The precise question appears at the end of this entry.

With all the recent advances in understanding infinitesimals, we still don't fully understand why Leibniz's definition of $\frac{dy}{dx}$ as literally a ratio works the way it does and seems to explain numerous facts including chain rule.

Note that Robinson modified Leibniz's approach as follows. Suppose we have a function $y=f(x)$. Let $\Delta x$ be an infinitesimal hyperreal $x$-increment. Consider the corresponding $y$-increment $\Delta y=f(x+\Delta x)-f(x)$. The ratio of hyperreals $\frac{\Delta y}{\Delta x}$ is not quite the derivative. Rather, we must round off the ratio to the nearest real number (its standard part) and so we set $f'(x)=\text{st}\big(\frac{\Delta y}{\Delta x}\big)$. To be consistent with the traditional Leibnizian notation one then defines new variables $dx=\Delta x$ and $dy=f'(x)dx$ so as to get $f'(x)=\frac{dy}{dx}$ but of course here $dy$ is not the $y$-increment corresponding to the $x$-increment. Thus the Leibnizian notation is not made fully operational.

Leibniz himself handled the problem (of which he was certainly aware, contrary to Bishop Berkeley's allegations) by explaining that he was working with a more general relation of equality "up to" negligible terms, in a suitable sense to be determined. Thus if $y=x^2$ then the equality sign in $\frac{dy}{dx}=2x$ does not mean, to Leibniz, exactly what we think it means.

Another approach to $dy=f'(x)dx$ is smooth infinitesimal analysis where infinitesimals are nilsquare so you get equality on the nose though you can't form the ratio. On the other hand, Leibniz worked with arbitrary nonzero orders of infinitesimals $dx^n$ so this doesn't fully capture the Leibnizian framework either.

Question: In an algebraic-geometric or other algebraic or analytic context (with suitable limitations on $f$), is there a way of assigning a precise sense to the Leibnizian generalized equality using global considerations?

Note. Related material can be found at this MO post.

  • $\begingroup$ In elementary algebraic geometry, morphisms are defined by polynomials and derivatives are taken formally. Perhaps you can be more specific what you mean by "algebraic-geometric context". Could you give an example to which you wish to assign a precise sense of this concept? $\endgroup$
    – RghtHndSd
    Aug 5, 2014 at 20:50
  • $\begingroup$ @RghtHndSd, in the popular article Mumford, David. Intuition and rigor and Enriques's quest. Notices Amer. Math. Soc. 58 (2011), no. 2, 250–260, Mumford analyses Enriques' proof using deformation theory based on infinitesimal deformations. I was hoping this approach could be used to mimick Leibniz's procedures. To give an elementary example, let $y=f(x)$ and $z=g(x)$. Then $dy=f'(x)dx$ and $dz=g'(x)dx$ where the equalities can be interpreted as saying that the ratio of the two sides is infinitely close to 1. If we now wish to claim that $dy+dz=f'(x)dx+g'(x)dx$ we have a problem, because... $\endgroup$ Aug 6, 2014 at 8:08
  • $\begingroup$ ... you can't add approximate equalities. $\endgroup$ Aug 6, 2014 at 8:12
  • $\begingroup$ "Infinitesimal deformations" means something rather specific - namely lifting with respect to a nilpotent ideal. In your example, what are x, y, z? What space is this? Affine space? Over what base? Complex numbers? What is the definition of "dy", etc. $\endgroup$
    – RghtHndSd
    Aug 6, 2014 at 15:29
  • $\begingroup$ My $x$ and $y$ are Leibnizian so I can't vouch for what he meant but one thing is clear: they are not the independent and dependent variables. This distinction was introduced later. If we choose $x$ to be the independent variable and $dx$ its increment then $dy$ should be the corresponding $y$-increment, but perhaps this is not the right thing to do if one wants to formalize Leibniz. $\endgroup$ Aug 6, 2014 at 15:33

2 Answers 2


I will just try to describe the closest things that I have seen, mainly focussing on synthetic differential geometry but with a mention of algebraic geometry at the end. The wishlist is:

  1. $f(x+dx)=f(x)+f'(x)dx$
  2. in particular, it should allow addition
  3. it should generalize to higher derivatives

A basic idea is to use nilpotent elements. If you have a polynomial $f=\sum_{n=0}^N c_nx^n,$ then $f(x+\epsilon)=\sum_{n=0}^N c_n(x+\epsilon)^n=f(x)+\epsilon \sum_{n=0}^N nc_nx^n=f(x)+\epsilon f'(x)$ in the ring $\mathbb C[x,\epsilon]/(\epsilon^2).$ The technicalities come from trying to apply this local definition to varieties or manifolds.

The Wikipedia smooth infinitesimal analysis article only mentions nilsquare infinitesimals, but in synthetic differential geometry higher order differentials can be defined. Kock's book on SDG (http://home.math.au.dk/kock/sdg99.pdf) defines subsets $$D_n=\{x\in R \mid x^{n+1}=0\}$$ where $R$ is the base ring - like the reals but with infinitesimals. (Beware that $D_n$ is not an ideal - $(\epsilon_1+\epsilon_2)^2$ is not necessarily zero even if $\epsilon_1^2=\epsilon_2^2=0.$ This algebraic fact is an important point in that AMS article of Mumford you mentioned.)

These can be collected into the nilpotent ideal $$D_\infty=\bigcup_{n\geq 0}D_n.$$ SDG postulates that the restriction $f|_{D_\infty}:D_\infty\to R$ of any map $f:R\to R$ is given uniquely by a formal Taylor series near zero. You can compose with translations to get Taylor series near other points.

These formal Taylor series compose in the usual way, giving the chain rule. The space of maps $D_n\to R$ becomes a free, finitely generated $R$-module. If $f(0)=0,$ then $f:R\to R$ restricts to an $R$-linear map from $D_1$ to $D_1,$ which is just the familiar notion of the derivative of a differentiable map as a linear map of tangent spaces. Using translation to handle possibly non-zero $x$ and $f(x),$ this formalizes $f(x+dx)=f(x)+f'(x)dx$ - take $dx$ to be an indeterminate element of $D_1.$

Geometrically, a map $g:D_n\to R$ is an element of the $n$-th order jet space around zero. In particular, $g:D_1\to R$ is a tangent vector at $g(0).$ These jet spaces are ordinary objects used in "non-synthetic" differential geometry.

For manifolds you can use addition in local co-ordinates. You can't reliably add non-infinitesimal quantities, because the result might fall outside the parameterization. And for anything except tangent vectors, the addition is not geometric: it is not parameterization-invariant. An example for $2$-jets is that the $\mathbb R\to\mathbb R$ curves $t\mapsto t$ and $t\mapsto -t$ add to zero as $\infty$-jets in the usual co-ordinates, but if we change co-ordinates locally via $x'=x+x^2$ these curves become $t\mapsto t+t^2$ and $t\mapsto -t+t^2,$ which don't add to zero.

In algebraic geometry there is a simple construction of the "Zariski tangent space" as the dual of the cotangent space. And there are constructions of $n$-jets. I think these can be packaged up into a single object using maps of completed local rings, or using formal schemes, but that's going way outside my comfort zone. You might be interested in:


When phrased as "equality up to negligible terms", it sounds as if you're describing the use of a congruence relation. For example, in algebra, one of the common ways to define the derivative of a polynomial over a field $F$ is by the identity

$$ f(x + \epsilon) \equiv f(x) + \epsilon f'(x) \pmod{\epsilon^2} $$

in the polynomial ring $F[x, \epsilon]$. Recall that the congruence $$ a \equiv b \pmod c $$ is equivalent to saying $b-a$ is a multiple of $c$ in the relevant setting. In particular, $$ c \equiv 0 \pmod{c} $$ so this is a systematic way of "neglecting" $c$. For example, if $f(x) = x^2$, then $$ (x + \epsilon)^2 = x^2 + 2 x \epsilon + \epsilon^2 \equiv x^2 + 2 x \epsilon \pmod{\epsilon^2} $$ and thus we see $f'(x) = 2x$.

One thing to pay attention to is that, while the equation is neglecting 'infinitesimals' of second order and above, this equation nonetheless defines $f'$ exactly, since we are insisting that it is a function of $x$ alone.

This can be generalized slightly to rational functions: we can define $$ u \equiv v \pmod{\epsilon^n}$$ to mean that in the factorization of $u-v$ into polynomials (negative exponents allowed!), the exponent on $\epsilon$ is at least $n$.

For example, if we have $f(x) = 1/x$,

$$ \frac{f(x+\epsilon) - f(x)}{\epsilon} = -\frac{1}{x^2} + \frac{\epsilon}{x^2(x + \epsilon)} \equiv -\frac{1}{x^2} \pmod{\epsilon} $$

  • $\begingroup$ Thanks for this. Leibniz though does not specify with respect to what term he is writing down his generalized relation of equality. For example, division of a relation by epsilon (in the case when both sides are divisible by epsilon) would result in another relation where the term with respect to which one is taking the "congruence" has to be changed (say, from $\epsilon^2$ to $\epsilon$). This works fine mathematically but does not really have any source in Leibniz's texts themselves. Note that a suitable limit interpretation can also work fine mathematically. $\endgroup$ Jan 21, 2018 at 11:14

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