# Suff. condition for exact differential - PROOF?

1. Jul 18, 2012

### Trave11er

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

2. Jul 18, 2012

### 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.

3. Jul 21, 2012

### Trave11er

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

4. Jul 21, 2012

### HallsofIvy

5. Jul 21, 2012