Differential — Mathematically conform?

Question

In calculus, I know that one defined the differential quotient $$\frac{dy}{dx} := \lim\limits_{h \rightarrow 0}{\frac{y(x+h)-y(x)}{h}}$$ I learned that it is not a quotient, but can be treated as one in many cases which you can prove like $$ \frac{dy}{dx} = {\frac{dx}{dy}}^{-1} \quad or \quad \frac{dy}{dx} \frac{dx}{dt} = \frac{dy}{dt} $$ For examples like that, it seems more intuitive and is easier to understand — but in a mathematically conform way.

As far as this, no problem — until I reach some content in my book saying things just like $$ dU = d\vec{r} \cdot \operatorname{grad} U \quad \text{or even} \quad d\vec{r} \times \stackrel{\rightarrow}{A} = 0 $$

This confuses me in two ways:

1. Math is the only science where it is essential to define everything which appears in an equation/expression etc. When I see the term $dy$, I ask myself “How is this defined?”. Each of the terms $\frac{dy}{dx}$, $∫fdx$ etc. have a concret definition, wheras $dy$ doesn't seem to have one — intuitively, one supposes to say $dx := \lim\limits_{h\rightarrow 0}{\left(x – (x+h)\right)}$, which would exactly be zero. According to what Wikipedia says, it is defined as $df(x,Δx):=f'(x)Δx$, which would not accord to the differential with one parameter as always used. Therefore, WP says $df(x):=f'(x)dx$ which is not appropriate because one cannot define a new operator under usage of this new operator itself.
2. When some new content is introduced in a book with these expressions, even if I understand the intuitive sense or meaning of this equation, I feel like not to have understood a single word (or variable), because 90% of my thoughts ask how I should evaluate the equation/expression mathematically and that it is not legitimate to approve such knowledge based on wrong or unclear axioms, which results in a 2h-long bafflement.

Could you please make this topic a little more clear for me?

Answer 1

Let me denote your y with an $F:\mathbb{R}^n\rightarrow \mathbb{R}$ (like function).

The following $\Delta x$ is just a quantity that is intended to be very small in taylor expansion :

$$F(x+\Delta x)=F(x)+F'(x)\Delta x+o(\Delta x)$$ This is an approximation around $F(x)$.

But $dx$ is a very different thing and is about differentiation : The differential of F (F approximated by a linear function) at point x is the linearization on F around x. F being from $\mathbb{R}^n$ to $\mathbb{R}$, dF(x) (the differential of F at point x) is a linear operator. To be more precise (Frechet) : F is differentiable at point x if there exist a (continuous but ok in finite dimension) linear operator $L:\mathbb{R}^n \rightarrow\mathbb{R}$ such that :

$$F(x+\Delta x)=F(x)+L(\Delta x) +o(\Delta x)$$

Then we denote L=dF(x).

You probably know that if $G:\mathbb{R}^n\rightarrow \mathbb{R}$ is linear, $e_i$ being the standard basis of $\mathbb{R}^n$, you can define the dual basis $dx_i$ in ${\mathbb{R}^n}^*$ (linear forms = linear operators from $\mathbb{R}^n$ to $\mathbb{R}$) and decompose G onto this basis : $$G(x_1,\dots,x_k)=\sum_i g_i dx_i$$

L being a linear operator from $\mathbb{R}^n$ to $\mathbb{R}$, you can decompose it on this dual basis, which is denoted most of the time $dx_i$ according to the variable. With only one variable, you get : $$dF(x)=\alpha dx$$ Where $dx$ is the dual of the standard basis (vector "1") of $\mathbb{R}$ (And is a function !!), and $\alpha$ a (fixed given $x$ !) is a number. In one dimension you have $\alpha=F'(x)$.

I hope it partially answer your question about the $d$

Answer 2

IMO the best way to understand $\text{d}f$ is abstractly: it's simply a new sort of object. You can model the concept -- e.g. look up "differential form" in any text on differential geometry -- but that introduces a lot of detail that obscures understanding.

The important features are things like you can multiply a differential by a scalar: $x \text{d}y$ makes sense. You can add differentials. You have identities $\text{d}(x+y) = \text{d}x+\text{d}y$, $\text{d}(xy) = x\text{d}y + y\text{d}x$, and $\text{d}a=0$ whenever $a$ is a constant. One particularly notable feature is if $y=f(x)$, then $\text{d}y = f'(x) \text{d}x$; simple calculations with differentials look very much like implicit differentiation. Another is the relation between differentials and integration.

Alas, I've never seen a textbook simply introduce differentials in a way that says "here, these are what you can do with them!", except in the specific case of Kahler differentials in abstract algebra, and I know I wouldn't do it justice if I tried myself. :(