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This MSE question made me wonder where the Leibnitz notation $\frac{d^2y}{dx^2}$ for the second derivative comes from. It does not arise immediately as the obvious generalization of $\frac{dy}{dx}$. Did Leibnitz use it himself? Or was it introduced later?

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Leibniz did use this notation for instance in his paper Supplementum geometriae practicae, Acta Eruditorum, April 1693, p. 179 (Google Books link):

enter image description here

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    $\begingroup$ So this is an answer. Note that the colon indicates division. $\endgroup$ – Gerald Edgar Jan 17 '16 at 18:50
  • $\begingroup$ Is this paper available online anywhere? $\endgroup$ – Michael Bächtold Dec 15 '17 at 10:17
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    $\begingroup$ @michael-bächtold Link added. $\endgroup$ – Viktor Blasjo Dec 15 '17 at 14:16
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    $\begingroup$ ...and note that the overline above the $y$ actually went over the $d$, as can be seen in the google Books link. But it should not go over the 2. That was an alternative notation to parenthesis, so $\overline{dy}^2=(dy)^2$, $\endgroup$ – Michael Bächtold Apr 26 at 8:42
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The differential symbold $dx$ is due to Leibniz.

He introduced also "iterated" differentials; see :

Moreover, to introduce higher-order differentials, first-order differentials have to be conceived as variables ranging over an ordered sequence; if only a single $dx$ is considered, $ddx$ does not make sense. The following quotation from Leibniz ["Monitum de characteribus algebraicis", 1710] illustrates this:

Further, $ddx$ is the element of the element or the difference of the differences, for the quantity $dx$ itself is not always constant, but usually increases or decreases continually.

See also The Early Mathematical Manuscripts of Leibniz (J.M. Child ed., 1920 - also Dover reprint): manuscript of an answer to Bernhard Nieuwentijt, page 144-on:

$dx, ddx, dv, ddv, dy, ddy \ldots$

We have to note that Leibniz has $xx$ for $x^2$; see page 151 :

Then, since $y-xx : a \ldots$

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  • $\begingroup$ So we have found the numerator in the form $ddx$. Where do we find (for the first time) the denominator? $\endgroup$ – Gerald Edgar Jan 17 '16 at 14:23
  • $\begingroup$ @GeraldEdgar - see e.g. page 156 : $ddy/ddx$. $\endgroup$ – Mauro ALLEGRANZA Jan 17 '16 at 16:17
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The accepted answer leaves no doubt that Leibniz was the first to write $d^2y/(dx)^2$ for the second derivative. But since I've found so many misleading justifications for this notation online, I feel that something additional needs to be said about it.

Most justifications in the links above are along the lines of: "by formal manipulation" or "too obviously" $$ \frac{d}{dx} \frac{d}{dx} =\frac{d^2}{dx^2}. \tag1 $$ But Leibniz, the Bernoullis or Euler would not have approved of this without reservation. Not even if the equation was written in the form $$ \frac{d \left(\frac{dy}{dx} \right)}{dx} = \frac{d dy}{(dx)^2}, \tag2 $$ which is closer to the standard of the time.

To explain let me make a simple analogy first. No one today would claim that the following is correct $$ \frac{\log \frac{\log y}{\log x}}{\log x} = \frac{\log \log y}{(\log x)^2}, \tag3 $$ and everyone can spot the error.

Analogously, for Leibniz, $d$ was an operator (he might not have called it that way, but he knew it acted on variables just like $\log$) and he knew the quotient rule for $d$. So he might have approved of the following general equation

$$ \frac{d \left(\frac{dy}{dx} \right)}{dx} = \frac{d^2y}{(dx)^2} - \frac{dy\cdot d^2x}{(dx)^3}. \tag4 $$ The reason the second term on the right disappeared, was because an additional assumption was often made: it was assumed that the differential of the differential of $x$ is zero (i.e. $d^2x=0$), or put differently: $dx$ was assumed constant.

This can be seen in the 1693 article of Leibniz quoted by @ViktorBlasjo, a line above $ddx:\overline{dy}^2$, where he writes

posita $dy$ constante

It can also be found in Eulers Institutiones Calculi Differentialis (1743) § 131.

Now we will proceed under the assumption that $x$ increases uniformly, so that the first differentials $dx, dx^I , dx^{II},\ldots$ are equal to each other, so that the second and higher differentials are equal to zero. We can state this condition by saying that the differential of $x$, that is $dx$, is assumed to be constant. Let $y$ be any function of $x$; ...

And it can be found in Lacroix's Traité du calcul différentiel et du calcul intégral (1797) p.96

Pour la simplifier nous observons que l'accroissement $dx$ étant regardé invariable, $f'(x)dx$ se change en $f'(x+dx)dx$ ...

Summarizing: for Leibniz, Euler and others the equation $$ \frac{d \left(\frac{dy}{dx} \right)}{dx} = \frac{d^2y}{dx^2} \tag5 $$ was only true under the additional assumption that $dx$ is constant.

This leaves a question for me, which hopefully someone else can answer: when and why did mathematicians forget this additional assumption and simply adopt the notation $\frac{d^2}{dx^2}$ for what should actually be written as $\left(\frac{d}{dx}\right)^2$?

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    $\begingroup$ Agreed, and your calculation (4) is literally in Bézout (1767, end of §18). To your concluding question, an answer is hard to pinpoint but Lagrange’s differential-free Fonctions analytiques (which has (4) as $(y′/x′)′/x′$, p. 60) must have been influential. This is discussed in Bos (1974, esp. §5 “Euler’s Program to Eliminate Higher Order Differentials”) and Domingues (2008, §§3.1.1 and 3.2.4). $\endgroup$ – Francois Ziegler Oct 27 '18 at 6:18
  • $\begingroup$ As to whether $d^2/\,dx^2$ should actually be written $(d\,/\,dx)^2$, I think they were equal by the same convention used e.g. for Riemannian $\smash{ds^2}$. Bézout spells it out on the previous page: “To denote the square of $dx$, one should naturally write $\smash{(dx)^2}$; but for simplicity one writes $\smash{dx^2}$, which cannot cause confusion, and be mistaken for the differential of $\smash{x^2}$, which we agreed [§7] to denote thus $\smash{d(x^2)}$.” $\endgroup$ – Francois Ziegler Oct 27 '18 at 6:25
  • $\begingroup$ Thanks for those additional pointers @FrancoisZiegler. I'm not sure if I can follow the reasoning in your second comment. Even if we write $\frac{d^2}{(dx)^2}$ for $\frac{d^2}{dx^2}$, I don't see how to arrive there from $\left(\frac{d}{dx}\right)^2$ by using conventions for orders of operations. Take the similar example $\left(\frac{\log}{\log{x}} \right)^2$, which I think is different from $\frac{\log^2}{(\log x)^2}$ with all conventions I can think of. $\endgroup$ – Michael Bächtold Oct 27 '18 at 12:19
  • $\begingroup$ You’re right. Equality only follows from what Bézout says under the assumption $ddx=0$ (no second term in (4)). $\endgroup$ – Francois Ziegler Oct 27 '18 at 12:42
  • $\begingroup$ A small pointer which I should follow up on: Max Stegemann in his popular Grundrisse der Differential- Integralrechnung 4th Edition 1880 still mentions the requirement dx constant. $\endgroup$ – Michael Bächtold Feb 25 at 16:45
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I think $\frac{d^2y}{dx^2}$ comes from multiplying $\frac{dy}{dx}$ by $\frac{d}{dx}$. In the Notation(https://en.m.wikipedia.org/wiki/Abuse_of_notation#Derivitive) multiplication signifies iteration.

(Disclaimer; This is a very rough response. There have not been any other answers yet, I will look for the notation in a textbook.)

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  • $\begingroup$ Thanks for the answer! That is a very reasonable explanation, but the MSE question that I have linked in the OP raises a good point: even assuming that one can multiply differential terms with impunity, it seems that in this notation one 'factor' $d$ is missing from the denominator. $\endgroup$ – Federico Poloni Jan 16 '16 at 8:35
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    $\begingroup$ @Federico Poloni I guess at least physicians view $dx$ as a quantity itself, rather than a product of d*x, the second differential being compared to the square of the small increment. $\endgroup$ – VicAche Jan 16 '16 at 14:16
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    $\begingroup$ The denominator is the square of $dx$. Since $dx$ is not a product, just write it $dx^2$. No need for $(dx)^2$. $\endgroup$ – Gerald Edgar Jan 16 '16 at 16:14
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    $\begingroup$ @VicAche, you mean physicists, not physicians (= medical doctors). $\endgroup$ – KCd Dec 18 '17 at 23:12
  • $\begingroup$ @KCd true that :) can't edit or flag myself for edition right? $\endgroup$ – VicAche Dec 19 '17 at 13:00
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$\frac{d^2}{dx^2}$ $y$ = $\frac{d^2y}{dx^2}$ is too obviously built from $\frac{d}{dx}$$\frac{d}{dx}$ $y$ = $(\frac{d}{dx})^2$ $y$ to deserve any further explanation.

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    $\begingroup$ I don't see how this contributes to answering the historical part of the question. If your claim is that Leibniz thought like this, you would have to back that up. I personally doubt he thought of $d/dx$ as an object in itself. $\endgroup$ – Michael Bächtold Jun 12 '17 at 8:24

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