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I Are the functions for mixed derivative always equal?

  1. Nov 5, 2016 #1
    Hi all, I understand that the mixed partial derivative at some point may not be equal if the such mixed partial derivative is not continuous at the point, but are the actual functions of mixed partial derivatives always equal? In other words, if I simply compute the mixed partial derivatives without using the limit definition or in the point, do I get the same functions which are discontinuous only at the critical point?
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  3. Nov 5, 2016 #2

    Simon Bridge

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    How would you go about computing any derivative without using the limit definition?
    The limit definition is what a derivative is...

    Can you express your question as a mathematical statement.
    ie. ##\partial_{xy}f(x,y) \neq \partial_{yx}f(x)## says the mixed partials of function f are not the same if the order of differentiation is reversed.
    Is this what you mean by "the mixed partial derivative at some point may not be equal"?

    ... a function of mixed partials would be like ##f(g_1,g_2):\; g_1=\partial_{xy}g(x,y),\; g_2=\partial_{yx}g(x,y)##, is that what you mean?

    I am not clear on the question here:
    Are you asking: if ##g(x,y)## has discontinuity point p, do ##g_1## and ##g_2## also have discontinuities at point p?
  4. Nov 5, 2016 #3


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    Last edited: Nov 5, 2016
  5. Nov 5, 2016 #4
    Sorry about the confusion. What I want to ask is that, for example, is it possible to have a function f(x,y) such that fxy = x while fyx = y ?
  6. Nov 6, 2016 #5

    Simon Bridge

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    See how much clearer it is when you use maths as a language?
    Take a look at the references in post #3 and see if you can find one for that specific situation.
  7. Nov 6, 2016 #6
    Yes I have looked at both pages but both pages only talk about whether mixed second derivatives are equal at a specific point. I want to know whether fxy and fyx will always look the same in their function forms...
  8. Nov 6, 2016 #7


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    Well, that would require them to be equal at every point, so ...
  9. Nov 6, 2016 #8


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    That means that ##f_x## and ##f_y## are continuous, hence they must be equal. So, that is impossible.
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