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I have only taken basic real analysis, so to me $\dfrac {dy}{dx}$ is just a formal symbol which stands for the limit of the difference quotient. Now, I know that $dy = y'(x) dx$ has some intuitive meaning historically, but if nowdays it's just another representation for the equation $\dfrac {dy}{dx} = y'(x)$ (which does have a formal meaning), why does it even exist in modern mathematics?

Ovi
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  • I think it's partly because "infinitesimal intuition" seems very clear sometimes. Thinking about tiny changes in $x$ and the corresponding tiny changes in $y$ is a clear way to understand intuitively many ideas in calculus. – littleO Jul 28 '18 at 03:36
  • $dy=y'dx$ is not an "intuitive meaning". It is the precise relationship between the differential of $y$ and the differential of $x$. The one that you are calling an equation, $\frac{dy}{dx}=y'$, is the one that is saying nothing. It is just two names, from different notations, for the same thing put on both sides of an equal sign. –  Jul 28 '18 at 03:45
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    This is a question that has been asked many many times. Search in the site. Better instead, pick up Calculus on Manifolds by Spivak. It is a short book. You will get there what is a differential, why $dy=y'dx$, or $dy=\frac{dy}{dx}dx$, or $dy=D_xy dx$, or $dy=\dot{y}dx$ (whichever notation you want for the derivative of $y$ with respect to $x$), etc. –  Jul 28 '18 at 03:55
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    @nextpuzzle I am aware that differentials have to do with something called "differential forms", and I know that the symbol $\dfrac {dy}{dx}$ has some meaning in those terms as well. However, in the context of basic real analysis the definition of $\dfrac {dy}{dx}(t)$ is $\lim_{x \to t} \dfrac {f(x) - (t)}{x-t}$. Is there a corresponding definition in basic real analysis of $dx$? – Ovi Jul 28 '18 at 04:00
  • The definition of $\frac{dy}{dx}$ is the same that you are saying everywhere. –  Jul 28 '18 at 04:03
  • Yes, $dx$ has a corresponding definition in basic real analysis. Search it. You can find it almost everywhere. And come back if you didn't understand it. –  Jul 28 '18 at 04:04
  • If you google "Calculus Manifolds Spivak filetype:pdf" you find a book that will explain it clearly. Chapter 2 directly. –  Jul 28 '18 at 04:06
  • Very related question: Is $\frac{\mathrm{d}y}{\mathrm{d}x}$ not a ratio? – Mike Pierce Jul 28 '18 at 04:14
  • There is one link. Just a warning. Don't necessarily believe that votes in an internet website mean much. For example, Brendan Cordy's answer has 55 votes at the moment, while it doesn't make any sense. The top two answers are good. Of course, don't believe me either. –  Jul 28 '18 at 04:26
  • @nextpuzzle Thanks for the advice; I have searched for a basic real analysis definition of $dx$, but so far have only found either intuitive explanations which I alreayd know, or definitions for $\delta x$. However, I have to go to sleep now and then work in the morning, so I will conitnue searching tomorrow and get back to you. – Ovi Jul 28 '18 at 04:32
  • I think it's chapter 4 of Spivak's Calculus on Manifolds that gives a definition of $dx$, etc. – littleO Jul 28 '18 at 06:25
  • Let me tell you what is $dx$ on $\mathbb{R}$ such that you can compare with Spivak. Fix a point $x=a\in\mathbb{R}$. All the objects that we will talk about are associated to that point. Differentials are just linear functions. In the case of $dx=adx$, at the point $a$, is just the identity linear function, $dx(h)=h$. By definition it is the differential of the coordinate function $x$. Lets verify it: $\lim{h\to0}\frac{|(x+h)-x - dx(h)|}{|h|}=\lim_{h\to0}\frac{|(x+h)-x - h|}{|h|}=0$. It works. –  Jul 28 '18 at 11:39
  • Recall from Spivak that the differential of a function $f$ at a point $x=a$ (he calls it derivative, denotes it $Df(a)$, but let me use the notation that other people use, such that you can connect them) is a linear function $df$ (this notation doesn't show the dependence on $a$, but it does depend) such that $\lim_{h\to0}\frac{|f(x+h)-f(x)-df(h)|}{|h|}=0$. If you compare this limit with the definition of derivative (not Spivak's derivative, but the one you studied in Calculus) $f'(a)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$, you realize that the linear function $df(h)$ is $f'(a)h=f'(a)dx(h)$. –  Jul 28 '18 at 11:46
  • And there you go, you get the formula $df=f'(a)dx$. –  Jul 28 '18 at 11:47
  • @nextpuzzle Thank you very much! I will also read Spivak's book – Ovi Jul 29 '18 at 17:09

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