In proving the product rule, how do we know to add and subtract f(x+h)g(x) from the numerator in the derivative definition? I watched two YouTube videos to try to get a proof that makes sense, but in both videos, the authors said something to the effect of "add and subtract f(x+h)g(x)" without a good explanation as to how to come up with that step (in this video, for example, the author just says "this will make everything work out here".)
 A: Let's look at the limit
$$
\lim_{h \to 0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}=\lim_{h \to 0} \frac{\Delta(fg)}{h}
$$
If we're trying to relate this to the derivatives of $f$ and $g$, we'll need to find some way to relate the terms in the numerator to the expressions $\Delta f=f(x+h)-f(x)$ and $\Delta g = g(x+h)-g(x)$.
How could $f(x+h)g(x+h)$ occur as part of some expression that included $\Delta f$ and/or $\Delta g$? Well, it's the first term in $f(x+h)\Delta g$:
$$
f(x+h)\Delta g=f(x+h)[g(x+h)-g(x)]=f(x+h)g(x+h)-f(x+h)g(x)
$$
So, if the numerator included a $-f(x+h)g(x)$ term, we could combine it with $f(x+h)g(x+h)$ to get something involving $\Delta g$. This suggests that it's worth trying to add and subtract $f(x+h)g(x)$: certainly subtracting it will let us do something useful to $f(x+h)g(x+h)$, and maybe we'll luck out and adding it will let us do something useful to $f(x)g(x)$. (And, as it happens, it does!)
A: In contemplating
$$(1)\ \ \ \ f(x+h)g(x+h) - f(x)g(x)$$
you might curse a bit and pose yourself the question "wouldn't life be simpler if it was
$$f(x+h)g(x) - f(x)g(x)$$
instead?" A hallowed principle of mathematical skullduggery is to just blatantly insert what you want to see, and then fess up and correct for it. So we write $(1)$ as 
$$[f(x+h)g(x) - f(x)g(x)] + [f(x+h)g(x+h) - f(x+h)g(x)].$$
In other words [what we want to see] + [correction]. There is no way to know ahead of time if this nefarious scheme is going to work; the correction may be just as difficult to think about as the original. But sometimes it works just fine thank you, and this is one of those times.
A: For $h\ne0$ one has, identically in $x$ and $h$, 
$$\eqalign{{f(x+h)g(x+h)-f(x)g(x)\over h}&={f(x+h)-f(x)\over h}\ g(x)+f(x)\ {g(x+h)-g(x)\over h}\cr&\qquad+\ {f(x+h)-f(x)\over h}\ {g(x+h)-g(x)\over h}\ h\ .\cr}$$
Now let $h\to0$.
