Has anybody ever considered "full derivative"? 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:
$$D_{full}[f(x)]=\frac{f(x+\varepsilon)-f(x)}{\varepsilon}$$
Now we can compute full derivatives of polynomials in closed form:
$$D_{full}[a]=0$$
$$D_{full}[ax]=a$$
$$D_{full}[x^2]=2x+\varepsilon$$
$$D_{full}[x^3]=3 x^2+3 \varepsilon  x+\varepsilon ^2$$
etc. 
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:
$$\frac{f(x+\varepsilon)-f(x)}{\varepsilon}=f(x)$$
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 
$$\int_{full} a \,dx=ax$$
$$\int_{full} x \,dx=\frac{x^2}{2}-\frac{\varepsilon  x}{2}$$
$$\int_{full} x^2 \,dx=\frac{x^3}{3}-\frac{\varepsilon  x^2}{2}+\frac{\varepsilon ^2 x}{6}$$
$$\int_{full} a^x \,dx=\frac{\varepsilon  a^x}{a^{\varepsilon }-1}$$
$$\int_{full} \sin x \,dx=-\frac{1}{2} \varepsilon  \sin (x)-\frac{1}{2} \varepsilon  \cot \left(\frac{\varepsilon }{2}\right) \cos (x)$$
etc.

Note also that we can define full derivative in a more symmetric way:
$$D_{sym}[f(x)]=\frac{f(x+\varepsilon)-f(x-\varepsilon)}{2\varepsilon}$$
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
$$f(x)=\left(\sqrt{\varepsilon ^2+1}+\varepsilon \right)^{x/\varepsilon }$$
 A: 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.
A: 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:
$$D(f(x)) = \frac{f(x + \epsilon) - f(x)}{\epsilon}$$
