \frac{n-x}{(1-p)^2}\end{align}

    in the 2nd term the first minus sign comes from \frac{d}{dp}\frac{1}{p}=-\frac{1}{p^2} the second comes from the -p that's part of the denominator.
    Thanks from alyosha2
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2009
    Posts
    109

    Re: Mysterous negative sign

    OK, got it. Thanks.
    Last edited by alyosha2; January 31st 2014 at 05:59 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    2,484
    Thanks
    961

    Re: Mysterous negative sign

    Quote Originally Posted by alyosha2 View Post
    But why doesn't \frac{d}{dp}(\frac{x}{p}-\frac{n-x}{1-p}) = -\frac{x}{p^2}-(\frac{(n-x)}{(1-p)^2}(-1))? where does that extra minus sign come from in front of the second term?
    that's the form of the derivative.

    \frac{d}{dp}\left(\frac{n-x}{1+p}\right)=-\frac{n-x}{(1+p)^2}

    when you change p to (-p) you have another factor of -1 that has to be included

    \frac{d}{dp}\left(\frac{n-x}{1- p}\right)=-\frac{n-x}{(1-p)^2}(-1) =\frac{n-x}{(1-p)^2}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    May 2009
    Posts
    109

    Re: Mysterous negative sign

    Jut realised I didn't actually understand the answer given.

    It's just a really basic use of the chain rule of differentiation. Chain rule - Wikipedia, the free encyclopedia

    The second term is being treated as a composite function u = (1 - p) and so the -1 is just the differential of (1 - p).

    Disgustingly simple.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: October 23rd 2012, 08:05 PM
  2. Replies: 6
    Last Post: February 11th 2010, 07:08 PM
  3. is the negative sign part of x?
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: November 21st 2009, 06:25 AM
  4. How to force unary minus (negative sign)?
    Posted in the LaTeX Help Forum
    Replies: 7
    Last Post: July 30th 2009, 05:04 AM
  5. Negative Sign & Degree Symbol?
    Posted in the LaTeX Help Forum
    Replies: 3
    Last Post: June 12th 2009, 08:08 PM

Search Tags


/mathhelpforum @mathhelpforum