How is path dependence related to the differential? An example would help  here, but my problem is attempting to understand intuitively the difference between an exact and inexact differential. Maybe one of my problems may be the very definition of differential.
I assume differential is not the same as derivative. It is the change in y but how then can this be integrated? Don't you need dy/dx ? and what does it mean for it's integral to be path dependent vs. one which may be path independent.  
The books are jumping around using "state" instead of path so I assume they are the same here. Not sure though. If the integral turns out to be path independent it is said to be exact. Wish I could get some intuition here with some simple problems with concrete example. 
 A: For concreteness, let us work with the space $X=\mathbb R^2$. Then a smooth $1$-differential (or form) is just an expression of the form 
$$\omega:=f(x,y)dx+g(x,y)dy$$ where $f,g$ are smooth functions on $X$. Just specifying an $f$ and a $g$ will determine a $1$-differential. 
We say that $\omega$ is closed if $g_x=f_y$ (partial derivatives) are equal. 
We say that $\omega$ is exact if for every simple closed path $C\in X$, $$\int_C f(x,y)dx+g(x,y)dy=0$$ 
This property when attributed to the vector field $\langle f,g\rangle$ is sometimes called path independance or being conservative. This is equivalent to being able to find a function $F(x,y):\mathbb R^2\to \mathbb R$ such that $F_x=f,F_y=g$. There is no magic here. This is just a path integral you would do in a calculus class. 
As we are assuming $X=\mathbb R^2$, it turns out that closed $1$-forms and exact $1$-forms are the same thing! (This is true in any contractible space, if you know what that means).
Exercise: If we work instead with the space $X=\mathbb R^2\setminus\{0\}$, the plane with the origin removed,  consider  the differential $$\frac{-y}{x^2+y^2}dx +\frac{x}{x^2+y^2}dy.$$
Show that this is closed but it's integral around the unit circle is not zero therefore it is not exact.
