Difference between the difference operator in recursion equations vs the derivation opertor. The derivation operator is $D=d/dx$, where we have: $D(f(x)g(x)) = fDg+gDf$.

Now while reading a book on difference equations, it occured to me that for the discrete operator, i.e $\Delta(y_n) = y_{n+1}-y_n$, we don't have the same identity, i.e. we have: $\Delta(z_n y_n) = z_{n+1}y_{n+1}-z_n y_n$, and $z_n\Delta y_n + y_n \Delta z_n = z_n(y_{n+1}-y_n)+y_n(z_{n+1}-z_n) = z_n y_{n+1}+y_nz_{n+1}-2z_ny_n$.

What other differences are there between these operators?
I prefer those that aren't obvious, although if you were to ask me before calculating I would have said the same identity should be also true for the discrete case.
They are pretty similar but at the same time slightly different due to the fact that:

$$\Delta y=y(x+1)-y(x)=\frac{y(x+1)-y(x)}{(x+1)-x}$$

The solution to the equation:

$$y=Dy$$

Is as we know $ce^x$.

The solution to the equation:

$$y=\Delta y$$

Is on the other hand,

$$y=c2^x$$

Given $f(x)=Dy$ then to find $y$ we integrate (a constant also appears when we integrate):

$$y= \int f(x) dx$$

Given $f(x)=\Delta y$ then to find $y$ we sum and add a corresponding constant for example,

$$\sum_{i=0}^{x-1} f(x)=\sum_{i=0}^{x-1} (y(i+1)-y(i))=y(x)+y(0) \implies y(0)+\sum_{i=0}^{x-1} f(x)=y(x)$$

We also have a pretty similar but different taylor series:

$$y(x)=\sum_{i=0}^{\infty} \Delta^i(0){x \choose i}$$

If we take $\Delta^0(0)=1$ and $\Delta^i(0)$ to represent $\Delta^i(f(x))$ when $x=0$.

The product rule  similar as mentioned in the comments illustrates similarity but slight difference , but "the chain rule" is very different and more complex: Chain rule for discrete/finite calculus.

The product rule can also be written as,

$$\Delta(f(x)g(x))=g(x)\Delta(f(x))+f(x)\Delta(g(x))+\Delta(f(x))\Delta(g(x))$$

Here is the quotient rule which is again slightly different:

$$\Delta(\frac{f(x)}{g(x)})=\frac{g(x)\Delta f(x)-f(x)\Delta g(x)}{g(x)(g(x)-\Delta g(x))}$$
The Chain Rule is trivial for differences: $$\Delta (g\circ f)(c)= \Delta g\, (f(c)).$$