Meaning behind differentials So I think I understand what differentials are, but let me know if I'm wrong.

So let's take $y=f(x)$ such that $f: [a,b] \subset \Bbb R \to \Bbb R$.  Instead of defining the derivative of $f$ in terms of the differentials $\text{dy}$ and $\text{dx}$, we take the derivative $f'(x)$ as our "primitive".  Then to define the differentials we do as follows:

We find some $x_0 \in [a,b]$ where there is some neighborhood of $x_0$, $N(x_0)$, such that all $f(x)$ in $\{f(x) \in \Bbb R \mid x \in N(x_0)\}$ are differentiable.  Then we choose another point in $N(x_0)$, let's call it $x_1$, such that $x_1 \ne x_0$.  Then let $dx = \Delta x = x_1 - x_0$.  Now this $\Delta x$ doesn't actually have to be very small like we're taught in Calculus 1 (in particular it's not infinitesimal, it's finite).  In fact, as long as $f(x)$ is differentiable for all $x \in [-10^{10}, 10^{10}]$ we could choose $x_0 = -10^{10}$ and $x_1 = 10^{10}$.

Then we know that $\Delta y = f'(x_0) \Delta x + \epsilon(\Delta x)$, where $\epsilon(\Delta x)$ is some nonlinear function of $\Delta x$.  If $f(x)$ is smooth, we know that $\epsilon(\Delta x)$ is equal to the sum of powers of $\Delta x$ with some coefficients, by Taylor's theorem.  But of course, $\epsilon(\Delta x)$ won't be so easy to describe if $f(x)$ is only once differentiable.  So we define $dy$ as $dy = f'(x_0) dx$: that is, $dy$ is the linear part of $\Delta y$.  This has the very useful property that $\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \frac{dy}{dx} = f'(x_0)$.  This is then not a definition of the derivative, but a consequence of our definitions.

It can be seen from this $dy$ really depends on what we choose as $dx$, but $f'$ is independent of both.  

This definition can be extended to functions of multiple variables, like $z = f(x, y)$ as well, by letting $\Delta x = dx,\ \Delta y=dy$ and defining $dz$ as $dz = \frac{\partial f(x_0, y_0)}{\partial x}dx + \frac{\partial f(x_0, y_0)}{\partial y} dy$.  So $dz$ is the linear part of $\Delta z$.  Does all of the above look correct?

If so, then where I'm having a problem is: 1) how then do we define the derivative of $f(x)$ if not by $f'(x_0) = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$? 2) how do we apply this definition of $dx$ to $\int_a^b f(x)dx$?  It seems like the inherit arbitrariness of $dx$ is really going to get in the way of a good definition of the integral.
$\mathrm{d}y$ depends only on $y$: it doesn't depend on any choice of $x$ or anything else: that's one of the big advantages to differentials (as opposed to, say, partial derivatives).

A differential is a gadget that expresses how something varies. There are three main things you can do with such a gadget:


You can compare two differentials: e.g. if $x$ and $y$ are dependent on one another in a differentiable way, then they are multiples of each other. e.g. if $y = f(x)$, then $\mathrm{d}y = f'(x) \mathrm{d}x$.
Given a differential, you can ask if it has an antiderivative: e.g. $2x \mathrm{d}x$ is the differential (often called the "exterior derivative") of $x^2$.
You can compute a (path) integral to 'add up' along a path all of the variations the differential expresses. e.g. $\int_0^1 2x \mathrm{d}x$ means we 'accumulate' all of the variations $2x \mathrm{d}x$ as we go from $x=0$ to $x=1$. And as we know $2x \mathrm{d}x = \mathrm{d}(x^2)$, our intuition is satisfied in the sense that accumulating how $x^2$ varies from $x=0$ to $x=1$ works out to $1^2 - 0^2$.


You can also ask the differential to give you an ordinary number expressing a variation along a (tangent) vector. A common notation for this is, e.g. in $(x,y)$ coordinates, to let the symbol $\partial/\partial x$ and $\partial/\partial y$ denote vectors, and for a differential $\omega$, the notation $\frac{\partial}{\partial x} \omega$ means ordinary number that $\omega$ yields for a variation by the vector $\partial/\partial x$.

e.g. we have
$$ \frac{\partial}{\partial x} \mathrm{d}x = 1
\qquad \qquad \frac{\partial}{\partial x} \mathrm{d}y = 0
\qquad \qquad \frac{\partial}{\partial y} \mathrm{d}x = 0
\qquad \qquad \frac{\partial}{\partial y} \mathrm{d}y = 1$$

This is consistent with the notation for partial derivatives you've learned, in that, e.g.,

$$ \frac{\partial}{\partial x} f = \frac{\partial}{\partial x} 
\mathrm{d} f $$

where the left hand side is the meaning taken from introductory multivariable calculus, and the right hand side is the meaning I describe above. (usually first introduced in differential geometry)

Incidentally, I think partial derivative notation is absolutely terrible, and I avoid using it whenever possible. I also think differentials are more intuitive than partial derivatives as well, and I prefer to do all of my calculus in terms of differentials these days. A convenient analog to $f'$ for multivariable functions is to let, e.g., $f_1$ denote the derivative of $f$ in its first argument, $f_2$ denote the derivative in the second argument, and so forth. So I would prefer to write

$$ \mathrm{d}f(x,y) = f_1(x,y) \mathrm{d}x + f_2(x,y) \mathrm{d}y $$

rather than anything resembling the traditional notion of partial derivatives. If I want derivatives in the direction where $y$ is held constant, I express that as setting $\mathrm{d}y = 0$ rather than resorting to partial derivatives.

This use of combining vectors with differentials is related to the (unfortunately common) mistake / abuse of notation that you often see, where the notation $\mathrm{d}x$ is treated an actual change in $x$, rather than as a gadget that can tell you what the change in $x$ is.
Differentials are infinitely small changes in x or y. For instance, the concept of the integral is the sum of the areas of an infinite number of rectangles under a curve. The height of each is f(x) and the width is dx.