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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    $\begingroup$ See en.wikipedia.org/wiki/Q-derivative $\endgroup$ – kjetil b halvorsen Mar 25 '15 at 10:20
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    $\begingroup$ 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). $\endgroup$ – Hayden Mar 25 '15 at 10:23
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    $\begingroup$ 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. $\endgroup$ – user117644 Mar 25 '15 at 10:35
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    $\begingroup$ 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 $. $\endgroup$ – Mark S. Mar 25 '15 at 19:07
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    $\begingroup$ Interesting how, with the symmetric variant, your version of $e$ is $(\sqrt{\varepsilon^2+1}+\varepsilon)^{1/\varepsilon}=e-\dfrac{e\varepsilon^2}6 +\dfrac{4e\varepsilon^4}{45}-\dotsb$ $\endgroup$ – Akiva Weinberger Mar 26 '15 at 20:04

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.

  • $\begingroup$ Does it mean time scales? $\endgroup$ – Anixx Mar 26 '15 at 19:42
  • $\begingroup$ 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. $\endgroup$ – Eliad Mar 27 '15 at 9:38

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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    $\begingroup$ 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. $\endgroup$ – Kevin Carlson Mar 25 '15 at 23:04
  • $\begingroup$ 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. $\endgroup$ – user117644 Mar 25 '15 at 23:53

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