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Calculus instructors are mostly very careful to explain that $\frac{\mathrm{d}y}{\mathrm{d}x}$ is not a fraction, and multiplying both sides of an equation by $\mathrm{d}x$ is nonsense, wrong, or evil.

However, I really want to be able to say that "$y$ is the sum of all the little pieces of $y$," i.e. that $y = \int \mathrm{d}y$. And I'd really like to say that when I integrate $\frac{\mathrm{d}y}{\mathrm{d}x} \mathrm{d}x$, it's the same as integrating $\mathrm{d}y$. And when I separate variables, I really like splitting up the $\frac{\mathrm{d}y}{\mathrm{d}x}$.

This comes to the question: what is a good example of the harm that we will cause if we allow students to think $\frac{\mathrm{d}y}{\mathrm{d}x}$ is a fraction? Are we worried that students will think that $\mathrm{d}y$ and $\mathrm{d}x$ are numbers?

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    $\begingroup$ This question has been asked before on MO: mathoverflow.net/q/73492/22971 ("How misleading is it to regard dy/dx as a fraction?") $\endgroup$
    – Benjamin Dickman
    Aug 28, 2014 at 2:03
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    $\begingroup$ (Though the answers at a similar MSE query linked from MO - math.stackexchange.com/q/21199/37122 - may prove to be more useful.) $\endgroup$
    – Benjamin Dickman
    Aug 28, 2014 at 2:08
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    $\begingroup$ There's no reason why you can't think of dx and dy as one forms on xy space. You can't divide one forms but if you have a relation like dy = 2xdx then you can think of that as picking out a one-dimensional subspace defined by the one form dy - 2xdx. On that subspace dy and dx are still one forms but indistinguishable from scalars so you can form their ratio. This doesn't work with partials because those aren't ratios of one-forms but inner products of one forms with basis vectors. $\endgroup$
    – Francis Davey
    Aug 28, 2014 at 9:44
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    $\begingroup$ Can't you say that integrating $\dfrac{dy}{dx} dx$ is the same as integrating $dy$ "because there's a theorem that says so", rather than saying that integrating $\dfrac{dy}{dx} dx$ is the same as integrating $dy$ "because $\dfrac{dy}{dx} dx$ is a fraction and the $dx$s cancel"? $\endgroup$
    – Steve Jessop
    Aug 28, 2014 at 12:46
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    $\begingroup$ Just a small note that people have tried to rigorize the use of $dy$ and $dx$ as infinitesimals, calling it non-standard analysis. I don't really know anything about it, but I guess it's at least a disclaimer that there is a whole other Wikipedia article about criticism of non-standard analysis. $\endgroup$
    – MGA
    Aug 28, 2014 at 15:14

7 Answers 7

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If $y$ is a highly oscillating function of $x$, it can become quite unclear exactly what is going on with domains of integration if one tries to write out in precise terms the kind of "integration identity" you would like to express (in terms of definite integration). My high school physics teacher used exactly the kind of reasoning your are suggesting (with integrating $(dy/dx)dx = dy$ to get $y$) as a way of suggesting that the Fundamental Theorem of Calculus is almost a triviality, or a matter of defining notation (without carefully thinking through why the notation is well-posed, etc.).

Since we know that the Fundamental Theorem of Calculus is a real theorem, the danger of putting too much emphasis on the fraction notation is that it deludes students into thinking that FTC has no content, or is some kind of triviality. For most students the goal can't be to make them appreciate the subtleties of definitions and proofs of theorems, but it would be good to at least make sure they understand that the fractional formalism gives "right answers" because it is well-designed notation masking the content of some real theorems (a fact which is hard for them to accept if they think the derivatives literally are fractions). That isn't to say one should completely divorce the intuition behind the notation from the properties one wants it to satisfy, of course.

Overall, the fractional notation does work really well (for Chain Rule, u-substitution, derivative of inverse functions, etc.), so I wouldn't belabor the point after making one attempt with an example of non-invertible $y$ that the possibility to compute "as if" $y$ were a variable in its own right is an amazing feature of Leibniz' notation.

That being said, I think the situation with the multivariable chain rule mentioned in one of the other answers is a very good illustration of why too much blind faith in the fractional notation without carefully thinking through why it is working will lead students into traps when they move on to multivariable calculus (and the reality is that most students cannot be expected to think very carefully about the notation when they haven't mastered or been taught the precise definitions).

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  • $\begingroup$ The other answers are also very good, but I really think the earliest key point is that integrating $\dfrac{dy}{dx}dx$ to get $y$ is not a triviality (it's a rather important theorem!), so I'm accepting this one. $\endgroup$
    – Chris Cunningham
    Aug 29, 2014 at 17:09
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    $\begingroup$ @ChrisCunningham I think it is important to realize that the idea of the fundamental theorem of calculus is kind of trivial. A formal proof might be a bit tricky, but it is fundamentally a very intuitive notion. Worth making sure students realize that. $\endgroup$
    – Steven Gubkin
    Aug 30, 2014 at 2:10
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    $\begingroup$ @StevenGubkin: The idea of the proof of Serre's GAGA theorem comparing complex-analytic and complex-algebraic geometry is also kind of trivial...as is the idea of the proof of effectivity of faithfully flat descent for morphisms of schemes...and so are the ideas of the proofs of lots of other deep results once someone first has the idea! It does not help students new to a mathematical topic to tell them that the basic idea behind a key theorem is "kind of trivial". Hindsight is 20-20, but foresight is not. I believe that your advice, while well-intentioned, will not help students. $\endgroup$
    – user52824
    Aug 30, 2014 at 3:21
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    $\begingroup$ @user52824 I think most students (at least at my school) do not have even an intuitive understanding of FTC. Conveying some kind of intuition about FTC is important. If this intuition is "$\frac{dy}{dx}$ is the little change in $y$ divided by the little change in $x$, so $\frac{dy}{dx}dx$ is the little change in $y$, so if I add all of these up I get the change in $y$ over the whole interval", I think that is great. They should understand it is not a formal proof, but it is still better than just memorizing a formula. $\endgroup$
    – Steven Gubkin
    Aug 30, 2014 at 15:10
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    $\begingroup$ Then we are in complete agreement! I always avoid words like "trivial" and "easy" when talking to students. $\endgroup$
    – Steven Gubkin
    Aug 31, 2014 at 0:47
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A big issue I've seen is in understanding the multivariable chain rule. I've had many students over the years argue that $$\frac{\partial f}{\partial t} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial t}$$ implies $$\frac{\partial f}{\partial t} = 2\frac{\partial f}{\partial t}, \text{ so } \frac{\partial f}{\partial t}=0$$ since the $\partial x$ and $\partial y$ cancel.

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    $\begingroup$ The problem here isn't that the student is reasoning using infinitesimals. You can make up an example with $f(x,y)$ a bilinear function, and $x(t)$ and $y(t)$ linear functions, convert all the $\partial$s to $\Delta$s, and the mistake in the calculation is the same mistake, even though everything is now a real number and no calculus is involved. The mistake has nothing to do with calculus. The mistake is a logical mistake about not understanding that one variable is kept constant while another one is varied. $\endgroup$
    – user507
    Aug 28, 2014 at 3:16
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    $\begingroup$ Indeed the problem isn't that the student is reasoning using infinitesimals. The student in fact isn't doing that. The problem is that the student is treating the partial derivative notation as a fraction, which is what was asked for except that "partial" isn't what was in the question. $\endgroup$
    – Steve Jessop
    Aug 28, 2014 at 12:36
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    $\begingroup$ I am highly sympathetic to this answer (and it's the first one that sprang to mind), but I think that there may be an important subtlety: as Ben Crowell points out below (matheducators.stackexchange.com/a/4288/2070), there is a rigorous sense in which $\text dy/\text dx$ really is a fraction, so that no logical contradiction can result from treating it as such. (I think that there is no rigorous sense in which $\partial z/\partial x$ and $\partial z/\partial y$ are simultaneously fractions; but do not know enough of infinitesimals to be confident saying so.) $\endgroup$
    – LSpice
    Aug 29, 2014 at 23:28
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    $\begingroup$ (The punnish confusion between being "really a fraction", i.e., genuinely such an object, but not "a real fraction", i.e., a fraction of real numbers, was inadvertent.) $\endgroup$
    – LSpice
    Aug 29, 2014 at 23:29
  • $\begingroup$ Btw: when Jacobi popularized the standard notation for partial derivatives, he was well aware of the real problem highlighted by Ben Crowell above. Unfortunately, Jacobi's suggestion for fixing the notation probably lead to even more confusion. $\endgroup$
    – Michael Bächtold
    Mar 3, 2018 at 14:09
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When Leibniz invented the notation, he considered $dx$ and $dy$ to be infinitesimal numbers and $dy/dx$ to be their quotient. Non-standard analysis (NSA) has essentially vindicated that view. One has to distinguish between the quotient $dy/dx$ and its standard part, which is the derivative. In fact, Leibniz made this distinction [Blaszczyk 2012], although it was not carefully observed by later users of his notation.

Keisler is a very nice, free freshman calc book that develops calculus from the NSA point of view.

There was a period ca. 1820-1965 when this stuff was not clearly understood, and it was believed, incorrectly, that limits were the only possible way to give calculus a rigorous foundation. Even during this period, scientists and engineers never stopped using infinitesimals. Freshman calc students need to be literate in the practices of science and engineering, so a disservice is done to them when the discredited belief is inculcated upon them that these practices are somehow wrong, ignorant, or logically suspect.

This comes to the question: what is a good example of the harm that we will cause if we allow students to think $\dfrac{dy}{dx}$ is a fraction? Are we worried that students will think that $dy$ and $dx$ are numbers?

There is no such harm. Students should of course be told that $dx$ and $dy$ are not real numbers. In old-school terminology, they're infinitesimal numbers. In the termninology of NSA, they're hyperreal numbers.

Blaszczyk et al., "Ten Misconceptions from the History of Analysis and Their Debunking," http://arxiv.org/abs/1202.4153

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    $\begingroup$ Indeed. And it is still very widely believed, incorrectly, that only the Cauchy-Weierstrass approach can legitimize calculus. And caricatures of Leibniz (and Euler's) viewpoints on infinitesimals are still widely taken as accurate. Berkeley's objections had substance, but the whole thing was inconclusive even at the time. $\endgroup$
    – paul garrett
    Aug 28, 2014 at 19:11
  • $\begingroup$ I don't think that Berkeley's objections really had much substance. He combined a popular notion about mathematics (though from mathematicians) that mathematics never leaves off even the tiniest portion from its answers, with the notion of treating higher-order infinitesimals as zero, and established a contradiction. This is exactly analogous to what we now call "taking the standard part", but Berkeley did not or chose not to understand, because he believed in other mathematician's insistence that no piece, no matter how small, be left behind. $\endgroup$
    – johnnyb
    Mar 6, 2018 at 18:09
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On the same line of Santiago Canez's answer, an identity that appears all over thermodynamics is:

$$\frac{\partial x}{\partial y}\bigg|_z\frac{\partial y}{\partial z}\bigg|_x\frac{\partial z}{\partial x}\bigg|_y=-1$$

(where there exists $f(x,y,z)=0$)

The first approach of many people is to apply the chain rule, thereby not understanding where the minus sign came from.

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I think the problem is understanding the line between intuition and rigor.

The Riemann rearrangement theorem? Intuitive operation (you can just move around terms right?) unintuitive result.

Similar for things like $1+2+3+\cdots = -1/12$.

So with derivatives, it's like to think they are like fractions since $dx$ and $dy$ are like infinitescimals (and per Ben Crowell's answer even that viewpoint can be rigorized), but numerous theorems, such as the chain rule, are required to vindicate this view. Perhaps show a proof of the chain rule guided by our understanding of them as fractions, but remember to emphasize the technicalities it takes to underpin this.

Also note that the hypotheses are essential. These rules don't work when things aren't differentiable, for instance.

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    $\begingroup$ I'd not want to tell students that there was a line between intuition and rigor, because I think that is a false distinction. It'd be bad to get people flinching every time they need to "move terms around", because mostly they should do so. One simply wants a refined intuition, not a belief that some sort of "technicalities" are waiting to ensnare us. The goal is not "avoiding trouble" but doing something constructive. $\endgroup$
    – paul garrett
    Aug 28, 2014 at 19:09
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    $\begingroup$ @paulgarrett so thinking of dy/dx as a fraction is unrefined intuition? $\endgroup$
    – djechlin
    Aug 28, 2014 at 19:22
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    $\begingroup$ I do think that it is a good, but unrefined intuition, yes. (Btw, for me, "fraction" and "ratio" and ... are approximate synonyms, despite arguable technical differences. I don't have space in my head to keep track of "technical details" :) ) But/and, then, the little troubles that a too-naive interpretation of "fraction" engender must be dealt with sanely, not over-reacting by throwing the baby out with the bathwater, and all that metaphor... From the other side, I don't think "rule-based" mathematics is very effective/potent, however "rigorous" it may be. Rules are fundamentally artificial. $\endgroup$
    – paul garrett
    Aug 28, 2014 at 22:40
  • $\begingroup$ I think there are two disciplines of mathematics, but few people recognize the distinction. It is similar to the author/editor distinction in writing. The author does learn the rules of grammar, but not to the extent that an editor does. The goal of the author is creativity and exploration, and the goal of the editor is to rein the author in a bit and make sure they keep within the rules. Both of these are important functions of the English language. Some people do both, but they are recognized as distinct tasks, and we don't presume authors to always be editors. $\endgroup$
    – johnnyb
    Mar 6, 2018 at 18:59
  • $\begingroup$ Likewise, with math, some people explore while other people prove. Many people do both of course, but, requiring that everyone follow the same path seems odd, and, for the most part, just like we it is more productive to have more people writing than editing, it is more productive to have people thinking with mathematics, even if a "math editor" has to come along and point out some rule that is being violated. For instance, the Umbral Calculus led to many solutions long before it was ever proved why it worked. If we required mathematicians to all be editors we would have a lot less math. $\endgroup$
    – johnnyb
    Mar 6, 2018 at 19:03
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I am not sure that it has the same rigorous foundation as Santiago Canez's answer, but I think that, if one wanted to argue for the mathematical legitimacy of the notation $\frac{\text dy}{\text dx}$, then one could view it (pointwise) as a quotient of elements of a cotangent space $T^*$ that is 1-dimensional, for which therefore the scalar multiplication $\mathbb R \times (T^* \setminus \{0\}) \to T^*$ may be 'inverted' to give a division $T^* \times (T^* \setminus \{0\}) \to \mathbb R$. In this sense, it seems to me not so different from the common device of writing $Q - P$ for the vector $\vec{PQ}$ (i.e., viewing Euclidean space as an affine space under $\mathbb R^n$). Of course, such a viewpoint should never be explicitly mentioned in the calculus classroom; but I remember that it was actually advanced in a Monthly article that, unfortunately, I can no longer find.

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  • $\begingroup$ @Hurkyl, thanks. I've fixed it. $\endgroup$
    – LSpice
    Mar 10, 2018 at 23:20
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We are so careful because $dy/dx$ is the limit of a sequence of fractions. This limit is not "the final term" of the sequence and does not entail that all terms of the sequence exist. We should be even more explicit when teaching this matter since there are even some contemporary logicians confessing in public to believe in the idea that the terms of a sequence could be exhausted step by step until the index $\omega$ is reached - in contradiction with the successor axiom.

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    $\begingroup$ You are the only one on math educators that uses the term successor axiom, as a simple search reveals. Anything to do with ZFC which we usually associate with theories about ω ? $\endgroup$
    – user7145
    Mar 28, 2018 at 11:52
  • $\begingroup$ @Jan Burse: Usually we talk about Zermelo's successor function. Since it is anchored in an axiom, it suggests itself to call this part of the axiom of infinity the successor axiom. Others do this too: Two roads to the successor axiom rd.springer.com/article/10.1007/s11229-018-1752-5 mathoverflow.net/questions/66776/alternative-arithmetics $\endgroup$
    – user37237
    Mar 28, 2018 at 18:55
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    $\begingroup$ There is no successor axiom in Zermelos 1908 paper which is the founding of ZFC. The axioms all have different names. Also axioms do not have parts. Only if you join two axioms together by conjunction. But the axiom of infinity doesn't have such parts. Likely you are halucinating and/or you are highly confused about the axiom of infinity. $\endgroup$
    – user7145
    Mar 28, 2018 at 22:20
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    $\begingroup$ You can check yourself here: citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.85.3071 , (PA5) , we expect a successor relationship σ to be injective, but the axiom of infinity says nothing about a Zermelo successor being injective. Other axioms would do that. So your claim to read off a sucessor relationship from the axiom of infinity is baseless. $\endgroup$
    – user7145
    Mar 28, 2018 at 22:30

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