TITLE: Differentiating $y=x^{2}$
QUESTION [15 upvotes]: I am reading in a book about differentiating, but I am confused with one of the steps he takes.  We start with:
$$\begin{align}
y &= x^{2} \\
y + \mathrm{d}y &= (x + \mathrm{d}x)^2 \\
y + \mathrm{d}y &= x^2 + x\mathrm{d}x + x\mathrm{d}x + (\mathrm{d}x^2)
\end{align}$$
Now the author simplifies this to:
$$y + dy = x^2 + 2x\mathrm{d}x + (\mathrm{d}x^2)$$
I dislike how the middle term is simplified to $2x\mathrm{d}x$ instead of $2(x\mathrm{d}x)$, as I feel like it is more intuitive on what is going.  As in, $2$ of the term $x\mathrm{d}x$, instead of $2x\mathrm{d}x$.  But I fear writing it as $2(x\mathrm{d}x)$ may result in an incorrect distributive property.
Next, he omits the $(\mathrm{d}x^2)$:  $y + \mathrm{d}y = x^2 + 2x \mathrm{d}x$.
Subtract the original $y = x^2.$:
$$\mathrm{d}y = 2x \mathrm{d}x.$$
Now here is where I get confused:
$$\frac{\mathrm{d}y}{\mathrm{d}x} = 2x.$$
How can he just divide both sides by $\mathrm{d}x$!?  If the original term was $2$ of $x\mathrm{d}x$, wouldn't it have to be written out as $2x * 2\mathrm{d}x$, and thus divide both sides by $2\mathrm{d}x$ instead?
I think the root of my confusion is how to properly simplify:
$x\mathrm{d}x + x\mathrm{d}x$.
I trust that he is right, but I am looking for an explanation of why his simplification can work, and why $2(x\mathrm{d}x)$ would be incorrect.

REPLY [5 votes]: "If the original term was $2$ of $x \, dx$, wouldn't it have to be written out as $2x\cdot 2 \,dx$, and thus divide both sides by $2 \, dx$ instead?"
Don't let the infinitesimal stuff throw you for a loop.  I know the pure math guys are going to tear their hair out when I say this, but at your stage of the game you can just treat dx like a number.  Specifically, where you have $2(xdx)$, substitute 1 for $x$ and $0.1$ for $dx$.  Then you have $2(1*0.1)$.  Work it out.  Does this give you $2\times1\times2\times0.1=0.4$?  You probably see that it doesn't.  It gives you $2\times 1\times 0.1=0.2$.  In the same way, $2(xdx)=2\times x \times dx = 2xdx$.
My answer is basically the same as amWhy's earlier answer, but I thought a concrete example would be helpful.