Results 1 to 3 of 3
Like Tree1Thanks
  • 1 Post By jakncoke

Math Help - differentiable fn of three variables chain rule

  1. #1
    Member
    Joined
    Dec 2012
    From
    -
    Posts
    77

    differentiable fn of three variables chain rule

    "let f be a differentiable function of three variables, and define w(x,y,z) = f(x-y,y-z,z-x). use the chain rule to show dw/dx + dw/dy + dw/dz = 0". i am confused by this question because, in what i am meant to show, there is no f, i.e. it all comes from the w. so i don't know what part of what i am supposed to differentiate to show this. i also don't know what version of the chain rule i am meant to use. since i have notes on the chain rule that pertain to something entirely different. can someone help? thank you
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member jakncoke's Avatar
    Joined
    May 2010
    Posts
    387
    Thanks
    80

    Re: differentiable fn of three variables chain rule

     \frac{\partial{w}}{dx} = \frac{\partial{f}}{d(x-y)}}\frac{\partial(x-y)}{dx}+ \frac{\partial{f}}{d(y-z)}}\frac{\partial(y-z)}{dx} + \frac{\partial{f}}{d(z-x)}}\frac{\partial(z-x)}{dx} = [1, 0, -1]

     \frac{\partial{w}}{dy} = \frac{\partial{f}}{d(x-y)}}\frac{\partial(x-y)}{dy} + \frac{\partial{f}}{d(y-z)}}\frac{\partial(y-z)}{dy} + \frac{\partial{f}}{d(z-x)}}\frac{\partial(z-x)}{dy} = [-1, 1, 0]

     \frac{\partial{w}}{dz} = \frac{\partial{f}}{d(x-y)}}\frac{\partial(x-y)}{dz}  + \frac{\partial{f}}{d(y-z)}}\frac{\partial(y-z)}{dz} + \frac{\partial{f}}{d(z-x)}}\frac{\partial(z-x)}{dz} = [0, -1, 1]

    adding [1,0,-1] + [-1,1,0] + [0,-1,1] = [0,0,0]
    Last edited by jakncoke; February 26th 2013 at 09:02 PM.
    Thanks from learning
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Dec 2012
    From
    -
    Posts
    77

    Re: differentiable fn of three variables chain rule

    riiiiight i get it. cool, thanks a lot. it wasn't really presented like the way we've been studying so i wasn't sure how to approach it, but i get the idea now. i could see how the terms 'would' cancel using partials, quite obviously, but didn't know if that was what was required. nice work, thanks
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Chain Rule for 2 Variables
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: September 10th 2012, 03:49 AM
  2. Use of chain rule for two variables
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 13th 2011, 11:55 AM
  3. Use The Chain Rule To Prove f+g is Differentiable
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: June 19th 2010, 08:21 AM
  4. Chain rule with several variables
    Posted in the Calculus Forum
    Replies: 1
    Last Post: July 28th 2009, 06:31 AM
  5. Chain rule - 2 variables
    Posted in the Calculus Forum
    Replies: 4
    Last Post: May 30th 2009, 07:48 AM

Search Tags


/mathhelpforum @mathhelpforum