Product rule in discrete derivative in finite difference scheme. Suppose we are on real line and I want to discretize the usual derivative operator. Take a smooth function $u$ and step size $h$. Then I could define
$$
\Delta_+u(i) = \frac{u(i+1)-u(i)}{h}
$$
as the discrete form of derivative. Here $i$ is an integer.
Now my question is, when I try to compute $\Delta_+(u(i)\cdot v(i))$, should I have
$$
\Delta_+(u(i)\cdot v(i)) = \frac{u(i+1)v(i+1)-u(i)v(i)}{h} \tag 1
$$
or should I have a sort of product rule and obtain
$$
\Delta_+(u(i)\cdot v(i)) = \Delta_+u(i)\cdot v(i)+u(i)\Delta_+v(i).
$$
Which one is correct? Or which one is better?

In case someone stumbles upon this but wants to do a finite difference with a different step size. Given an arbitrary difference operator
$$ D_{h,x} [f] = \frac{f(x+h) - f(x)}{h} $$
Where that $h$ need not even be a number, it could be another explicit function. We have the following product rule
$$ D_{h,x} [f*g] = D_{h,x}[f]g + fD_{h,x}[g] + hD_{h,x}[f]D_{h,x}[g] $$
Setting $h=1$ gives you the formula you are looking for. And as a sanity check setting $h=0$ recovers the usual product rule from caculus.
