$\Delta$ operator is it as a differential operator? Consider the second law of dynamics written in terms of both momentum, momentum of a force and angular momentum,
$$\bbox[5px,border:3px solid #F5B041]{\overline{F}=\frac{\Delta\overline{p}}{\Delta t}} \tag 1$$
$$\bbox[5px,border:3px solid #A2B151]{\overline{M}=\frac{\Delta\overline{L}}{\Delta t}} \tag 2$$
In the proof of the $(2)$ we use the concept of $\overline{L}=\overline r \times \overline p$ where $\overline r$ it is the position vector and the $\overline{p}$ it is the momentum.
$$\Delta \overline L=\Delta (\overline r \times \overline p)$$
If I consider $\Delta$ as $d$ (differential operator) I know that using the calculus I will obtain$$\bbox[lightgray,5px,border:2px solid teal]{d(\overline r \times \overline p)=(d\overline r)\times \overline p \, + \,\overline r\times (d\overline p)} $$
In my textbook of Physics I have:
$$\bbox[orange,5px,border:2px solid gray]{\Delta(\overline r \times \overline p)=(\Delta\overline r)\times \overline p \, + \,\overline r\times (\Delta\overline p)} \tag 4$$

Because the operator $\Delta$ must have the same role as the differential operator $d$? Does $\Delta$ a laplacian operator? Is it possible to proof the $(4)$ without the calculus?

PS: My students, still they not use the calculus (derivates).
 A: No - here, $\Delta$ is certainly not the "Laplacian" operator!
[In the following, I'm not decorating vectors with overlines for reasons of laziness.]
In any event:  $\Delta (r\times p)$ equals, the first line by definition, and the next line by expanding out the terms in the cross product,
$$ \begin{align}\Delta (r\times p) &= (r+\Delta r ) \times (p+\Delta p) - r\times p\\&= r\times p + (r \times \Delta p) + (\Delta r \times p) + (\Delta r \times \Delta p)- r\times p.\end{align}  $$
Canceling (subtraction) the two $r\times p$ terms, on gets
$$ \Delta (r\times p) = (r \times \Delta p) + (\Delta r \times p) + (\Delta r \times \Delta p). $$
The above is almost a tautology...
Now, if, in your case, $\Delta r \times \Delta p =0$  [which is possible!], then
$$ \Delta (r\times p) = (r \times \Delta p) + (\Delta r \times p).$$
Otherwise, under the assumption that $(\Delta r \times \Delta p)$ is small compared to the other terms, one "only" has that
$$ \Delta (r\times p) \approx (r \times \Delta p) + (\Delta r \times p).$$
In that case, your text is at best taking a liberty in writing $=$ instead of $\approx$ [unless of course $\Delta r \times \Delta p =0$ in your case]. I am not sure that I approve! On the other hand, I have read an equation of Fermi which ended up equating an irrational number ($\pi$, say) to a rational number  ($3.1$, say) - and Fermi certainly knew what he was doing...  Fermi was only interested in maintaining accuracy of to $0.1$ (say)... For instance $ 5\times 0.1 + 0.1\times 10 = 1.5 $ is much bigger than $0.1\times 0.1 = 0.01$, which one might legitimately 'neglect'. But bear in mind what you mean.
I hope this helps...
