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How to prove differentiable everywhere?

  1. Sep 30, 2011 #1
    1. The problem statement, all variables and given/known data
    See photo, part b and c


    2. Relevant equations



    3. The attempt at a solution
    For part b
    It seems it is trival, in part a we have proved that [itex]f_{x}[/itex] and [itex]f_{y}[/itex] exist. Obviously, they are differentiable for x and y[itex]\neq[/itex]0

    For part c.
    It seems there are 2 method to do it.
    1. Use first principle.(i.e. take limit)
    2.Find [itex]f_{xy}[/itex] and [itex]f_{yx}[/itex]
    If they equal each other, then f is differentiable at(0,0)
     

    Attached Files:

  2. jcsd
  3. Sep 30, 2011 #2

    HallsofIvy

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    What limit are you talking about? "Differentiability" of a function of two variables is NOT defined by a limit. You might want to look up the definition of "differentiable" for functions of more than one variable.

    No, that's not true. There is a theorem that says "If a function is differentiable, on a region, then its mixed second derivatives are equal on that region", but the converse of that statement is not true.
     
  4. Sep 30, 2011 #3
    How about part b?
    You only talk about part c
     
  5. Sep 30, 2011 #4

    HallsofIvy

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    Yes, (b) is trivial- f is a quotient of two polynomials and the denominator is not 0 for any point other than (0, 0).
     
  6. Oct 2, 2011 #5
    But I saw from wiki that it is.
    If not, how do I prove differentiable?

    cb2be0dc4607423c38120e364c9d4a65.png
     
  7. Oct 2, 2011 #6
    I also wonder if [itex]f_{x}[/itex]=[itex]\frac{(2^xy^3-6x^2y^2-2x^4)}{(y^4+2x^2y^2+x^4)}[/itex] exist at (0,0).

    If not, it seems the question part a has some problems.
     
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