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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?

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    $\begingroup$ This is an intelligently posed question, but I have taken the liberty of removing some informal turns of speech that, unfortunately, encourage the reader not to take it as seriously. I also fixed some of your TeX and removed the "thanks", which is unnecessary here. $\endgroup$ – Ryan Reich Sep 13 '13 at 14:19
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    $\begingroup$ Cursing is a natural reaction to this kind of BS, I support the OP. $\endgroup$ – Git Gud Sep 13 '13 at 14:20
  • $\begingroup$ differential of $f$ at $x_0$ is defined as the linear function approximating $f(x)-f(x_0)$ (namely $f'(x_0)(x-x_0)$). This makes most of the expressions you worry about perfectly well defined, in particular $df/dx$ is a quotient. It's somewhat trickier with the $dx$ in $\int f dx$. $\endgroup$ – user8268 Sep 13 '13 at 14:33
  • $\begingroup$ But if it's an approximation function, woldn't make this kind of writing make it unneccessarily hard to completely understand within the mathematical context in comparison to just taking diff'quotients? $\endgroup$ – Lukas Juhrich Sep 13 '13 at 14:39
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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 $d$

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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. :(

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  • $\begingroup$ But the identities you listed are also explainable with diff'quotients, with the advantage that you can prove (or, in my case, even understand satisfyingly) them much easier. So why are they used so often? Are there cases in which one cannot understand the facts as easy in this more ‘pure’ form (my opinion ^^)? I don't think so, because there are a lot of things much harder to understand in math. $\endgroup$ – Lukas Juhrich Sep 13 '13 at 19:46

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