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Suff. condition for exact differential - PROOF?

  1. Jul 18, 2012 #1
    ...is that second partial derivatives are equal - how to prove it?
     
  2. jcsd
  3. Jul 18, 2012 #2
    If [itex]M(x,y) dx + N(x,y) dy[/itex] is exact, that means it equals [itex]dF[/itex] for some function [itex]F(x,y)[/itex]. But the chain rule says that [itex]dF=\frac{\partial F}{\partial x}dx+ \frac{\partial F}{\partial y}dy[/itex], which implies that [itex]M(x,y) = \frac{\partial F}{\partial x}[/itex] and [itex]N(x,y) = \frac{\partial F}{\partial y}[/itex]. Now Clairaut's theorem states that (for well-behaved functions) the mixed 2nd partial derivatives are equal, i.e. [itex]\frac{\partial^{2}F}{\partial x \partial y}=\frac{\partial^{2}F}{\partial y \partial x}[/itex], which in terms of M and N becomes [itex]\frac{\partial N}{\partial x}=\frac{\partial M}{\partial y}[/itex]. That's the condition for exact differentials.

    I hope that helps.
     
  4. Jul 21, 2012 #3
    Ok, then the question is how to probe the Clairaut's theorem.
     
  5. Jul 21, 2012 #4

    HallsofIvy

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  6. Jul 21, 2012 #5
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