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Is the following criterion:

$$ \frac{\partial^2 f}{\partial x\partial y} = \frac{\partial^2 f}{\partial y\partial x} $$

Equivalent to:

$$ \frac{\partial^2 \ln f}{\partial x\partial y} = \frac{\partial^2 \ln f}{\partial y\partial x} $$

Because $\ln$ function is a bijection?

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  • $\begingroup$ Your criterion assesses closedness of $df$, not exactness. It is necessary but not sufficient. $\endgroup$ – Mark Fantini Jul 15 '14 at 21:04
  • $\begingroup$ I don't understand what "assessing exactness of $df$" means here. The form $df$ is exact because it is $df$; there is nothing else to say about this. $\endgroup$ – user147263 Jul 16 '14 at 4:56
  • $\begingroup$ I have edited the question. $\endgroup$ – jlandercy Jul 16 '14 at 12:31

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