When differentiating we usually take a limit and drop the infinitesimal terms.

But what if not to drop anything?

First, we extend the real numbers with an infinitesimal element $\varepsilon$ which has its own inverse $1/\varepsilon=\omega$.

And define the full derivative of a function formally as follows:


Now we can compute full derivatives of polynomials in closed form:

$$D_{full}[a]=0$$ $$D_{full}[ax]=a$$ $$D_{full}[x^2]=2x+\varepsilon$$ $$D_{full}[x^3]=3 x^2+3 \varepsilon x+\varepsilon ^2$$


We also can find a function that remains invariant against full differentiation. It is not exponent with base $e$ though. To find it we solve the equation:


The solution is a set of functions

$$C (\varepsilon +1)^{\frac{x}{\varepsilon }}$$

of which the most simple is

$$(\varepsilon +1)^{\frac{x}{\varepsilon }}$$

We can call it "full exponent" and re-define trigonometric and inverse trigonometric functions accordingly. For instance, full logarithm, sine and cosine become

$$\operatorname{flog}\,\,x=\frac{\varepsilon \ln(x)}{\ln(\varepsilon + 1)}$$

$$\operatorname{fsin}\,\,x=\frac{ (1+i\varepsilon)^{x/\varepsilon }-(1-i\varepsilon )^{x/\varepsilon }}{2i}$$

$$\operatorname{fcos}\,\,x=\frac{ (1+i\varepsilon)^{x/\varepsilon }+(1-i\varepsilon )^{x/\varepsilon }}{2}$$

etc (these full sine and full cosine satisfy the equation $f''=-f$ with full derivative).

The same expressions for differentiation occurs in time scale calculus with a scale parameter. I wonder whether anybody ever considered such operation of "full differentiation" either in the framework of non-standard analysis or time scales or otherwise and whether it has any established name?

Note that we can also in a similar way define its inverse operator, "full integral" that would be

$$\int_{full} f(x)dx=\varepsilon \lim_{t\to x/\varepsilon} \sum_t f(\varepsilon t)$$

where $\sum_t$ is indefinite sum.

Thus we get

$$\int_{full} a \,dx=ax$$

$$\int_{full} x \,dx=\frac{x^2}{2}-\frac{\varepsilon x}{2}$$

$$\int_{full} x^2 \,dx=\frac{x^3}{3}-\frac{\varepsilon x^2}{2}+\frac{\varepsilon ^2 x}{6}$$

$$\int_{full} a^x \,dx=\frac{\varepsilon a^x}{a^{\varepsilon }-1}$$

$$\int_{full} \sin x \,dx=-\frac{1}{2} \varepsilon \sin (x)-\frac{1}{2} \varepsilon \cot \left(\frac{\varepsilon }{2}\right) \cos (x)$$


Note also that we can define full derivative in a more symmetric way:


With this definition some formulas become simplier:

$$D_{sym}[e^x]=\frac{e^x \sinh (\varepsilon )}{\varepsilon }$$

$$D_{sym}[\sin x]=\frac{\sin (\varepsilon ) \cos (x)}{\varepsilon }$$

$$D_{sym}[1/x]=\frac{1}{\varepsilon ^2-x^2}$$

The invariant function for this operation, playing the role of exponent will be

$$f(x)=\left(\sqrt{\varepsilon ^2+1}+\varepsilon \right)^{x/\varepsilon }$$

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    See https://en.wikipedia.org/wiki/Q-derivative – kjetil b halvorsen Mar 25 '15 at 10:20
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    In non-standard (hyperreal) analysis, the derivative is defined as the standard part of your "full derivative". So in a sense, yes, it has been considered, but more interest seems to be on the "real part". It doesn't have a name, so far as I'm aware (judging from Keisler's book). – Hayden Mar 25 '15 at 10:23
  • @kjetil b halvorsen this is not q-derivative, totally different thing. – Anixx Mar 25 '15 at 10:27
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    It works perfectly well if you don't drop the infinitesimals - but you just end up studying a curve approximated by a series of connected straight lines. It's easiest to do this with smooth infinitesimal analysis. –  Mar 25 '15 at 10:35
  • @Hayden in usual non-standard analysis they usually do not introduce distinguished elements $\varepsilon$ and $\omega$, thus they cannot uniquely define full derivative. – Anixx Mar 25 '15 at 11:01
  • @Anixx This is true; through taking the standard part one removes any dependencies on the choice of $\epsilon$. You could think of your process as giving a family of functions, indexed by the set of infinitesimals. Then the ambiguity no longer appears (of course, I'm not sure if anyone actually does this in practice). – Hayden Mar 25 '15 at 12:58
  • @Hayden I am about introducing one element, $\varepsilon$, similarly to how complex $i$ introduced. One can argue such system would be undefinable, but then complex numbers are also undefinable because $i$ is indistinguishable from $-i$. By the way, in the context of surreals, hyperreal numbers can be considered a subfield $No(\omega)$. Then this $\omega$ has definite meaning: it is considered equal to the first infinite ordinal. – Anixx Mar 25 '15 at 13:02
  • @Anixx I understand, but I just mean it is natural to extend your definition to look at the set of all such "full derivatives". And yes, sadly I don't think one could expect an isomorphism interchanging any two infinitesimals. – Hayden Mar 25 '15 at 13:15
  • http://math.stackexchange.com/questions/822664/could-we-assign-a-numerical-value-to-an-infinitesimal –  Mar 25 '15 at 13:49
  • A side comment that has no bearing on this question: @Anixx, you can take $No(\omega)$, but as I said in answer to http://math.stackexchange.com/questions/1193422/why-hyperreal-numbers-are-built-so-complicatedly/ that won't give you the hyperreals. – Mark S. Mar 25 '15 at 18:38
  • @Mark S. $No(\omega)$ is a hyperreal system, subfield of surreals http://ohio.edu/people/ehrlich/Unification.pdf – Anixx Mar 25 '15 at 18:41
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    I think you misunderstand the notation in that PDF. $No(\omega)$ is actually the dyadic rationals thanks to the tree rank (see Theorem 15). $ No(\omega_1) $ is a hyperreal system assuming CH, but that has a lot of different flavors of infinitesimals, not just what you can get with reals and $\omega $. – Mark S. Mar 25 '15 at 19:07
  • @Mark S. so basically if to add to rationals $\varepsilon$ as in this post we get Levi-Civita field? – Anixx Mar 25 '15 at 19:15
  • @Anixx you're working with the minimal hyperreal field $\mathbb{R}(\epsilon)$, in which you can indeed uniquely define this "full derivative." – Kevin Arlin Mar 25 '15 at 20:57
  • @Kevin Carlson is this field truly hyperreal? – Anixx Mar 25 '15 at 20:58
  • @Anixx You're right, I shouldn't call it hyperreal, which means your approach is also not identical to that of nonstandard analysis. – Kevin Arlin Mar 25 '15 at 23:02
  • @Anixx If you add an infinitesimal to rationals, you get formal Laurent series with rational coefficients. The Levi-Civita field is different in that it allows real coefficients and rational powers of $\epsilon$. But to define your full derivative in cases that aren't polynomials, you need a method for defining things like $sin(\epsilon)$. Hyperreal fields make this work perfectly, but http://www.physics.umanitoba.ca/~khodr/Publications/RS-Overview-offprints.pdf suggests that the Levi-Civita field is probably good enough, at least for analytic functions (I haven't thought about it much). – Mark S. Mar 26 '15 at 01:07
  • @Mark S. what a problem in defining $\sin \varepsilon$? It just can be represented as a series or in closed form... Where the problem is? – Anixx Mar 26 '15 at 09:16
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    Interesting how, with the symmetric variant, your version of $e$ is $(\sqrt{\varepsilon^2+1}+\varepsilon)^{1/\varepsilon}=e-\dfrac{e\varepsilon^2}6 +\dfrac{4e\varepsilon^4}{45}-\dotsb$ – Akiva Weinberger Mar 26 '15 at 20:04
  • @columbus8myhw I think e can be expressed in closed form by modifying the formula. – Anixx Apr 01 '15 at 13:18
  • @Anixx Is that not "closed form"? $(\sqrt{\varepsilon^2+1}+\varepsilon)^{1/\varepsilon}$? (Also, weirdly, that's an even function of $\varepsilon$. I didn't expect that.) – Akiva Weinberger Apr 01 '15 at 13:19
  • @columbus8myhw it is not equal to e. – Anixx Apr 01 '15 at 13:41
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    @Anixx $\displaystyle\lim_{\varepsilon\to0} (\sqrt{\varepsilon^2+1} +\varepsilon)^{1/\varepsilon}=e$ – Akiva Weinberger Apr 01 '15 at 13:42
  • @columbus8myhw this is not closed form... – Anixx Apr 01 '15 at 14:05

2 Answers2


As I understand it, this is just the same as h-calculus. The h-derivative is defined as,

$$ D_{h} = \dfrac{f(x+h) - f(x)}{h} $$

, where $h\ne 0$. [1] has a small chapter on it.

[1] Kac, V., & Cheung, P. (2002). Quantum calculus. Springer Science & Business Media.

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  • Does it mean time scales? – Anixx Mar 26 '15 at 19:42
  • h-Calculus is a "subset" of Quantum calculus, which is in turn an example of time-scale calculus. This might also be helpful: ČERMÁK, J., & NECHVÁTAL, L. (2010). On (q, h)-analogue of fractional calculus. Journal of Nonlinear Mathematical Physics, 17(1), 51-68. – Eliad Mar 27 '15 at 09:38

Here's an example of retaining the infinitesimals - although they may be increments with a numerical value:

$$y = x^3$$

$$D(x^3) = 3x^2 + 3\epsilon x + \epsilon^2$$

$$D^2(x^3) = 6(x + \epsilon)$$

$$D^3(x^3) = 6$$

$D^n$(x) just means the nth derivative of x. This kind of thinking allows us to do something unusual - setting the slope of the linear segment. For example, set D($x^3$) to 5 and x to 2. The line with slope 5 that intersects the curve above x = 2 also intersects the curve in two other places. With those substitutions we have:

$$5 = 12 + 6\epsilon + \epsilon^2$$

Solving the quadratic for epsilon yields:

$$\epsilon = -1.586$$ $$\epsilon = -4.414$$

Meaning the horizontal distances between the right intersection and the two to its left correspond to those values. This can be confirmed by plotting the curve and line and measuring the distances.

NB The derivatives are worked out using:

$$D(f(x)) = \frac{f(x + \epsilon) - f(x)}{\epsilon}$$

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    I don't understand what the point of this is. You're just computing where a certain line intersects $x^3$, ignoring the intended interpretation of $\varepsilon$ as an infinitesimal. – Kevin Arlin Mar 25 '15 at 23:04
  • I'm answering the question that was asked. If you 'not to drop anything' you get linear approximations to the curve and you can do things like this. If he intended another interpretation he should have asked a different question. –  Mar 25 '15 at 23:53
  • Plong the fong into long song. – Matt Samuel May 10 '20 at 00:50