Exact differentials in thermodynamics I was reading the book "Thermodynamics and Statistical Mechanics" of Allis and Herlin and I have a doubt in one part of the first chapter. Consider the differential $$df = A(x,y)dx + B(x,y) dy.$$The book said that, if the integral of $df$ doesn't depend on the path of integration, $f$ is a state function and $df$ is a exact differential, and we can write $$A(x,y) = \left( \frac{\partial f}{\partial x} \right)_{y}, \ \ \ B(x,y) = \left( \frac{\partial f}{\partial y} \right)_{x}.$$ Why is this not true to an no exact differential?
 A: Do note that 
$$\frac{\partial A(x,y)}{\partial y} = \frac{\partial B(x,y)}{\partial x}$$
is a necessary condition for EXACT differentials.
This is due to the Clairut's theorem where if we consider the definition of the total differential:
$$df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy$$
let $A(x,y) = \frac{\partial f}{\partial x}$ and $B(x,y) = \frac{\partial f}{\partial y}$
then (Clairut's theorem applies here):
$$\frac{\partial A(x,y)}{\partial y} = \frac{\partial}{\partial y} \left(\frac{\partial f}{\partial x}\right) = \frac{\partial}{\partial x} \left(\frac{\partial f}{\partial y}\right)=\frac{\partial B(x,y)}{\partial x}$$
An inexact differential is defined by NOT as a straight forward differential of any function. This would imply that
$$A(x,y) = \frac{\partial f}{\partial x}$$ and $$B(x,y) = \frac{\partial f}{\partial y}$$
would not make any sense because there is not a common $f$ which you can differentiate to get $A(x,y)$ and $B(x,y)$
Hence you cannot write $A(x,y) = \frac{\partial f}{\partial x}$ and $B(x,y) = \frac{\partial f}{\partial y}$
