> Can you multiply by dx?
Maybe. But can you multiply by ∂x?
Maybe. But can you multiply by ∂x?
∂x would have a problematic definition, as it would require selecting from its components based on information not supplied. E.g., let x = y+z. Then dx = dy + dz. But ∂x (by extension) = ∂y + ∂z, but at least one of these terms on the right is identically zero, depending on whether y or z is held constant. So ∂x doesn't have a meaning.
Like, I get “dx”, but I cannot put my finger to it!!! This might be because the Precise Definition of the Limit phrases it as “x approaches a”; it is as though we are “sent” to the land of dx, but not told what it is as an atomic concept!
Sure it does. There is no need to know about smooth manifolds or differential forms to understand the differential of a function of one variable at a point and the meaning of dy = f’(x)dx.
dy = f'(x)dx is just a definition for notional convenience, primarily employed when doing u or u-v substitution. My point is that dx in single variable calculus is notation. It is not an intrinsic object. dx is an intrinsic object as a differential form on a smooth manifold. Of course, the real line R is a 1-manifold, so dx does have that meaning, but you need to understand what a differential form is to know that.
One doesn't necessarily need the full generality of smooth manifolds though. Harold Edwards' Advanced Calculus: A Differential Forms Approach and Advanced Calculus: A Geometric View teach differential forms for Euclidean manifolds.
1) The differential of a function (at a point), dy, is not notation, it is a concept.
2) The differential of the function y = x, dx, is not, then, a notation, either; and, since the derivative is 1, dx = 1 Δx = Δx = x - x0.
3) You can argue, of course, that using dx instead of Δx in dy = f'(x)dx is "notation," but I think the above shows that it is more than that.
However, and this is very amusing to me, it turns out that the process of automatic differentiation (see https://en.wikipedia.org/wiki/Automatic_differentiation, the section on dual numbers) works in exactly the same way. Just replace all of the primed symbols (u', v', etc.) with du, dv, etc. and dual numbers are isomorphic to infinitesimals (if I'm using that term correctly.)