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

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