A couple of posts ([1], [2]) on matheducators.SE seem to suggest that Leibniz originally got the wrong form for the product rule, perhaps thinking that $(fg)'=f'g'$. Is there any actual historical evidence for this?

It seems particularly hard to believe that he would have made the hypothesis $(fg)'=f'g'$. We would then have $x'=(1x)'=(1')(x')=0$. And presumably anyone inventing calculus would take $(x^2)'$ to be a prototypical problem, and would realize pretty early on that $(x^2)'\ne (x')(x')=1$. It's also pretty trivial to disprove this conjecture based on dimensional analysis or scaling.

There is some discussion on this Wikipedia talk page, with some sources cited, but it appears to be inconclusive.

  • 3
    $\begingroup$ I checked Stewart and it says "By analogy with the Sum and Difference Rules, one might be tempted to guess, as Leibniz did three centuries ago, that the derivative of a product is the product of the derivatives." with no citation whatsoever. I suppose this might be a mild joke. $\endgroup$ – Bombyx mori Sep 21 '14 at 3:04
  • 31
    $\begingroup$ It is completely natural to wonder if the derivative of a product is given by that (false) rule. Nobody is saying Leibniz thought it might be true for any extended amount of time. According to p. 254 of "The Historical Development of the Calculus" by C. H. Edwards, Leibniz wrote about his search for a product rule on November 11, 1675. He asked himself if $(uv)' = u'v'$ and quickly dismissed it by the example you gave: $u = v = x$. He did not know a correct product rule at the time. By July 11, 1677 he had the product and quotient rules (see p. 255 of the book by Edwards). $\endgroup$ – KConrad Sep 21 '14 at 4:06
  • 11
    $\begingroup$ @Bombyxmori: Stewart's text is not a work of scholarship, so I would not expect him to studiously document his historical remarks. $\endgroup$ – KConrad Sep 21 '14 at 4:10
  • 3
    $\begingroup$ @KConrad It's a work of something, that much I agree... $\endgroup$ – Yemon Choi Sep 22 '14 at 13:09
  • 3
    $\begingroup$ In one of his more outrageous moments, Arnold did comment that Leibniz got the product rule wrong, as a way of "proving" Newton's alleged superiority over Leibniz. I always thought this kind of thing is due to Arnold's inability to absorb the hyperreals and consequent insecurity. $\endgroup$ – Mikhail Katz Dec 4 '14 at 17:35

In the manuscript "Determinationum progressio in infinitum" (pp. 668-675 of Sämtliche Schriften und Briefe, Reihe VII, Band 3, Teil C, available in pdf here), Leibniz writes on p. 673 (with "$\sqcap$" in place of "$=$"):

$$ \odot = \overline{dt}\int\frac{a^2}{a^2 + t^2}. \quad\text{Hence}\quad \overline{d\odot} = \frac{a^2}{a^2 + t^2}\overline{d\overline{dt}} $$

This amounts to asserting that $d[uv] = dv\,du$ where $u=dt$ and $v=\int\frac{a^2}{a^2+t^2}$; and thus differentiating the product wrong, as the editors comment in footnote 14. On p. 668 they take this as grounds to date the manuscript early November 1675, since by November 11 he was pointing out this error (in "Methodi tangentium inversae exempla", quoted by Edwards in KConrad's comment above).

Addendum: The first time Leibniz gets his general rule right appears to be in "Pro methodo tangentium inversa et aliis tetragonisticis specimina et inventa" (dated 27 November 1675; pp. 361-371 of the same Sämtliche Schriften, Reihe VII, Band 5, Teil B; English translation here), where he writes on p. 365:

Therefore $d\overline xy = d\overline{xy}-xd\overline y$. Now this is a really noteworthy theorem and a general one for all curves.

  • $\begingroup$ KConrad's comment says that on Nov 11, 1675, "He asked himself if (uv)′=u′v′ and quickly dismissed it by the example you gave: u=v=x." Is that correct? Your answer makes it sound like the dates are all much fuzzier, perhaps by years, and we have no way of knowing whether it took Leibniz five minutes or a year to discard the hypothesis. Am I correct in understanding that this was all material from an unpublished personal notebook? $\endgroup$ – Ben Crowell Sep 21 '14 at 13:48
  • 3
    $\begingroup$ @BenCrowell The mentioned source, with the error you asked about, gets dated between end of October and November 11th (both 1675). I do not know why you think this is that fuzzy. The end of Oct based on the fact that some notation is used already that was introduced then, and before November 11th due an error being present there (the one you ask about) that was recognized as such at least by Nov 11th. And, yes, AFAIU it is a draft. Anyway, the claimed error seems to be there as an error. $\endgroup$ – user9072 Sep 21 '14 at 15:36
  • 1
    $\begingroup$ Just to be absolutely sure: is the bar here used to group together symbols, just like we do with parentheses nowadays? $\endgroup$ – Jules Lamers May 2 '18 at 11:54
  • 2
    $\begingroup$ @JulesLamers Yes, it’s a vinculum. $\endgroup$ – Francois Ziegler May 2 '18 at 14:38
  • $\begingroup$ @FrancoisZiegler Nice, and cool that this notation survived in e.g. our notation for square roots $\endgroup$ – Jules Lamers May 3 '18 at 9:39

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.