I have two real functions, $z$ and $y$, initially defined for discrete domain. The function $z$ is related to $y$ and to $\Delta x$ by $$z(x)=y(x+\Delta x)-y(x).$$
These functions are dimensioneless, but $x$ have a dimension of lenght.
Well, an author states that, making the intervals smaller, we get:
$$z(x)=y'(x)$$
However, $z$ is dimensioneless and $y'$ has dimension $[\text{lenght}]^{-1}$.
I did the calculus: $$\dfrac{z(x)}{\Delta x}=\dfrac{y(x+\Delta x)-y(x)}{\Delta x}$$ $$\lim_{\Delta x\to 0}\dfrac{z(x)}{\Delta x}=\lim_{\Delta x\to 0}\dfrac{y(x+\Delta x)-y(x)}{\Delta x}$$ $$\dfrac{z}{dx}=\dfrac{dy}{dx}$$
This matches the units, but does what mean $\dfrac{z}{dx}$?
Many thanks.
If $y$ is differentiable and $\Delta x$ is small, then $$y(x+\Delta x)-y(x)\approx y'(x).{\Delta x}$$. This would give you $$z(x)\approx y'(x).\Delta x$$