(This article needs to be polished; the idea is correct, but the explanation is unnecessarily complex.)

The calculus perspective: dx and $\Delta x$

This is a direct and well-known perspective. $\Delta x$ is a finite change in x. $dx$ is an infinitesimal change in x. If we accept limit(infinite), the derivative definition:

\[\frac{d f}{d x}=\lim _{\Delta x \rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{\Delta f}{\Delta x}\]

So $df / dx$ is the limit of the ratio $\Delta f / \Delta x$. In a sense, “d” is the infinitesimal version of “$\Delta$”. So “dx” means “an infinitely small piece of x”.

But next let me rigorously introduce a differential geometry perspective. Why? Because it is more fundamental, it reflects the cognitive model: what is the intrinsic truth we believe? How do we describe it?

The differential geometry perspective

The general idea: $dx$ is a 1-form that extracts the $x$-component of a tangent vector. $dx(v)$ gives the components of $v$ in the $x$-coordinate system. In $\mathbb{R}^n$ with standard coordinates, $dx_i$ is literally the projection onto the $i$-th coordinate where $i\in \lbrace n\rbrace$.

Firstly, we need to clarify that $x$ is not number but defined as a function :

\[x:\Omega\rightarrow \mathbb{R}^n.\]

There are many ways to interpret this. Here we view $x$ as a coordinate map (a diffeomorphism from $\Omega$ onto an open subset of $\mathbb{R}^n$). A coordinate map assigns to each point $p \in \Omega$ an n-tuple of real numbers:

\[p \mapsto\left(x_1(p), x_2(p), \ldots, x_n(p)\right)\]

We can think that $x$ describe $\Omega$ using $n$ real numbers, but these numbers are independent, so $x$ is an $n$-dimensional coordinate of $\Omega$.

Then we define $v$ as a tangent vector at point $p\in\Omega$.

Let’s take our general idea a step further: $dx$ is a function whose input is $v$ and output is $n$ real numbers : $[dx_1(v),dx_2(v),dx_3(v)\cdots dx_n(v)]$. Each $dx_i(v)$ is a component of $dx(v)$.

Recall that a vector encodes both magnitude (a scalar) and direction. Here $v$ is a geometric object that exists intrinsically on the manifold; it is coordinate-independent. $dx(v)$ is a representation of this object $v$ using the $x$-coordinates.

Because $x$ is a coordinate, we introduce the notation ${\partial}/{\partial x_i}$ as the tangent vector pointing the $x_i$ direction. Geometrically, for example, $\partial / \partial x_1$ represents the velocity of a curve where only $x_1$ is changing:

\[\gamma(t)=\left(x_1+t, x_2, x_3, \ldots, x_n\right)\]

The tangent vector to this curve is $\partial / \partial x_1$.

Back to the general idea: $dx(v)$ is a representation of $v$ in x-coordinate. How do we represent $v$ using $dx(v)$? We use tangent vector ${\partial}/{\partial x_i}$ as the basis (it’s feasible because the $\partial/\partial x_i$ are linearly independent), component $dx_i(v)$ as the coefficient.

Remember $dx(v)$ is a representation of $v$, their information content is the same. The vector $v$ and its components $[dx_1(v),dx_2(v),dx_3(v)\cdots dx_n(v)]$ contain equivalent information. We can reconstruct $v$ from its components:

\[v=d x_1(v) \frac{\partial}{\partial x_1}+d x_2(v) \frac{\partial}{\partial x_2}+\cdots+d x_n(v) \frac{\partial}{\partial x_n}=\sum_i^n d x_i(v) \frac{\partial}{\partial x^i}\]

Here, $dx_i(v)$ is the speed of $v$ in $x_i$ component, ${\partial}/{\partial x_i}$ is the direction of $v$ in $x_i$ component.

The basis 1-forms $\lbrace dx_i\rbrace$ are dual to the basis tangent vectors $\lbrace\partial/\partial x_j\rbrace$: $dx_i(\partial/\partial x_j) = \delta_i^j$. This duality is the same one that appears in the pushforward/pullback correspondence.

Recall $v$ is an intrinsic vector, so it’s coordinate-independent; if $f$ is a (smooth) function, $v[f]$ means the speed (scalar, real number) of $f$ if we move along $v$, it’s called the directional derivative of $f$ along $v$. Another notation of $v[f]$ is $df(v)$, they are exactly the same meaning.

For $f=f(x)$, the differential $df$ is:

\[d f=\frac{d f}{d x} d x\]

Caveat: $\frac{d f}{d x}$ is not a fraction, it’s not a cancellation in $\frac{d f}{d x} d x$. Instead, $df/dx$ is defined as the unique scalar function such that:

\[d f=\left(\frac{d f}{d x}\right)\cdot d x.\]

We will further explain it later.

Now apply both sides to $v$ :

\[d f(v)=\frac{d f}{d x} \cdot d x(v)\]

holds for every vector $v$ , then we can drop the $v$ and write:

\[d f=\frac{d f}{d x} \cdot d x\]

This is like in linear algebra: if two linear maps give the same output for every input, they are the same map. And $f^\prime(x)$ is just another notation for $df/dx$, we get the common equation in calculus:

\[d f=f^{\prime}(x) d x\]

For multivariable functions, this generalizes to $df = \sum_i (\partial f/\partial x_i) \, dx_i$, which is exactly the multivariable chain rule expressed in the language of differential forms.

An example

For example, $f=x^2+y^3$, in this case $f$ is a smooth map:

\[f:\mathbb{R}^2\rightarrow\mathbb{R}\]

but not a coordinate map. For $\mathbb{R}^2$, the coordinate map is simply:

\[\begin{aligned} & (x, y): \mathbb{R}^2 \rightarrow \mathbb{R}^2 \\ & p \mapsto(x(p), y(p)) \end{aligned}\]

which is the identity map. Because each point is already labeled by its coordinates. Then $df=2xdx+3y^2dy$ is the changing rate of $f$ in 2d Euclidean space. We choose $p=(2,3)\in\mathbb{R}^2$, at this point, the $df$ becomes:

\[\left.d f\rvert_{(2,3)}=2(2) d x+3(3)^2 d y=4 d x+27 d y\]

Then we choose a tangent vector. (In Euclidean space any vector is a tangent vector, but other non-trivial spaces, e.g. $S^2$, we must choose vectors tangent to the sphere). Pick any vector, say $v=1 \partial / \partial x+2 \partial / \partial y$. Apply $v$

\[d f(v)=4 \cdot d x(v)+27 \cdot d y(v)=4(1)+27(2)=58\]

It means: in $\mathbb{R}^2$, for the function $f = x^2 + y^3$, at the point $p=(2,3)\in\mathbb{R}^2$, if we move along the vector $v=1 \partial / \partial x+2 \partial / \partial y$, the function $f$ changes at a rate of 58.

Why $df = f’(x)\,dx$ is not a cancellation

In informal calculus, people write things like:

\[\frac{d y}{d x}=f^{\prime}(x) \quad \Rightarrow \quad d y=f^{\prime}(x) d x\]

and say “multiply both sides by dx.” But this isn’t algebraic multiplication! The first $\frac{d y}{d x}=f^{\prime}(x)$ is just notation. Then we (algebraically) multiple $dx$ to both side:

\[\left(\frac{d f}{d x}\right)\cdot d x=f^\prime(x)\cdot dx\]

By definition

\[\left(\frac{d f}{d x}\right)\cdot dx=df\]

Thus

\[df=f^\prime(x)dx\]

The deeper point: $df$ and $dx$ are both 1-forms, and $f’(x)$ is the scalar function relating them. The “$\cdot$” in $f’(x) \cdot dx$ is scalar multiplication of a 1-form, not division of differentials.

Why do this? Because $f^{\prime}(x)$ is a scalar, not a 1-form, so it cannot directly act on a vector. We need $df=f^\prime(x)dx$ to get a 1-form that can. Now we can apply both sides to a vector $v$:

\[d y(v)=f^{\prime}(x) \cdot d x(v)\]

to calculate the changing rate of $f$ in vector $v$.