Can one prove that the product rule depends on continuity of differentiable functions? In calculus textbooks, the theorem asserting continuity of differentiable functions often appears just before the product rule.  The proof of the product rule usually says
\begin{align}
& \underbrace{\frac{f(x+h)g(x+h) - f(x)g(x)} h}_{\text{This}\to(fg)'(x)\text{ as }h\to0.} \\[12pt]
= {} & \overbrace{f(x+h)}^{\begin{smallmatrix}\text{This}\to f(x) \\ \text{ as }h\to0.\end{smallmatrix}}\ \ \overbrace{\left(\frac{g(x+h)-g(x)} h \right)}^{\text{This}\to g'(x)\text{ as }h\to0.} + g(x)\overbrace{\left(\frac{f(x+h)-f(x)} h\right)}^{\text{This}\to f'(x)\text{ as }h\to0.}.
\end{align}
The continuity of differentiable functions is used to prove the part that says $f(x+h)\to f(x)$ as $h\to0$.
My question is whether it can be demonstrated that all proofs of the product rule must rely on this fact or something in some reasonable sense equivalent to it?
(Just how it would enter the proof of the product rule by logarithmic differentiation I haven't finished thinking through yet.)
PS: That I was momentarily guilty of a stupid mistake is what caused me to think of posting this here.  In proving the quotient rule, we have
$$
\frac{\frac{f(x+h)}{g(x+h)} - \frac{f(x)}{g(x)}} h = \frac{\left(\frac{f(x+h) - f(x)} h \right)g(x) - f(x)\left(\frac{g(x+h)- g(x)} h \right)}{g(x+h)g(x)}.
$$
Looking at the numerator and losing sight of the denominator, I thought: lo and behold! We don't need continuity of differentiable functions for this one.  Upon further cogitation, I realized what I'd missed.
 A: You are using a much stronger fact, both $f$ and $g$ are, per assumption, differentiable in $x$ and thus $$f(x+h)=f(x)+hf'(x)+o_f(h)$$ and $$g(x+h)=g(x)+hg'(x)+o_g(h).$$ 
Which means that the functions must be automatically continuous in $x$.

One can of course also directly use those Weierstrass decompositions to prove the product rule,
$$
f(x+h)g(x+h)=f(x)g(x)+h(f(x)g'(x)+f'(x)g(x))+(h^2f'(x)g'(x)+f(x)o_g(h)+g(x)o_g(h))
$$
where the last term is again of size $o(h)$.

For the inverse function and thus the quotient rule one can similarly compute
$$
\frac1{g(x+h)}=\frac1{g(x)+hg'(x)+o_g(h)}=\frac{g(x)-hg'(x)}{g(x)^2+o(h)}=\frac1{g(x)}-\frac{g'(x)}{g(x)^2}+o(h)
$$
with application of binomial formulas resp. geometric series.
A: $\newcommand{\st}{\operatorname{st}}$I don't know if this counts, but the following is the approach taken by Keisler's Elementary Calculus: An Infinitesimal Approach. The important thing to note here is that $\st()$ denotes the standard, or non-infinitesimal part of a number, and $\st(a \cdot b) = \st(a) \cdot \st(b)$. Also note that the derivative of a function is defined in this case as $$f'(x) = \frac{df}{dx} = \st\left(\frac{f(x + \Delta x) - f(x)}{\Delta x}\right)$$ 
Where $\Delta x$ is infinitesimal.
Proof of the Product Rule.
Let $y = uv$, then $y + dy = (u + \Delta u)(v + \Delta v) = uv + u \ \Delta v + v \ \Delta u + \Delta u \ \Delta v$, or $$\Delta y = u \ \Delta v + v \ \Delta u + \Delta u \ \Delta v$$
(since $y = uv$). Then $$\frac{\Delta y}{\Delta x} = \frac{u \ \Delta v + v \ \Delta u + \Delta u \ \Delta v}{\Delta x} = u \frac{\Delta v}{\Delta x} + v \frac{\Delta u}{\Delta x} + \Delta u \frac{\Delta v}{\Delta x}$$
Taking the standard part, $$\frac{dy}{dx} = u \frac{dv}{dx} + v\frac{du}{dx} + 0$$
Or just $$\frac{d}{dx}[u v] = u \frac{dv}{dx} + v \frac{du}{dx}$$
Maybe this wasn't exactly what you were looking for, but this is a proof of the product rule without appealing to continuity (in fact, continuity isn't even discussed until the next chapter). Of course, this is if you're comfortable with nonstandard analysis. I await other answers that address your question in terms of standard analysis.
