Who invented the Leibnitz notation $\frac{d^2y}{dx^2}$ for the *second* derivative?

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?

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

• So this is an answer. Note that the colon indicates division. – Gerald Edgar Jan 17 '16 at 18:50
• Is this paper available online anywhere? – Michael Bächtold Dec 15 '17 at 10:17
• @michael-bächtold Link added. – Viktor Blasjo Dec 15 '17 at 14:16
• ...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 $\overline{dy}^2=(dy)^2$, – Michael Bächtold Apr 26 at 8:42

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 $y-xx : a \ldots$

• So we have found the numerator in the form $ddx$. Where do we find (for the first time) the denominator? – Gerald Edgar Jan 17 '16 at 14:23
• @GeraldEdgar - see e.g. page 156 : $ddy/ddx$. – Mauro ALLEGRANZA Jan 17 '16 at 16:17

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 $$\left(\frac{d}{dx}\right)^2$$?

• 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). – Francois Ziegler Oct 27 '18 at 6:18
• 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 $\smash{d(x^2)}$.” – Francois Ziegler Oct 27 '18 at 6:25
• 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. – Michael Bächtold Oct 27 '18 at 12:19
• You’re right. Equality only follows from what Bézout says under the assumption $ddx=0$ (no second term in (4)). – Francois Ziegler Oct 27 '18 at 12:42
• 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. – Michael Bächtold Feb 25 at 16:45

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.)

• 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. – Federico Poloni Jan 16 '16 at 8:35
• @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. – VicAche Jan 16 '16 at 14:16
• The denominator is the square of $dx$. Since $dx$ is not a product, just write it $dx^2$. No need for $(dx)^2$. – Gerald Edgar Jan 16 '16 at 16:14
• @VicAche, you mean physicists, not physicians (= medical doctors). – KCd Dec 18 '17 at 23:12
• @KCd true that :) can't edit or flag myself for edition right? – VicAche Dec 19 '17 at 13:00

$\frac{d^2}{dx^2}$ $y$ = $\frac{d^2y}{dx^2}$ is too obviously built from $\frac{d}{dx}$$\frac{d}{dx}$ $y$ = $(\frac{d}{dx})^2$ $y$ to deserve any further explanation.

• 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. – Michael Bächtold Jun 12 '17 at 8:24