What does dx = dy mean formally? I have a circle with circumference $X$ and charge density $\lambda$ and have to find a potential at distance $a$ from the center. I understand how to do that, the physics way is to write $E_x = \dfrac{K\lambda a l}{(a^2+x^2)^{\dfrac{3}{2}}},$
and then say that this implies $\mathbb d E_x = \dfrac{K\lambda a}{(a^2+x^2)^{\dfrac{3}{2}}}\mathbb d l,$  and then integrate with respect to $\mathbb d l$ from $0$ to $2 \pi x$.
But what does $\mathbb dE = ... \mathbb d l$? mean mathematically
I tried to approach it with taking the total derivative of $E_x$ which would work only if I viewed $E_x$ as function of only $l$, because then I get
$\mathbb dE = \dfrac{\partial}{\partial l} (\dfrac{K\lambda a l}{(a^2+x^2)^{\dfrac{3}{2}}}) \mathbb d l = \dfrac{K\lambda a}{(a^2+x^2)^{\dfrac{3}{2}}} \mathbb d l .$
If I would take this approach (which I doubt works every time), then I have no idea how to integrate total differentials.
Am I on the right track at least? Or is there some part of math I am missing?
 A: I'm not 100% familiar with your example so I will answer with an equally valid example. Say you want to compute the area of the disk of radius $R$ and you only know the formula for the length of a circle of radius $R$ ($2\pi R$) and calculus.
One way is to draw the annulus between the disk of radius $r+\delta r$ and $r$ and compute its area. Here $\delta r$ is just a small quantity. I will be more precise later.
The area of this annulus is approximately given by
$$
\delta A = 2 \pi r \delta r .
$$
Note that strictly speaking the formula is wrong, and in fact it is only valid up to linear order in $\delta r$. To be more precise we should write
$$
\delta A = 2 \pi r \delta r + O(\delta r^2) .
$$
which tells us that the error of the first formula is of order $\delta r^2$. If we knew already the formula for the area of the disk we could write
$$
\delta A = \pi \big[ (r +\delta r)^2 -r^2 \big] = \pi (2r \delta r +\delta r^2 ), 
$$
but we don't.
Now to find the area of the disk of radius $R$ you can simply integrate $ \delta A$ from $0$ to $R$. To be more precise what we are doing here is really evaluating the integral as a limit of Riemann sums.
$$
A(R) = \int_0^R dr 2 \pi r = \lim_{\delta r\to 0} \sum_j  2\pi r_j \delta r
$$
In this form it does not matter whether we use the first or second expression (or third) for $\delta A$ because the term $O(\delta r^2)$ contributes zero to the sum in the limit $\delta r\to 0$. (If you want to be explicit you can choose $\delta r = R/N$, $r_j= j \delta r$, $j=0,1,\ldots,N-1$.
Now we learn that the area is
$$
A(R) = \pi R^2
$$
You can indeed differentiate and  obtain
$$
\frac{dA}{dr} (r) = 2 \pi r 
$$
which you can "interpret" as
$$
dA = 2\pi r dr. 
$$
This very powerful interpretation, is allowed because, after Leibniz, we write the derivative as the limit of the incremental ratio.
In this context this is more than sufficient. In differential geometry one can say that $dr$ (and $dA$) are 1-forms but I would say this is beyond the scope of the question (one needs a definition of forms in the first place).
A: In the way you used it, it seems like $df = g\,dl$ means $\frac{\partial f}{\partial l} = g$. If $f$ is a function of more than one variable, this is not equivalent to the interpretation of $df$ as a one form.
