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Please forgive me for asking such a fundamental question.

I know for an exact differential $f(x,y)=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy$, we have $$\frac{\partial^2f}{\partial x\partial y}=\frac{\partial^2 f}{\partial y\partial x}.$$

Now if a differential $U(x,y)dx+V(x,y)dy$ is exact, then we certainly have $$\frac{\partial U}{\partial y}=\frac{\partial V}{\partial x}. \tag{1}$$

My question is, how to use Eq.(1) to prove that the differential $U(x,y)dx+V(x,y)dy$ is exact?

Wein Eld
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  • Hint: if ${\partial f\over\partial x}=U(x,y)$, then what is ${\partial^2f\over\partial y\partial x}$? – amd Jun 08 '17 at 21:51
  • I know Eq(1) is a necessary condition. What I want is to show that Eq(1) is also a sufficient condition. – Wein Eld Jun 08 '17 at 22:17
  • In general, it isn’t. You also need conditions on the domain. – amd Jun 08 '17 at 22:54

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