How to understand the notion of a differential of a function In elementary calculus (and often in courses beyond) we are taught that a differential of a function, $df$ quantifies an infinitesimal change in that function. However, the notion of an infinitesimal is not well-defined and is nonsensical (I mean, one cannot define it in terms of a limit, and it seems nonsensical to have a number that is smaller than any other real number - this simply doesn't exist in standard analysis). Clearly the definition $$df=\lim_{\Delta x\rightarrow 0}\Delta f =f'(x)dx$$ makes no sense, since, in the case where $f(x)=x$ we have that $$ dx=\lim_{\Delta x\rightarrow 0}\Delta x =0.$$
All of this leaves me confused on how to interpret expressions such as $$df=f'(x)dx$$ Should it be seen simply as a definition, quantifying the first-order (linear) change in a function about a point $x$? i.e. a new function that is dependent both on $x$ and a finite change in $x$, $\Delta x$, $$df(x,\Delta x):=f'(x)\Delta x$$ then one can interpret $dx$ as $$dx:=dx(x,\Delta x)= \Delta x$$ such that $$\Delta f=f'(x)dx+\varepsilon =df +\varepsilon$$ (in which $\varepsilon$ quantifies the error between this linear change in $f$ and the actual change in $f$, with $\lim_{\Delta x\rightarrow 0}\varepsilon =0$).
I feel that there must be some sort of rigorous treatment of the notion of differentials since these kind of manipulations are used all the time, at least in physics?!
I've had some exposure to differential geometry in which one has differential forms, in particular $1-$forms which suggestively notationally "look like" differentials, for example $$\omega =df$$ but as I understand it these are defined as linear maps, members of a dual space to some vector space, $V$, which act on elements of $V$, mapping them to real numbers. Furthermore, the basis $1-$forms are suggestively written as what in elementary calculus one would interpret as an infinitesimal change in x, $dx$. But again, this is simply symbolic notation, since the basis $1-$forms simply span the dual space and are themselves linear maps which act on elements of $V$.
I've heard people say that differential forms make the notion of a differential of a function mathematically rigorous, however, in my mind I can't seem to reconcile how this is the case, since at best they specify the direction in which the differential change in a function occurs, via $$df(v)=v(f)$$ (since $v(f)$ is the directional derivative of a function $f$ along the vector $v$).
If someone could enlighten me on this subject I'd really appreciate it.
 A: A differential is a $1$-form. At each point a $1$-form gives a linear functional in a tangent space, and the level sets of a linear functional are parallel affines subspaces. Now $df$ in $\mathbb R^3$  gives a sort of a ruling in each tangent. For a given $p$, the level sets of $df_p$ are what we can see if we zoom in on the level sets of $f$ in a neighborhood of $p$, like in the picture bellow. 
$\hskip1in$ 
where on the left should be the level sets of $f$. 
A: I think that the biggest problem comes from the subtlety of the notation.
First of all, there is a general formula for the derivative, and that is usually represented as $df\over dx$ or $f^\prime(x)$.
There is also the notion of a derivative at a point $x_0$, and that is represented as ${df \over dx}|_{x=x_0}$,  When you see the manipulation of differentials, it is usually in this context.  (And it is all the more confusing because, when differentials are used by themselves, there is never any mention of the $x_0$ that links $dy$ and $dx$.)
For instance, take $y \equiv f(x) = x^2$ and  $x_0 = 0$, and consider the discrete sequence $x_n = {1 \over 2^n}$.  (We see that $x_n \rightarrow 0.)  $
For the  $k$th term of this sequence, we can (exercise) make the finite difference $\Delta x_k = x_{k+1} - x_k = -{{1}\over {2^k}}$. Then (another exercise)  $\Delta y_k = -{3\over 4}\cdot {1\over{2^{2k}}}$.
We see that $\Delta y$ and $\Delta x$ both go to 0 as $k\rightarrow \infty$, but the above exercise shows that the speed of $\Delta y$ toward 0 is dictated by $\Delta x$.  So, it makes sense to split $dx$ and $dy$ apart, so long as we keep track of how they are related at the point at which $dy/dx$ is evaluated. 
Hope that made sense--I kind of think that I might be rambling.
A: Differentials are not really the same as differential 1-forms, this is something that I think confuses a lot of people because the notation is the same despite them being fundamentally different mathematical objects. This can be seen best when looking at how $\iint dx\wedge dx =0\neq\iint dxdx$ despite $\iint dx\wedge dy = \iint dxdy$ 
Not thinking of differential 1-forms as differentials requires a new understanding of them. Differential forms are mathematical objects that fundamentally deal with space and geometry which is why a differential 1-form can be thought of as the level sets of a vector and a differential 2-form can be thought of as a signed area. All the differentials in the differential-forms are (The dx part in fdx) are bases of the co-vector space of differential forms. 
Using this framework we can see that the exterior derivative of a function is just how much a change in the direction of one of its coordinates effects each basis in this co-vector space. 
A quick example would be 
$$f(x)=x  
\\ \text{Then taking the exterior derivative}
\\ df=f_xdx=dx$$
All this is saying is that as you change the function f in the x-direction in the vector space f is defined in the dx basis in its co vector space changes by the same amount, which in this case is the classical single variable derivative of the function f. 
So instead of thinking of differential 1-forms as being tiny changes in variables think about them as bases in a co-vector space that lives alongside the vector space in which a function is defined. 
