What is the simplest, yet still rigorous, way to define $\text{d}x$?

Question

While reading calculus books, I see sections on differentials which refer to "infinitessimals" in a very loose way, alluding to the fact that this view on calculus is not the standard, but makes a lot of intuitive sense. However, the next "level" of rigor I've seen is in differential geometry books, where differentials are well-defined in a standard way, but require a depth of knowledge just to understand that definition.

My question is: is there a middle ground here, where we can rigorously and formally define the concept of a differential $\text{d}x$ for an independent variable $x$, that is accessible to students fresh out of undergraduate or first-year-graduate analysis?

I emphasize independent because I've heard many times that a differential is defined as $\text{d}y = f^\prime(x)\text{d}x$, but without defining $\text{d}x$ this definition is pointless.

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Edit: I think what I'm looking for is basically a simpler explanation of $\{\text{d}x^i\}$ as a dual basis for $\{e_i\}$, without talking about tangent and cotangent spaces and, if possible, without talking about bases at all. When we say $\text{d}x^i(e_j) = \delta_{ij}$, what is the definition of $\text{d}x^i$ that is used?

Answer 1

I emphasize independent because I've heard many times that a differential is defined as $\text{d}y = f^\prime(x)\text{d}x$, but without defining $\text{d}x$ this definition is pointless.

One way would be to define the (first) differential $\mbox{d}\,f$ of a function $f$ as a map: $$\mbox{d}\,f:(x,\Delta x) \mapsto f'(x) \Delta x$$ Note that this mapping takes a point $x$ and an increment $\Delta x$. The number which is associated with a point $x$ and increment $\Delta x$, is how much the $y$-value changes on the tangent line (the linearisation of the function), which is an approximation of the change in the real function value.

The explicit dependency of $\mbox{d}\,f$ on $x$ and the increment $\Delta x$ is often omitted and for $y=f(x)$, this also leads to the following notation: $$\mbox{d}y = f'(x) \Delta x$$

Applying this to the function $f(x)=x$, which is differentiable everywhere with $f'(x) = 1$, leads to the following relation where we take $y=x$: $$\mbox{d}y = f'(x) \Delta x \to \mbox{d}x = 1 \Delta x \implies \mbox{d}x = \Delta x$$ This can be seen as a motivation to replace $\Delta x$ by $\mbox{d}x$, giving the more common: $$\mbox{d}y = f'(x) \, \mbox{d}x$$ As a bonus, this nicely "agrees" with the Leibniz notation for derivatives: $$\frac{\mbox{d}y}{\mbox{d}x} = f'(x)$$

Answer 2

My preferred way is through nonstandard analysis, which requires some medium-duty logic to set up but then allows you to treat $\mathrm{d}x$ simply as an infinitesimal. There are textbooks of introductory analysis which do it this way, such as Pétry's Analyse Infinitésimale: une présentation non standard. It's within the grasp of an undergraduate, though to do it properly (justifying that one can rigorously make infinitesimals exist) is not trivial.

Answer 3

The notation $dx$ suggests that we had a coordinate map $x:M\to \mathbb R$, $p \mapsto x(p)$.

Then $dx(p)$ is the map derivative, a push-forward between tangent spaces, or, when $M$ (and $\mathbb R$) is a normed vector space, $dx_p$ is called the Fréchet derivative. $M$ itself could be another copy of $\mathbb R$, but holding on to this generality allows me to emphasize the point $p\in M$.

$dx_p$ is a linear map mapping (tangent) vectors in $M$ to vectors (increments $\Delta x$ sometimes, or simply a real number - that's why it's also called a [basis] 1-form) in $\mathbb R$.

You then consider a map $f:\mathbb R\to\mathbb R$ and a composite map $y:M \to \mathbb R$, $y=f\circ x$, and use the chain rule to find $dy(p)=df(x(p))\circ dx(p).$

A confusing moment happens when we decide to drop $p$, as if reusing $x$ to also denote the target element in $\mathbb R$, $x=x(p)$(!?), and write $dy=df(x)\circ dx.$

Finally we introduce notation $f'(x)=df(x)$ and this map being simply a scalar multiplication for a given $x(p)$ we replace the composition by the multiplication, $dy=f'(x)dx.$