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"Can there exist a $C^2$ function $f(x,y)$ with $f_x = 2x-5y$ and $f_y=4x+y$"?

Given this question, am I simply to take the second derivative of these functions to prove the equivalence of the mixed partial derivative?

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    $\begingroup$ What can you say about $f_{xy}$ and $f_{yx}$? $\endgroup$
    – Steve Kass
    Oct 13, 2018 at 0:28
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    $\begingroup$ Well, certainly if the two second mixed partials are not the same it would be a sign of a problem... $\endgroup$ Oct 13, 2018 at 0:28
  • $\begingroup$ lol, thanks for the hint Paul. $\endgroup$
    – Jaigus
    Oct 13, 2018 at 0:29
  • $\begingroup$ :) ..... ..... .. $\endgroup$ Oct 13, 2018 at 0:49
  • $\begingroup$ It is true, though, to be more proper, that there is a non-trivial condition for the two second derivatives to be equal pointwise. Called Clairault's theorem. But, distributionally, by taking Fourier transform, we see that the two mixed partials of any tempered distribution are equal, because Fourier transform turns "partial derivative with respect to $x$/$y$" into "multiplication by (some constant times) $x$ or $y$, and these obviously commute... But that is a different context... $\endgroup$ Oct 13, 2018 at 0:52

2 Answers 2

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With $f_x = 2x-5y$ and $f_y= 4x+y$, there is no $C^{2}$ function for which these relations hold.

Simply check, $f_{xy} = -5$ and $f_{yx}=4$ which are different.

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Given that in the mixed partial derivative the order in which you take the derivatives is not important, you could try do compute $f_{xy}=\frac{f_x}{\partial y}$ and $f_{yx}=\frac{f_y}{\partial y}$ and compare them. If they are indeed equal then $f_x$ and $f_y$ may come from the same primitive.

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  • $\begingroup$ It took me too long, :-) $\endgroup$
    – Patricio
    Oct 13, 2018 at 0:41

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