Suff. condition for exact differential - PROOF?

[Trave11er]

...is that second partial derivatives are equal - how to prove it?

 

[lugita15]

If $M(x,y) dx + N(x,y) dy$ is exact, that means it equals $dF$ for some function $F(x,y)$. But the chain rule says that $dF=\frac{\partial F}{\partial x}dx+ \frac{\partial F}{\partial y}dy$, which implies that $M(x,y) = \frac{\partial F}{\partial x}$ and $N(x,y) = \frac{\partial F}{\partial y}$. Now Clairaut's theorem states that (for well-behaved functions) the mixed 2nd partial derivatives are equal, i.e. $\frac{\partial^{2}F}{\partial x \partial y}=\frac{\partial^{2}F}{\partial y \partial x}$, which in terms of M and N becomes $\frac{\partial N}{\partial x}=\frac{\partial M}{\partial y}$. That's the condition for exact differentials.

I hope that helps.

 

[Trave11er]

Ok, then the question is how to probe the Clairaut's theorem.

 

[HallsofIvy]

I would suggest you start by looking it up. It should be given in any Mathematical Analysis text and in any good Calculus text. Or google it on the internet.

There are several theorems with that name you want:
http://en.wikipedia.org/wiki/Symmetry_of_second_derivatives
or
http://www.sju.edu/~pklingsb/clairaut.pdf

 

[lugita15]

Here are two more links explaining the proof:

http://homepage.smc.edu/nestler_andrew/math11/m11clairaut.pdf

http://calculus.subwiki.org/wiki/Clairaut's_theorem_on_equality_of_mixed_partials#Proof