Product rule, help me understand this proof I see this proof of the product rule:

Two questions:


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*When we divide by $\Delta x$, why do we put the $\Delta x$ under the other $\Delta$? Why does it not go under the constant v or u?

*In the last part, why does the $\Delta x$ go under the $\Delta v$ and not the $\Delta u$? Is there a reason?
 A: 
  
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*When we divide by $\Delta x$, why do we put the $\Delta x$ under the other $\Delta$? Why does it not go under the constant v or u?
  

If you’re asking why we write $u\frac{\Delta v}{\Delta x}$ instead of $\frac{u}{\Delta x}\Delta v$, the answer is that they are algebraically the same; $\Delta x$ is not a differential but rather a standard number, so you can move it around like one. I’m sure you remember from algebra that $a(b/c)=(a/c)b$, and that same rule applies here—just set $a=u$, $b=\Delta v$ and $c=\Delta x$. Also, putting the incremental changes together as a ratio is more reminiscent of Leibniz notation and slope calculations, which the authors are using to try to make the calculus derivation clearer.
If you’re interested in a further example, when you get into more advanced calculus, you will see people separate the differentials and even put them on separate sides of equalities. For example, the following are equivalent statements:
$$\frac{dy}{dx}=f’(x) \iff dy=f’(x)\,dx$$
In fact, this becomes a critical technique in what is called $u$-substitution, which is kind of like the chain rule for integrals.
The moral: in early calculus, it’s okay to move your $\Delta x$ and $dx$ around with algebra, as long as you understand the meaning of the statement. (When you get very high up, people start to say this is abuse of notation, which is somewhat true, at least in most fields—again, it’s all about understanding the meaning of the statement.)


  
*In the last part, why does the $\Delta x$ go under the $\Delta v$ and not the $\Delta u$? Is there a reason?
  

Simply because it doesn’t matter, and you would get the same result anyways:
$$\left( \lim_{\Delta x\to0}\frac{\Delta u}{\Delta x}\right) \left( \lim_{\Delta x\to0}\Delta v\right) = \frac{du}{dx}\cdot0=0$$
A: 1) I think $\Delta x$ is under the other $\Delta$ purely because it looks nicer that way.  I don't see why it couldn't be written under v or u.  Also, keep in mind that v and u aren't constants, they are functions $u = f(x)$ and $v = g(x)$.
2) Once again, the order doesn't matter.  The way the book has it set up would yield $0\cdot{dv\over dx}$, and the other way would yield $0\cdot{du\over dx}$, both of which are just $0$.
