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Determine whether a function with these partial derivatives exist

  1. Dec 14, 2012 #1
    1. The problem statement, all variables and given/known data
    Determine whether a function with partial derivatives [tex]f_x(x,y)=x+4y[/tex] and [tex]f_y(x+y)=3x-y[/tex] exist.


    3. The attempt at a solution

    The method I've seen is to integrate [tex]f_x[/tex] with respect to x, differentiate with respect to y, set it equal to the given [tex]f_y[/tex] and show that it can't be possible.

    So after integrating [tex]f_x[/tex], we get [tex]f(x, y) = \frac{1}{2}x^2+4xy+g(y)[/tex]

    Then differentating that w.r.t y gives

    [tex]f_y(x,y)=4x+g'(y)[/tex]

    So

    [tex]3x-y = 4x+g'(y)[/tex]
    [tex]g'(y)=-x-y[/tex]

    Why would this be a contradiction? Is it because g'(y) can only be in terms of y and not x? Couldn't x be treated as a constant?

    Would integrating both derivatives and showing that they're not equal be a valid method?

    So with respect to x would be [tex]f(x,y)=\frac{1}{2}x^2+4xy+g(y)[/tex] and with respect to y would be [tex]f(x,y)=\frac{3}{2}x^2-\frac{1}{2}y^2+g(x)[/tex]
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 14, 2012 #2

    Dick

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    If your function is sufficiently well behaved, then you ought to have [itex]f_{xy}=f_{yx}[/itex]. See http://en.wikipedia.org/wiki/Symmetry_of_second_derivatives Does that hold? And you don't really mean [itex]f_y(x+y)[/itex], right? And sure, integrating them both and showing the partials are not equal would work, but it might be more than you need to show.
     
    Last edited: Dec 15, 2012
  4. Dec 16, 2012 #3
    Thanks.

    I'm not sure what [tex]f_y(x+y[/tex] is but it's probably not what I mean!

    I'm still not 100% sure why line 9 (tex) is a contradiction..
     
  5. Dec 16, 2012 #4

    Dick

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    I'm guessing line 9 is g'(y)=(-x-y). Sure, that's a contradiction. If x=0 then that's g'(y)=(-y) and if x=1 that's g'(y)=(-1-y). They can't both be true. The equation can't be true for all x and y if you have an x one side and not the other.
     
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