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The "school identities with derivatives", like $$ (x^2)'=2x $$ are not identities in the normal sense, since they do not admint substitutions. For example if we insert $1$ instead of $x$ into the identity above, the appearing equality will not be true: $$ (1^2)'=2\cdot 1. $$ That is why when explaining this to my students I present the derivative in the left side as a formal operation with strings of symbols (and interpret the identity as the equality of strings of symbols).

This however takes a lot of supplementary discussions and proofs which look very bulky, and I have no feeling that this is a good way to explain the matter. In addition, people's reaction to this my question makes me think that there are no texts to which I could refer when I take this point of view.

I want to ask people who teach mathematics how they bypass this difficulty. Are there tricks for introducing rigor into the "elementary identities with derivatives" (and similarly with integrals)?

EDIT. It seems to me I have to explain in more detail my own understanding of how this can be bypassed. I don't follow this idea accurately, in detail, but my "naive explanations" are the following. I describe Calculus as a first-order language with a list of variables ($x$, $y$,...) and a list of functional symbols ($+$, $-$, $\sin$, $\cos$, ...) and the functions which are not defined everywhere, like $x^y$, are interpreted as relation symbols (of course this requires a lot of preparations and discussions, that is why I usually miss these details, and that is why I don't like this way). After that the derivative is introduced as a formal operation on terms (expressions) of this language, and finally I prove that this operation coincide with the usual derivative on "elementary functions" (i.e. on the functions which are defined by terms of this language).

Derek Elkins suggests a simpler way, namely, to declare $x$ a notation of the function $t\mapsto t$. Are there texts where this is done consistently? (I mean, with examples, exercises, discussions of corollaries...)

@Rebellos, you identity $$ \frac{d}{dx}(x^2)\Big|_{x=1}=2\cdot 1 $$ becomes true either if you understand the derivative as I describe, i.e. as an operation on expressions (i.e. on terms of the first order language), since in this case it becomes a corollary of the equality $$ \frac{d}{dx}(x^2)=2\cdot x, $$ or if by substitution you mean something special, not what people usually mean, i.e. not the result of the replacement of $x$ by $1$ everywhere in the expression (and in this case you should explain this manipulation, because I don't understand it). Anyway, note, that your point is not what Derek Elkins suggests, since for him $x$ means a notation of the function $t\mapsto t$, it can't be substituted by 1).

Sergei Akbarov
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    I really think you are mis-interpreting what "inserting $x=1$" means for the specific expression. Actually, it is $$\frac{\mathrm{d}}{\mathrm{d}x}(x^2) \bigg|_{x=1} = 2\cdot 1 = 2$$ which is indeed correct. This is the first step for explaining to someone that $f'(1) \neq (f(1))'$. – Rebellos Jan 25 '19 at 20:20
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    Why don't you just tell them the true identity, which is $f'(x) = 2x$, for $f:x\mapsto x^2$ ? (or you could see $(x^2)' = 2x$ as an equality between polynomials - not polynomial functions; and be done with it; but that's probably not the right thing to do if you're having this kind of difficulty with the students) – Maxime Ramzi Jan 25 '19 at 20:21
  • @Rebellos the writing $\frac{d}{dx}(x^2)$ means exactly the operation with strings of symbols, not an operation with functions. – Sergei Akbarov Jan 25 '19 at 20:23
  • @Max of course I give them the accurate definition of derivative and give the examples (including the one the you suggest). But I can't avoid writings like $(x^2)'$ since they simplify calculations. – Sergei Akbarov Jan 25 '19 at 20:26
  • Related. – Git Gud Jan 25 '19 at 21:39

4 Answers4

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There are at least a few approaches to this.

At a conceptual level the key is that differentiation acts on functions, not numbers. This leads to the first approach. Often notation like $\text{D}f$ is used where $f$ is a function. From this perspective, $(x^2)'$ is like $\text D(x\mapsto x^2)$. This makes $x$ a bound variable. We can't just replace it with a number. Naively done that would produce something like $\text D(1\mapsto 1^2)$ which is nonsense. We could interpret $(1^2)'$ as $\text D(x\mapsto 1^2)$ which would work fine and produce the constantly $0$ function as desired. Modulo some technicalities, this is closest to what's happening in the standard approach.

An alternative, more algebraic approach is to consider things like derivations. Here, instead of having functions, we have algebraic terms. $x$ ceases to be a variable and is instead a special constant for which we assert the equality $\text D x = 1$. In this case, it wouldn't make sense to naively replace $x$ with $3$, say, because that would be like saying "let $5$ be $7$." Alternatively, an interpretation that mapped $x$ to $3$ would likely be unsound. You could make a sound interpretation of $x$ as the identity function though and interpret the derivation operator as the $\text D$ operator from the first approach. Indeed, you could work with the algebraic notation but in terms of this particular interpretation of it. At any rate, this algebraic approach is very computational but not as flexible since you are limited to the terms of the algebra as opposed to being able to differentiate any (differentiable) function you can specify. On the other hand, this algebraic approach is applicable even in contexts where notions like limits don't make sense. This approach avoids having to deal with free/bound variables could likely be pretty jarring.

A kind of middle approach would be to restrict entirely to analytic functions and to work in terms of power series. If you want to go further, you could even identify power series with special (or, for formal power series, arbitrary) sequences of numbers. You could understand differentiating and integrating power series entirely as manipulations of sequences of numbers. More prosaically, you could work with the usual presentation of power series but emphasize manipulating coefficients. There's a lot of beautiful and useful power series manipulation calculations that could be introduced here that is useful for things like generating functions and Laplace/Z transforms.

There are other approaches as well, such as synthetic differential geometry or non-standard analysis. These also avoid some of the issues with bound variables, but introduce other issues and are less connected to standard practice in some ways. (Potentially, more connected to what mathematicians actually do in some ways.) I would say, more generally, that calculus is one of the first places in a typical (American, at least) math education that students have to start seriously grappling with notions of free and bound variables. Unfortunately, the notation of calculus makes this very confusing and opaque. I would recommend using notation that makes things more explicit (an extreme form is Structure and Interpretation of Classical Mechanics, but you don't need to go that far) and spend explicit time in the course teaching and practicing working with bound variables.

Derek Elkins left SE
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    +1 for the first paragraph. I'd suggest just a tiny modification to what OP is already doing, namely, write $(x \mapsto x^2)' = (x \mapsto 2x)$. To do this, you have to say that anything with a "mapsto" arrow, in this notation, automatically denotes a function on the reals. And you have to teach the students that it's equally valid to write $(x \mapsto x^2)' = (t \mapsto 2t)$, because the two right-hand-sides denote the same object (the "doubling" function on the reals). – John Hughes Jan 25 '19 at 21:35
  • Derek, @JohnHughes: writing $(x\mapsto x^2)'=(x\mapsto 2x)$ is another equation, not the one that I am asking about. I never saw somebody using this writing (apparently because it makes calculations bulky). But the writing $(x^2)'=2x$ is used almost everywhere (and it is indeed useful). – Sergei Akbarov Jan 26 '19 at 05:24
  • Derek, and your second approach is more or less what I am writing about: representation of derivative as a formal operation on terms of the language. This is the trick that I use to explain the situation, but in my opinion, it is again too bulky, because it requires two notions of derivative and a proof that they coincide on the functions defined by terms (i.e. on elementary functions). I thought maybe there are other, more simple tricks. – Sergei Akbarov Jan 26 '19 at 05:30
  • @SergeiAkbarov You don't need two notions of derivative (though that is probably the cleaner approach). You can, as I suggested, work within the semantics of the algebraic theory. That is, you can let $x$ be a different name for the identity function and $1$ mean the constantly $1$ function, and then it is just true that $\text Dx=1$ where $D$ is the differential operator from the first paragraph. That said, I haven't seen anyone try to seriously stick with this view, let alone teach it in an introductory class in lieu of a more standard presentation. – Derek Elkins left SE Jan 26 '19 at 05:48
  • Derek, if the solution is to make $x$ a notation of a concrete function, $t\mapsto t$, then the calculations become very unusual, for example, the chain rule for $\sin(x^2)$ becomes the following: $$ (\sin(x^2))'=(\sin x\circ x^2)'=\Big((\sin x)'\circ x^2\Big)\cdot (x^2)'=\Big(\cos x\circ x^2\Big)\cdot 2x=\cos x^2\cdot 2x. $$ This looks strange. – Sergei Akbarov Jan 26 '19 at 06:31
  • And I never saw this anywhere. – Sergei Akbarov Jan 26 '19 at 06:37
  • @SergeiAkbarov I agree it can look strange, though your example is a bit of a strawman. You could write $$(\sin(x^2))'=\sin'(x^2)(x^2)'=\cos(x^2)2x$$ which is almost identical to what you'd normally write. You don't have to explicitly write out composition, even though that is what's happening. You can even give the rules directly in terms of this notation. Any approach that avoids dealing with bound variables is going to look at least a little strange some of the time. I don't recommend this. Students getting comfortable with bound variables is important and should be attacked directly. – Derek Elkins left SE Jan 26 '19 at 07:20
  • Derek, it seems to me, the calculations can be much longer, in particular, we can't avoid $\circ$, look at this for example: $$ (\sqrt{1+x^2})'=(\sqrt{x}\circ (1+x^2))'=\Big((\sqrt{x})'\circ (1+x^2)\Big)\cdot (1+x^2)'=\Big(\frac{1}{2\sqrt{x}}\circ (1+x^2)\Big)\cdot 2x=\frac{x}{\sqrt{1+x^2}} $$ – Sergei Akbarov Jan 26 '19 at 08:47
  • Derek, and I don'tunderstand this idea of using bound variables without formal rules... I believe, there must be an explanation in terms of first order language, or I don't know in terms of the theory of real numbers. – Sergei Akbarov Jan 26 '19 at 08:59
  • You asked "which tricks to folks use?" I answered that I slightly modify the "school identities," which are nonsense, to make them interpretable in a sensible way, namely as statements about functions rather than statements about expressions. "I've never seen that" isn't really a good response. You might as well tell me that you want to keep using your screwdriver as a chisel, and just need to know how to sharpen it! – John Hughes Jan 26 '19 at 12:11
  • @JohnHughes, your suggestion, replacing $(x^2)'=2x$ by $(x\mapsto x^2)′=(x\mapsto 2x)$, makes the chains of calculations so heavy that it is not clear whether it is still possible to use them at all after that. Did you try to do this yourself? And how will you find the derivative of, say, $\sin ^2x$? Like this: $$ (x\mapsto \sin^2x)'=\Big((y\mapsto y^2)\circ(x\mapsto \sin x)\Big)'=\Big((y\mapsto y^2)'\circ(x\mapsto \sin x)\Big)\cdot(x\mapsto \sin x)'=... ? $$ My students will kill me when seing this "trick". – Sergei Akbarov Jan 26 '19 at 14:33
  • If your students don't want to understand what they're doing, that's their problem, not mine. Feel free to ignore my advice. BTW, the ellipsis in your chain is simply $((y \mapsto 2y) \circ (x \mapsto \sin x) \cdot (x \mapsto \cos x) = (x \mapsto 2 \sin x \cos x)$. That doesn't seem so bad to me, but opinions differ, of course. – John Hughes Jan 26 '19 at 14:37
  • @JohnHughes, the way I use to explain this -- giving two definitions of derivatives with proof that they lead to the same result on the class of elementary functions -- is rigor, and my students understand this. I was only wondering if there are other, simpler ways to explain this. What you suggest is correct, of course, but its benefits are doubtful for me, since it makes calculations several times longer. – Sergei Akbarov Jan 26 '19 at 14:48
  • John, and excuse me, BTW, $\sin^2x$ is not the same as $\sin 2x$. :) – Sergei Akbarov Jan 26 '19 at 15:02
  • The derivative of $\sin^2 x$ is indeed $2 \sin x \cos x$, which is exactly the result I arrived at. Perhaps my method is actually better than yours. :) [To be clear: I was replacing only the ellipsis in your comment (the "..."), not the whole computation.] – John Hughes Jan 26 '19 at 15:44
  • @DerekElkins, excuse me, how do people interpret integral in this language? $$\int f(x) d x=F(x)+C$$ And the partial derivatives like $$\frac{\partial(x^2y^3)}{\partial y}=3x^2y^2$$ – Sergei Akbarov Feb 10 '19 at 16:59
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The domain of the derivative is functions, not numbers. That's what at the bottom of this.

ncmathsadist
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One way to do it is to say $f’(x)$=2x, and define $f(x)=x^2$. This way, it should be relatively clear that x is not some placeholder variable, but rather the independent variable in an equation. Additionally, if you introduce the Chain Rule, it should become even more clear. If students still try to use function composition without using the Chain Rule, switching to only using Leibniz notation for a while should quickly stifle that urge. You can then introduce prime notation again after the desire to directly compose in functions has been suppressed.

H Huang
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There is no one right answer to this question. It depends on what the students are willing to accept. One approach mentioned comes from the Wikipedia article on Dual numbers. The key idea is that $\, f(x+\epsilon) = f(x)+f'(x)\,\epsilon\,$ where $\,\epsilon^2=0\,$ is postulated. Given this, then just use ordinary algebra $\,(x\!+\!\epsilon)^2 = (x\!+\!\epsilon)(x\!+\!\epsilon) = x^2\! +\! 2x\,\epsilon \!+\! \epsilon^2 = x^2\! +\! 2x\,\epsilon \, $ and therefore $\,(x^2)' = 2x.\,$ Notice that here you can do substitutions. For example, $\, x \to 3+\epsilon,\,$ and then $\,x^2 \to (3+\epsilon)^2 = 9+6\,\epsilon,\,$ and thus $\, (9+6\,\epsilon)' = 6\,$ where we define $\, (a+b\,\epsilon)' := b.$

Somos
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