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I am trying to find a function defined on a given domain on which it has \begin{align*} \begin{cases} \frac{\partial f}{\partial x} = \frac{1}{y} \\ \\ \frac{\partial f}{\partial y} = \frac{1}{x} \end{cases} \end{align*}

I tried many combinations but I always get the problem when I derive back and find out that my partial derivatives are not the right ones

Aleksandrov Topogorov
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  • Just integrate. – GReyes Nov 14 '23 at 19:57
  • I tried integrating but it's not sufficient since when I derive back it won't give me the partials that I need – Aleksandrov Topogorov Nov 14 '23 at 19:58
  • Find anti-derivatives one at a time, but keep in mind that the constant of integration will be a function of the OTHER variable. Just take care when you go on to the next variable's derivative. – Hypatia163 Nov 14 '23 at 20:19
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    Hint: A condition for $f$ to exist is $\frac{\partial^2f}{\partial x\partial y}=\frac{\partial^2f}{\partial y\partial x}$. – Gonçalo Nov 14 '23 at 20:20
  • @Etale7 In fact that is a great idea yet the problem persists whereas trying to integrate one at the time and make sure he constant depend only on one variable it doesn't work since when you take the respective derivatives after you finish all the work you find out that the latter do not coordinate with what you originally wanted ie: $ \frac{\partial f}{\partial x} = \frac{1}{y}
    \frac{\partial f}{\partial y} = \frac{1}{x}$     – Aleksandrov Topogorov Nov 14 '23 at 20:34
  • @Gonçalo do you suggest that f is a C2 function as a sufficient condition or a necessary one for it to exist since we can consider functions that are only C1 in other words I do not need anymore conditions of rigularity more then C1 and even if we wanted more that would be impossible because of the theorem of mixed partials ( our function clearly misses the test of mixed partials hence it cannot be C2 ) – Aleksandrov Topogorov Nov 14 '23 at 20:37
  • Does this answer your question? https://math.stackexchange.com/questions/3683647/conditions-for-a-vector-field-to-be-a-gradient – Gonçalo Nov 14 '23 at 20:43

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