Derivative $\Delta x$ and $dx$ difference This may seems like a dummy question but I need to ask it.
Consider the definition of derivative:
$$\frac{d}{dx}F(x) = \lim_{\Delta x->0}\frac{F(x+\Delta x) - F(x)}{\Delta x} = f(x)$$
Also:
$$f(x)\Delta x = F(x+\Delta x) - F(x) \tag{When $\Delta x$ gets closer to $0$}$$
I can also say that:
$$\frac{d}{dx}F(x) = f(x)$$
So:
$$dF(x) = f(x)dx$$ 
but $dF(x)$ can also be seen as $F(x+\Delta x) - F(x) \tag{When $\Delta x$ gets closer to $0$}$
So should $dx$ be considered $\Delta x \tag{When $\Delta x$ gets closer to $0$}$?
I think this is wrong because it's the same as saying $\lim_{\Delta x \to 0} \Delta x =dx$ when in true $\lim_{\Delta x \to 0} \Delta x =0$. Or maybe $\Delta x$ already means a change in $x$, so the limit of this change, aproaching infinity is gonna be $dx$. In this case, no problem, but and in cases that people use $h$ instead $\Delta x$?
I think i'm consufing it a lot. Sorry...
 A: Your question is very good. There's something called the "non-standard" numbers. Trying to define them, we would have the set $$\{ \alpha, \text{such that  } 0 < \alpha < x, \forall x \in \mathbb{R}\}$$
What happens, is that $\mathrm{d}x$ is in that set, while $\Delta x$ isn't. For instance, let's differentiate $y = f(x) = x^2$, think of $\mathrm{d}x$ as an infinitesimal disturbing in $x$ that causes another infinitesimal disturbing $\mathrm{d}y$ in $y$, that is:
$$y + \mathrm{d}y = (x+ \mathrm{d}x)^2 = x^2 + 2x \mathrm{d}x + {\mathrm{d}x}^2 \\ \mathrm{d}y = 2x \mathrm{d}x + {\mathrm{d}x}^2 \\ \frac{\mathrm{d}y}{\mathrm{d}x} = 2x + \mathrm{d}x $$
Then, you would ask: but isn't the derivative of $x^2$, $2x$? The fact that the differentials are smaller than any real number would justify neglecting the remaining $\mathrm{d}x$, we would take the standard part of the derivative we just calculated.
$$\frac{\mathrm{d}y}{\mathrm{d}x} = \operatorname{std}(2x + \mathrm{d}x) = 2x$$
In the same way we neglected the $\mathrm{d}x$ here, we would do the same to higher order differentials, like $\mathrm{d}x \mathrm{d}y$ (product), or ${\mathrm{d}x}^2$ (powers). I sugest you try and differentiate $x^3$ to feel this, and I hope this helps.
