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.
 A: $\mathrm{d}y$ 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.
A: Also, the definition of a derivative that I learned was $\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$...basically rise over run as run ($\mathrm{d}x$) approaches $0$ (thus the tangent line concept).
A: OP here.  I think I've figured this out:
This definition does seem to hold for differentiation and integration.
Differentiation
My worry here was that because $\Delta x = dx$ and $\Delta y$ is a function of $\Delta x = dx$, that $\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$ would also be dependent on $dx$, which would make the definition $f'(x) := \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$ a circular argument (as $dx$ was defined in terms of $f'(x)$) -- but in fact, it's really only $dy$ that's defined in terms of $f'(x)$.  $dx$ is just some arbitrary change in $x$, i.e., $dx = x_1 - x_0$.  We need $f'(x_0)$ to be defined in terms of $x_0$, but the inherit arbitrariness of $x_1$ would make anything defined explicitly in terms of $dx$ not-well-defined.  However, what the limit operation really does is remove the arbitrariness of $x_1$.  That is, the smaller we make $\| x_1 - x_0 \|$, the less "arbritrary" $x_1$ is.  In the limit, it has lost all of it's "arbitrariness".  So $\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$ does not actually depend on the value we initially select for $x_1$ at all.  So good, this definition works for differentiation.
Integration
I couldn't figure out here what some arbitrarily large interval $dx$ (weird to hear $dx$ described as "large" isn't it?) had to do with integration (and thus why it would be in the integrand).  But again, an arbitary interval under a de-arbitrarifying process (a limit) is exactly what we need.  In this case $\int_a^b f(x)dx$ really means something like $$\lim_{\| dx_i \| \to 0} \sum_{i=0}^n f(x_i)dx_i= \lim_{\max(x_{i+1}-x_i) \to 0} \sum_{i=0}^n f(x_i)(x_{i+1} - x_i)$$  where each $dx_i$ is a subinterval of $[a,b]$.  Notice that each $x_{i+1} \to x_i$ and thus $x_{i+1}$ loses it's "arbitrariness" in this limit.
In case you're thinking that these $dx_i$ are not defined the same as in my question, notice that in the definition I gave, $x_0$ was fixed by the problem -- our only variable that we were free to fix was $x_1$.  That's exactly how these $dx_i$ are defined: for instance, the first one, $dx_0 = x_1 - x_0 = x_1 - a$ for some arbitrary $x_1 \in (a,b]$.  Assuming $x_1 \ne b$, then to fill out our partition we must define another subinterval $dx_1 = x_2 - x_1$, where $x_1$ in this case is not arbitrary -- it's defined to be the end point of our $1^{st}$ (or $0^{th}$ maybe because of how I chose to define my partition) subinterval -- but $x_2$ is arbitrarily chosen from the interval $(x_1, b]$.  And so on.
NOTE: I don't particularly like the notation I used to "define" integration above (the $\lim_{\| dx_i \| \to 0} \dots$ part), one of several reasons is that it really doesn't tell you that in the limit, $n \to \infty$.  The wikipedia page on the Riemann integral doesn't have such an equation (they just describe it with words... ugh).  Do you guys know a better notation?
Conclusion
It seems that defining $dx:=\Delta x$ and $dy:=f'(x)dx$ does exactly what it needs to do.  The relationship between the two can be used to approximate small changes in $\Delta y$, neither differentiation nor integration is defined in terms of $dy$ and though they are both defined in terms of $dx = x_1 - x_0$, they are not dependent on any values we initially select for $x_1$, and probably one of the best things is that because $dx$ and $dy$ are finite, the relationship $f'(x) = \frac{dy}{dx}$ holds for any differentiable $f$.
NOTE: This definition of $dx$ and $dy$ does not seem to be the same as the one used in differential geometry -- as described by Hurkyl.  But then, I'm not entirely sure because I don't completely understand Hurkyl's answer.  If anyone knows of a good primer on the notations and concepts he's using -- suitable for someone who's gone through the calculus sequence and linear algebra only -- I would be grateful for a link.  However, even if the definitions are different, it doesn't mean mine is not useable -- in fact, unless you guys can come up with a situation where $dx = x_1 - x_0$ and $dy = f'(x)dx$ (where $dx$ and $dy$ can be arbitrarily large) don't do what they're supposed to, I'm just going to take them as my definition of them from now on.
A: 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.
