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I am readying in this website for a long time today similar questions and the answers provided about the issue I am asking now but I have to say I am more baffled than I was originally.

My original question would be:

I have read in many textbooks when they teach integration that they emphasize not to treat dx as a quantity but only as a notation that shows which is the variable of integrations. The same textbooks after a while they start talking about differentials and based on the following expressions

$$ df = f'(x)dx\quad df = f_{x}dx+f_{y}dy $$

they say that now that differentials have been introduced we can treat them as independent variables and use them freely as fractions.

Now I was always working with that idea, but after some digging I start seeing a lot of different answers, especially in here, and most of them were conflicting.

There were people mentioning Non-Standard Analysis and Differential Forms and were trying to justify that we can treat differentials as fractions. Also there were other people saying that every time we treat them as fractions we actually apply other theorems and so we don't actually treat them that way.

So the questions I have to ask now is:

  1. Can we actually treat differentials as fractions in EVERY framework, or is there today a debate about it?

    I don't want personal opinions or what we can do in a specific framework. I saw a lot of people mentioning that in single variable calculus you may treat them like that because you won't do mistakes. Can you actually treat them like that strictly mathematically in EVERY framework or you can just do it for ease but you actually apply different theorems behind the scenes?

  2. If differentials can't be treated in EVERY framework as a fraction then is it defined and proven that they can be always treated as a fraction in some frameworks?

  3. Given that we have defined differentials as

    $$ df = f'(x)dx\quad df = f_{x}dx+f_{y}dy $$

    then didn't we automatically defined than they can be used as independent quantities, so they can be seen as fractions also? Shouldn't we accept them as fractions from the moment of that definition existed?

Adam
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    The main question is, what do you understand as a differential? If you have a definition for this, then you begin to see that one may develop an algebra of differentials. – Allawonder Oct 24 '19 at 20:02
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    We don't know exactly what you've read. If you have questions raised by specific answers or questions already on the site, then you should link to them explicitly and explain what your confusion is or why it does not really answer your question. – Arturo Magidin Oct 24 '19 at 20:16
  • "Differentials as fractions" It doesn't look like that is the concern that gets discussed, but whether derivatives are fractions of differentials. Differentials are functions, you can make quotients of them, if you want. – conditionalMethod Oct 24 '19 at 20:17
  • @ArturoMagidin You can consider that I have literally read every single question in here that even remotely touches the subject of differentials. And I am not even kidding... – Adam Oct 24 '19 at 20:18
  • @conditionalMethod Another reason why the subject is confusing. Thank you :P – Adam Oct 24 '19 at 20:19
  • I wrote "It doesn't look like" to soften the phrase. I am actually telling that that is the case. So, if you read all the related posts, you should have noted that they are not about "differentials as fractions". So, none of your questions really make sense, and/or are not what you want to ask. – conditionalMethod Oct 24 '19 at 20:22
  • @ArturoMagidin I don't understand your objection. I have made 3 clear questions. The reason I said that was because I have both searched on google for math.stackexchange.com questions regarding differentials up to page 10 and also I have searched in here thought the search bar for the same issue so there is no point to link them as I have already read them. – Adam Oct 24 '19 at 20:22
  • @conditionalMethod That is the same thing. If you have a derivative then there is the question whether you can break it up as a fraction. Alternatively when you have differentials then there is the question whether you can treat them as a derivative after you take them as a fraction. Same thing from two different points of view. – Adam Oct 24 '19 at 20:25
  • @ArturoMagidin First of I never complained, secondly I have done that in a summary (because I can't write about the specific problem I have with the dozens of answers) saying that my issues are, mainly that people answered in a specific framework - and even then they don't agree with one another - and also that people answered mainly focused on single variable calculus. I have already done what you asked. My questions are clearly stated in my numbered list. – Adam Oct 24 '19 at 20:29
  • "I have already done what you asked." Let's see: I asked that if you think this is not a duplicate and that those questions confuse you or don't answer your questions, you link to them and say why they don't answer your questions. You have not done so. So, no, you have not already done what I asked. Again, you seem to think that "I explained my problem and carefully enunciated it" is a response to "this looks like a duplicate." It's not. – Arturo Magidin Oct 24 '19 at 21:43
  • @ArturoMagidin Nevermind this is getting way too boring. Instead of either answer the 3 question or just ignore them you just want to play games with words. You can't expect me to link 50 questions and point out every single misleading point in them. I told you the problem with all of them are that people answered in a specific framework and mainly focused on single variable calculus. I don't know how much easier I can state my problem with those questions. Literally these two are the problems in all of them. – Adam Oct 24 '19 at 22:19
  • Most debate among mathematicians is short lived, because one of them is right and the other is wrong and they both eventually figure out which is which. – Matt Samuel Oct 25 '19 at 00:05
  • @ArturoMagidin I dont want to improve my chances by linking unhelpful to me answers. I have clearly stated my questions, so whomever can answer them can. Every time I ask something there is someone else to link answers. In order to not waste anyone's time I already did my homework and read the similar questions but yet people still starting linking them. Do you think I would waste my time by writing such an extensive question if I found any other answer helpful? It is not really difficult to see it. – Adam Oct 25 '19 at 01:36
  • Since you don’t care to improve your chances, and you don’t care to say why previous ansewrs aren’t good enough. I’m just going to go ahead and add the fifth vote to close as duplicate. While you may be perfectly fine wasting other people’s time because you don’t want to “waste” yours, that is not the way to request free help. – Arturo Magidin Oct 25 '19 at 02:40
  • @ArturoMagidin "I don't care to say why the previous answers aren't good enough"? I mentioned the ACTUAL reasons at least 5 times. Let me mention them once more 1. people answered in a specific framework and 2. mainly focused on single variable calculus. Just don't say again I didn't say the reasons. The last thing I can actually do is link the answers and write the same 2 reasons below them. Also I didn't link them because I didn't want to waste other people's time as I wasted mine for reading those dozen unhelpful answers. Don't try to spin my words. Just vote and go. – Adam Oct 25 '19 at 02:44
  • I’ll say what I think is accurate, whether you like it or not. Generic statements don’t explain why specific answers aren’t good enough. “mainly focused” does not mean exclusively focused, so what is the problem? And “specific framework” doesn’t mean anything without an explanation of just what it is you mean by “framework”. Mathematics? Standard calculus? Nonstandard analysis? Anyway, feel free to get the last word. My vote is cast. – Arturo Magidin Oct 25 '19 at 02:49
  • @ArturoMagidin Great. – Adam Oct 25 '19 at 02:51

3 Answers3

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The reason the definition of a differential doesn't allow you to treat derivatives as fractions is because you're not automatically allowed to divide by everything you see in an equation. You can't divide by a three dimensional vector, for example, and this example is just like that. Differentials are vectors.

In nonstandard analysis in a single variable derivatives are the standard parts of quotients of infinitesimals. There's no disagreement about that. But the infinitesimals are not differentials.

While it's useful sometimes to treat derivatives as fractions, it can lead to error. There's a famous example, I forget the exact circumstances, where you have functions $u, v, w$ with $$\frac{\partial u} {\partial v} \frac{\partial v} {\partial w} \frac{\partial w} {\partial u} =-1$$ which proves that you can't treat multivariable derivatives as fractions and expect to always get the right answer.

Matt Samuel
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Short(ish) answer to these particular questions:

  1. Can we actually treat differentials as fractions in EVERY framework, or is there today a debate about it?

No, you can't ALWAYS treat differentials as fractions. But no, there is no debate today about it.

In some contexts - for example, differential geometry, nonstandard analysis - differentials (or infinitesimally small numbers) have a precise mathematical definition. In such contexts you can reason formally with $dx$ and $dy$ separate from $dy/dx$. That means the answer to your question (2) is "yes".

In many applications - particularly physics - thinking in terms of differentials is a very good way to turn intuition about a problem into mathematics.

You can see each of these responses in posts on this site.

Related: Why can't the second fundamental theorem of calculus be proved in just two lines?

Ethan Bolker
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  • You can reason formally with $dx$ and $dy$. However, $dy/dx$ is still non-sense. Differential forms, as far as I know, cannot be divided. If $y$ is a smooth function on $\mathbb{R}$ then $dy = y' d_1$ where $d_1$ is the standard one-form and $y'$ is the derivative. Furthermore, $dx = d_1$. However, what is the meaning of $dy/dx$? How do you divide one-forms? I never seen this before. Thus, even though you can reason with differentials separately, dividing them, as done in first-semester calculus, is still non-sense (correct me?). Which is why I refer to $dy/dx$ as non-sense in my answer. – Nicolas Bourbaki Oct 24 '19 at 21:58
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    @NicolasBourbaki There are contexts (e.g. nonstandard analysis) where the quotient $dy/dx$ makes sense. – Ethan Bolker Oct 25 '19 at 00:58
  • The question that was asked if $dy/dx$ can interpreted as a quotient of two differential forms. Not from the point-of-view of non-standard analysis. I see it said a lot, "you can make differentials rigorous using differential forms". But I disagree. You can make $dy$ and $dx$ rigorous and treat them separately. You cannot define $dy/dx$, however. Therefore, I stand by my statement that $dy/dx$ is a meaningless concept which is more of trick-of-notation, as I never never seen differentials divided in any manifolds book. – Nicolas Bourbaki Oct 25 '19 at 01:44
  • @NicolasBourbaki I don't think the OP's question was limited to differential forms. In any case I don't think we disagree about those. – Ethan Bolker Oct 25 '19 at 01:51
  • I am glad that we agree that $dy/dx$ cannot be defined rigorously using differential forms. I only bring this up because I seen people mention differential forms a lot whenever a student asks this common derivative-notation question. But in truth their responses do not really answer the question of what is $dy/dx$. It seems that the only rigorous way of making sense of that quotient is, like you said, non-standard analysis. Are you familiar with any other approach? – Nicolas Bourbaki Oct 25 '19 at 02:01
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This question comes a lot, and the best answer, to when you are new to learning calculus is to simply regard $\frac{dy}{dx}$ as non-sense which gives you a correct answer. Think of it as notation, with its own rules, that when you learn how to use it, will lead to the correct answer. Do not try to understand it.

I would suggest to learn calculus both the "non-sense" way and the precise way. The precise way you will avoid using this notation and instead rely on the properties of the derivative. This way you would be able to convince yourself that the trick-in-differential-notation leads to the answer that you would get if you did it precisely.

Nicolas Bourbaki
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  • Thank you for your answer, but I am not new to calculus and my whole point is that I try to understand it and not to pass an exam, or learn calculus. I have spent a lot of time thinking that is would be ok to just go with it, but after a while I even see papers trying to explain it (like the Extending the Algebraic Manipulability of Differentials) so I see that there is still a debate. – Adam Oct 24 '19 at 20:45
  • @Adam My answer was not trying to imply "do not think about what it means and just learn it". My answer was really trying to say, "differentials are mnemonic devices, which mathematically do not make any sense, but somehow when you use them you get the right answer in the end".

    When I learned calculus I had the same problem. Differentials never made sense to me. Until one day I had a huge revelation. They are not supposed to make sense. They are just symbols that give the right answers. That is all. And the struggle that everyone has in trying to explain something which has no meaning

     – Nicolas Bourbaki Oct 24 '19 at 20:55
  • @Adam Read Spivak's Calculus on manifolds, chapter 2. Very short. Once you know what differentials are and what derivatives are, then you will know. And it looks like Bourbaki should also read it, such that they know that differentials are not just mnemonic devices, but actual functions which importance becomes more apparent in several variables, and even more when doing calculus in manifolds other than $\mathbb{R}^n$. – conditionalMethod Oct 24 '19 at 21:01
  • @conditionalMethod Chapter 2 is as elementary as it gets what does it have to offer? Did you think that I don't know the definitions and theorems? – Adam Oct 24 '19 at 21:07
  • @conditionalMethod I read Spivak's book. I know what differential are. But they do not answer the question properly. Like I said $dy/dx$ is non-sense, just like Delta function is non-sense, and telling a first-semester calculus student to "go read Spivak" is a serious misunderstanding of pedagogy. – Nicolas Bourbaki Oct 24 '19 at 21:35
  • @Adam Is not that I think, I know that you don't know what a differential is. Just like Bourbaki doesn't know it either, just like they don't know what the delta function is. – conditionalMethod Oct 25 '19 at 00:13
  • @conditionalMethod You are insulting a lot of people without having any idea of their background. Wow. – Adam Oct 25 '19 at 00:17
  • @Adam If you knew, you wouldn't have asked this question AND you wouldn't have asked it wrong. You didn't even know what you wanted to ask. – conditionalMethod Oct 25 '19 at 00:19
  • @conditionalMethod And if you knew of my background and had read the comments in which I even link research papers asking the same question, you would just delete your comment in shame. Insulting people for any reason in forums is one of the best ways to downgrade yourself. – Adam Oct 25 '19 at 00:20
  • @conditionalMethod Okay, so prove me wrong. Give me a rigorous definition of what it means to be $dy/dx$. You can define $dy$, and you can define $dx$, but you cannot define a quotient of two differential forms. Therefore, $dy/dx$ is meaningless. The fact that you fail to realize this proves that you have no idea what you are talking about. Oh, and by the way, read Spivak in chapter 4, not chapter 2, but I am not surprised since you apparently never read his book. – Nicolas Bourbaki Oct 25 '19 at 01:41
  • @Adam Getting back to the topic at hand. I do not think that $dy/dx$ can be defined rigorous by using differential forms, since the quotient of differential forms is not a defined operation. ConditionalMethod has no idea what he is talking about if he fails to realize this simple point. – Nicolas Bourbaki Oct 25 '19 at 01:48
  • @NicolasBourbaki Thanks, I see too much controversy regarding differential forms and that fraction. – Adam Oct 25 '19 at 01:50
  • @Adam The fact that there is "controversy", should indicate that it is a non-rigorous concept. I should clarify, in case you are interested. You can give a precise definition for $dx$. And if $y$ is a function then $dy$ is a meaningful operation and would be equal to $y'\cdot dx$. So both $dx$ and $y'\cdot dx$ are meaningful. However, if you write $dy/dx = (y'\cdot dx)/(dx)$ then it becomes meaningless since these differentials cannot be divided. But if we are "sloppy", and just cancel, then we get $dy/dx = y'$. However, the cancellation is not justified. So it is just notation magic. – Nicolas Bourbaki Oct 25 '19 at 01:56
  • If you had only read that elementary book that I linked you would know that $df$ is a linear function that inputs an element of the tangent space and outputs a real number, likewise $dx$ is yet another linear functions that inputs an element of the tangent space and outputs a real number. Already from being two real functions on the same space you can divide them, which gives you yet another function on the tangent space. Now, all these functions also depend on the point at which the tangent space is taken. So, $df/dx$ is a function both of the increment and on the point. – conditionalMethod Oct 25 '19 at 02:16
  • When you take the quotient $df/dx$ the result is constant with respect to the element of the tangent space. So, it gives you a function of the point. The function of the point that it gives is the derivative. All of this is trivial. But you don't happen to know it well. – conditionalMethod Oct 25 '19 at 02:19
  • And that's enough time spent explaining things to cranks. – conditionalMethod Oct 25 '19 at 02:20
  • @conditionalMethod As I suspected, it does not answer my question that I asked you. You cannot define the quotient of two differential forms. What you are essentially doing is, writing $f'(p) = df(D_p)/d(D_p)$ where $D_p$ denotes the derivative at $p$ (so $D_p$ is a vector for the manifold $\mathbb{R}$ at $p$). If you eliminate the base point then you are saying $f' = df(D)/d(D)$ where $D$ is the vector field defined by the derivative. Therefore, you did not define $df/dx$, you rather defined $df(D)/d(D)$. Which you had to do because the quotient $dy/dx$ is not defined. – Nicolas Bourbaki Oct 25 '19 at 02:47