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Differentiability of xy function

  1. Oct 6, 2009 #1
    1. The problem statement, all variables and given/known data

    Dear all,

    How can I show that the function f(x,y)=xy is differentiable?

    Thanks

    Dimitris

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 6, 2009 #2

    Dick

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    What's your definition of 'differentiable' for a function of multiple variables?
     
  4. Oct 6, 2009 #3
    I have the definition of differentiability in the pdf file.

    View attachment question.pdf

    Thanks
     
  5. Oct 6, 2009 #4

    Dick

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    It can take several hours for an attachment to be approved. Sure you don't want to explain it in words?
     
  6. Oct 6, 2009 #5
    The definition is in that link http://en.wikipedia.org/wiki/Derivative

    at the 4.4 section

    "The total derivative, the total differential and the Jacobian"

    The first at 4.4 that appears

    Thanks
     
  7. Oct 6, 2009 #6

    Dick

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    Ok, so in that definition a and h are vectors, yes? Label the components. a=(a_x,a_y), h=(h_x,h_y). Now the formula says if you subtract the linear function f'(a)(h) from f(a+h)-f(a) you are supposed to be left with something that goes to 0 faster than h as h goes to 0. It might help to expand f(a+h)-f(a) in the case where f(x,y)=xy. Do you see a linear part in h?
     
  8. Oct 6, 2009 #7

    HallsofIvy

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    Well, what you give in the first response is NOT a definition of "differentiable". It certainly is not what is given in the Wikipedia site. I think you must have copied wrong- you have an "f(x,y)" that doesn't belong in there.

    Essentially, what Wikipedia says is that f(x,y) is differentiable at [itex](x_0,y_0)[/itex] if and only if there exist a vector [itex]u\vec{i}+ v\vec{j}[/itex] and a function [itex]\epsilon(x,y)[/itex] such that for for some [itex]f(x,y)= f(x_0,y_0)+ (u\vec{i}+ v\vec{j})\cdot ((x-x_0)\vec{i}+ (y- y_0)\vec{j})+ \epsilon(x,y)[/itex] and that [itex]\lim_{(x,y)\to (x_0,y_0)} \epsilon(x,y)/\sqrt{(x-x_0)^2+ (y-y_0)^2}= 0[/itex].

    In that case, the vector [itex]u\vec{i}+ v\vec{j}[/itex] is the "gradient" of f at [itex](x_0,y_0)[/itex].

    So I would recommend that you go ahead and find the gradient of f(x,y)= xy at [itex](x_0, y_0)[/itex] and use that to determine what [itex]\epsilon(x,y)[/itex] must be. If you can then prove that [itex]\epsilon(x,y)/\sqrt{(x-x_0)^2+ (y- y_0)^2}[/itex] goes to 0 as (x,y) goes to [itex](x_0,y_0)[/itex], you are done.
     
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