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.
Best Answer
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.