Talk:Chain rule

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The First Example states "(f ∘ g)(t) is the atmospheric pressure the skydiver experiences t seconds after his jump". It's not that simple, because the distance fell after t-seconds is a tiny bit less than that given by g(t) due to buoyancy. I expect you would need to use a differential equation to model the physics in the first example rather than a simple composite function.— Preceding unsigned comment added by MathewMunro (talk • contribs) 09:26, 14 February 2021 (UTC)Reply

I have removed this example as too technical here (this is not an article of physics), and totally irrealist (the aerodynamic force is much more important than buoyancy and is not considered). D.Lazard (talk) 10:19, 14 February 2021 (UTC)Reply

@MathewMunro: please sign all your talk page messages with four tildes (~~~~) — See Help:Using talk pages. Thanks.

Yes, I think user D.Lazard adequately took care of this . - DVdm (talk) 10:27, 14 February 2021 (UTC)Reply

If someone wants to re-do the recently deleted example, this might be useful: I found what is probably the simplest possible example of the Chain Rule: https://socratic.org/questions/5a6b377f7c0149554545a31c but of course you can't just plagiarise it. MathewMunro (talk) 11:02, 14 February 2021 (UTC)Reply

That source is a schoolbook example of an unreliable source. I also don't think the example should be redone, as, without a source likely having been a schoolbook example of wp:original research, it actually was a schoolbook example of a blatant wp:copyright violation of this (also unreliable) source. Good thing we got rid of it. - DVdm (talk) 13:17, 14 February 2021 (UTC)Reply

I can assure everyone that the example was not plagiarised because I invented it myself. Its appearance elsewhere is an example of other websites plagiarising Wikipedia rather than the other way around.

I wrote the example quite a long time ago. I think it's been sixteen years; it was before I had even decided to make an account. The example does not say that it is a physically correct description of falling through air, and indeed it was never intended to be. Rather, it was intended to demonstrate certain points which I found my students confused about. I thought the best way to do this was by means of an example where all the units involved were clear.

The specific points I wanted to demonstrate are in the last two paragraphs, which I'll quote here for reference. Notice the discussion of units! Also notice the discussion of various mistakes and how the units in those mistakes are nonsense!

For example, suppose that we want to compute the rate of change in atmospheric pressure ten seconds after the skydiver jumps. This is (f ∘ g)′(10) and has units of pascals per second. The factor g′(10) in the chain rule is the velocity of the skydiver ten seconds after his jump, and it is expressed in meters per second. ${\displaystyle f'(g(10))\!}$ is the change in pressure with respect to height at the height g(10) and is expressed in pascals per meter. The product of ${\displaystyle f'(g(10))\!}$ and ${\displaystyle g'(10)\!}$ therefore has the correct units of pascals per second.

Here, notice that it is not possible to evaluate f anywhere else. For instance, the 10 in the problem represents ten seconds, while the expression ${\displaystyle f'(10)\!}$ would represent the change in pressure at a height of ten meters, which is not what we wanted. Similarly, while g′(10) = −98 has a unit of meters per second, the expression f′(g′(10)) would represent the change in pressure at a height of −98 meters, which is again not what we wanted. However, g(10) is 3020 meters above sea level, the height of the skydiver ten seconds after his jump, and this has the correct units for an input to f.

I knew that some of my students would find this page. I hoped it would help them and possibly others. The fact that other websites have copied it is some evidence that it did!

Of course, Wikipedia is an encyclopedia not a textbook. In the intervening sixteen years, the rules on original research have become more strict, and my appreciation for those rules has also increased. Nowadays, I wouldn't feel quite so comfortable inventing an extended example like that and putting it onto Wikipedia.

Despite this, the example is still useful. What I would really like is if someone could find an example, with a citation to a textbook or paper, which clearly exhibited the same features: The equations involved should be simple, there should be some kind of obvious physical meaning, and the many mistakes that we all see when we teach calculus should result in physically meaningless quantities. That kind of example has encyclopedic value. Ozob (talk) 00:23, 15 February 2021 (UTC)Reply

@Ozob: there you go , if you invented it yourself, it is original research, and unless it is used/discussed/mentioned in the relevant literature to establish its worthiness to be included here, any discussion about its merit or validity is actually off-topic here. As you say, this is Wikipedia... - DVdm (talk) 11:04, 15 February 2021 (UTC)Reply

@DVdm: Well, of course it is. I said as much. But as I also said, I would like if someone were to find a similar example which could be cited. (Surely all it would require is perusing some textbooks...) A discussion of common chain rule mistakes is excellent content for an encyclopedia. Ozob (talk) 18:21, 15 February 2021 (UTC)Reply

It turns out that it's quite difficult to find examples that are of the kind I'm hoping for and that are also out of copyright. (I consider being out of copyright important here, since the article would essentially have to rework the example in full, and that might not be considered fair use.) As far as I can tell, calculus textbooks from a century or more ago drew their basic examples exclusively from analytic geometry. Everything is about curves and tangents. Physical applications, if they're included at all, are invariably in a later chapter, well after the main concepts have been introduced.

The best examples that I've found so far have been in Thompson's Calculus for the Practical Man, chapter IV. See [1]. I think problem (4), in particular (about a conical cup filling with water), would illustrate the same things that the skydiver example did; and it would be physically accurate and sourced. What does everyone else think? Ozob (talk) 23:03, 15 February 2021 (UTC)Reply

WP:Wikipedia is not a textbook, and the example was misplaced: placing a physical example before a mathematical definition may make sense in a textbook where readers are supposed to have there their first access to the subject. In the case of this article, readers may come here either by following a link or because they have encoutered the title somewhere. So they must have already heard of derivatives and function composition, and a motivating example before the definition is probably useless for them, and it may be confusing if it uses non-mathematical concepts that they do not master.

So, I strongly oppose this sort of example at this place. However, I do not oppose to add such an example later in the article, for example in a section "Example of application". D.Lazard (talk) 09:02, 16 February 2021 (UTC)Reply

That sounds fine to me. I think it would fit well in the "Applications" section, perhaps after the subsection entitled "Absence of formulas". Ozob (talk) 01:29, 17 February 2021 (UTC)Reply

Copyrights are about the actual text being used; if someone has put an example in a textbook, and you rewrite the same mathematics in your own words, there is no copyright issue. --JBL (talk) 13:03, 17 February 2021 (UTC)Reply

No; rewriting in our own words only creates a derivative work, to which the original copyright owner still has some rights. Ozob (talk) 04:20, 18 February 2021 (UTC)Reply

No: the idea behind a mathematical example (or a math problem) is not copyrightable. (Likewise cookbook recipes.) Problems only arise if you copy a collection of such things. --JBL (talk) 12:15, 18 February 2021 (UTC)Reply

I have a question about the example function g(x) in the First Proof. Currently the article states: ‘For example, this happens for g(x) = x^2*sin(1 / x) near the point a = 0.’ I think the example function should be a split function as follows: “g(x) = x^2*sin(1 / x) for x doesn’t equal 0, and g(x) = 0 for x = 0”. As it stands the function is undefined at the point where a = 0, and so not differentiable there. I think the point of the article would be valid if it used the split function I have suggested. — Preceding unsigned comment added by Matthew.howey (talk • contribs) 10:38, 29 March 2021 (UTC (UTC)

Please put new talk page messages at the bottom of talk pages and sign your messages with four tildes (~~~~) — See Help:Using talk pages. Thanks.

@Matthew.howey: Yes, the second part of the Q function mentions g(a), so that should probably be taken into account. - DVdm (talk) 11:36, 29 March 2021 (UTC)Reply

Fixed D.Lazard (talk) 11:40, 29 March 2021 (UTC)Reply

Hi – many thanks for your quick response and edit @DVdm: @D.Lazard:. Could I suggest it would be clearer if the value of g(x) at x = 0 is explicitly built into the definition of the function itself – perhaps: “this happens for the split function: g(x) = x²sin(1/x) if x ≠ 0, g(x) = 0 if x = 0”. Happy to make the edit but don’t want to do anything without agreement. Thanks.Matthew.howey (talk) 15:27, 29 March 2021 (UTC)Reply

I've proceeded to make the edit I suggested, obviously if anyone disagrees please amend or revert. Matthew.howey (talk) 20:40, 15 April 2021 (UTC)Reply

I think the current notation for partial derivatives (${\displaystyle D_{i}f}$) is unnecessarily pedantic. Thus I attempted to change it to what I believe is the most common notation (${\displaystyle {\frac {\partial f}{\partial g_{i}}}}$) [2], which was almost immediately reverted ([3]) with the reason "Partial derivative with respect to a function is not defined". While that is correct, it is also not what I wrote, but I agree that it should perhaps be very clearly stated that it is not to be read as that. I think the ${\displaystyle {\frac {\partial f}{\partial g_{i}}}}$ notation is by far the most common, and I don't think the article is helping anyone by not adhering to that. Does anyone have major objections to changing the notation, perhaps with the addition of a few sentences explaining how it should be read? QuarksAndElectrons (talk) 10:03, 27 October 2023 (UTC)Reply

Again, in the notation of partial derivatives, this is a variable that must appear in the denominator, not the name of a function. This is the reason for using the rather standard derivative#D-notation. Note that the reason is explained in the following sentences, and your edit makes confusing these explanations. D.Lazard (talk) 12:10, 27 October 2023 (UTC)Reply

And I am saying that it is incredibly common to either treat the variables and functions on equal footing in this type of situation (see eg. Terrence Tao Analysis II) or make it clear from context that ${\displaystyle \partial /\partial g_{i}}$ should be read as the derivative wrt. to the ith argument. I think that writing it with ${\displaystyle \partial }$s makes it much more readable.

I know that the D-notation exists, however i strongly disagree that it makes it simpler and clearer to use it. Especially since you end up with two different notations in the same equation in the following examples. QuarksAndElectrons (talk) 15:38, 27 October 2023 (UTC)Reply

Alternatively, how about writing ${\displaystyle f(y_{1},\ldots ,y_{k})=f(g_{1}(x),\ldots ,g_{k}(x))}$, so that the chain rule can be written:

${\displaystyle {\frac {d}{dx}}f(g_{1}(x),\dots ,g_{k}(x))=\sum _{i=1}^{k}\left({\frac {dy_{i}}{dx}}(x)\right){\frac {\partial f}{\partial y_{i}}}(g_{1}(x),\dots ,g_{k}(x)).}$ QuarksAndElectrons (talk) 16:00, 27 October 2023 (UTC)Reply

The problem is not making things simpler or clearer, it is to be mathematically correct. Your notation is a sort of jargon, that is, a shortcut that is clear for accustomed readers, but may be misleading or confusing for others. So, I still disagree with your change and its variant, which needs the introduction of k unneeded dependent variables.

Nevertheless, this page has 202 WP:watchers; let us wait the opinion of others. D.Lazard (talk) 16:15, 27 October 2023 (UTC)Reply

The explanation as given requires the reader to already know and understand the chain rule to reverse engineer the intended meaning.

At first glance "If a car travels twice as fast as a bicycle and the bicycle is four times as fast as a walking man, then the car travels 2 × 4 = 8 times as fast as the man." appears to be an irrelevant observation. Intuitively the explanation feels like a random fact thrown at the reader.

The necessary understanding needed to convey the chain rule intuitively is already given in the section above.

"In this case, the chain rule is expressed as ${\displaystyle {\displaystyle {\frac {dz}{dx}}={\frac {dz}{dy}}\cdot {\frac {dy}{dx}},}}$ and ${\displaystyle {\displaystyle \left.{\frac {dz}{dx}}\right|_{x}=\left.{\frac {dz}{dy}}\right|_{y(x)}\cdot \left.{\frac {dy}{dx}}\right|_{x},}}$ for indicating at which points the derivatives have to be evaluated."

"The rate of change of relative positions of the car and the bicycle is 2"

This is confusing the reader by conflating multiple concepts simultaneously. The rate of change of relative positions? No intuitive explanation is given for this. Relative positions with respect to what? Why relative positions in the first place? It sounds absurdly artificial and divorced from reality.

"The rate of change of positions is the ratio of the speeds, and the speed is the derivative of the position with respect to the time"

Again, it feels like someone is forced to come up with a fake example that is designed to sound "intuitive", but is in fact not and instead they messed around with the interpretation of the variables until it fits anyway. You could make the same argument by saying that pedestrians walk at 2m/s, bicycles travel at 4m/s and cars at 8m/s, then the derivative of the position of the pedestrian is his velocity, the same for the bicycle and car. But more importantly, the derivative of the car's absolute position with respect to the bicycle's reference frame is 4m/s and the derivative of the bicycle's absolute position with respect to the pedestrian is 2m/s. Hence we'd need to add not multiply with absolute positions, making the example maximally confusing. The reason why this doesn't work is that velocity is the derivative of position with respect to time. The relative positions are needed to encapsulate the passage of time with a single input, making it appear like a hack optimized for a specific carefully crafted scenario.

So here is my suggestion. Use something that is naturally based around ratios. Gears or clocks have fixed ratios. Seconds pass 60 times faster than minutes. Minutes pass 60 times faster than hours. Then f can be a function converting minutes to seconds aka y = 60x and g a function converting hours to minutes y = 60x. It would then be blatantly obvious that the two numbers need to be multiplied. The gradient of f(g(hours)) is measured in seconds. If having the same function (multiply by 60) twice is bad, then use days to hours to minutes with a 24 and 60 ratio respectively. — Preceding unsigned comment added by 134.169.7.133 (talk) 15:35, 20 June 2025 (UTC)Reply

Counterpoint: what's in the article is completely reliably sourced, and that's the only thing that counts on Wikipedia. Tito Omburo (talk) 22:40, 4 July 2025 (UTC)Reply

That's most certainly not "the only thing that counts" on Wikipedia. While we're at it though, it is not hard to find a wide variety of sources about the chain rule, some of which might be helpful for making accessible explanations.

@134.169.7.133: here's a published source which uses gears to explain the chain rule: JSTOR 26887043. –jacobolus (t) 14:11, 22 July 2025 (UTC)Reply

My friend will make a YouTube explanatory video following (more or less) your suggestion, 134.169.7.133! Thank you, it's a very nice suggestion! MathKeduor7 (talk) 13:12, 17 July 2025 (UTC)Reply

You might enjoy explanations from:

• SMP, Revised Advanced Mathematics, p. 298
• Calculus in Context, p. 157
• 3Blue1Brown on YouTube
• Holden (1967), JSTOR 27957718

–jacobolus (t) 14:30, 22 July 2025 (UTC)Reply

It's all very strange. Why do we need the word "chain rule" at all? It seems like the invention of the bicycle, when there are already cars. This is all called the derivative of a complex function, for example, ${\displaystyle f(g(h(x,y)))}$:

${\displaystyle df={\frac {\partial f}{\partial g}}{\frac {\partial g}{\partial h}}{\frac {\partial h}{\partial x}}dx+{\frac {\partial f}{\partial g}}{\frac {\partial g}{\partial h}}{\frac {\partial h}{\partial y}}dy}$ CalculatorOpt (talk) 07:29, 4 July 2025 (UTC)Reply

There's an answer to that on English StackExchange here; apparently the name comes from German Kettenregel, and was brought into English by the 1934 translation of Richard Courant's popular calculus textbook. It's somewhat off topic for this talk page, unless you think we should be discussing this in the article. –jacobolus (t) 18:45, 22 July 2025 (UTC)Reply

I think the proof via infinitesimals is not correct in case ${\displaystyle \Delta x=0}$. The correct proof is given on p86 here https://people.math.wisc.edu/~hkeisler/chapter_2b.pdf . However, I'm not very experienced with non-standard analysis and would prefer to leave it to someone more experienced to change the article. 270matt (talk) 06:46, 28 October 2025 (UTC)Reply

Agree, the proof is incorrect. Unfortunately, a correct proof rather depends on a bit more background. The derivative is a standard number L such that ${\displaystyle f(x+\Delta x)=f(x)+L\Delta x+\epsilon \Delta x}$, where ${\displaystyle \epsilon }$ is infinitesimal. Tito Omburo (talk) 15:57, 28 October 2025 (UTC)Reply

I removed this section. If @Tkuvho: wants to restore a corrected version, so be it. Tito Omburo (talk) 18:39, 28 October 2025 (UTC)Reply