calculus - Product rule proof involving logarithm and implicit differentiation really waterproof?

I know how to prove the product rule for differentiation by using the definition for derivatives, i.e you "add zero" and collect terms. Yesterday a friend showed me another proof that involves the logarithm, implicit differentiation and the chain rule but after looking at it for a while I started to ask myself some questions about the proof. I don't know if it's me thinking about it all wrong or if it's something that actually makes the proof less rigorous. Anyway the proof goes like this (as most of you already know):

Let
$$
y = f(x)g(x)
$$
and take the logarithm of both sides

$$

\ln y = \ln f(x) + \ln g(x).
$$

Now use implicit differentiation with respect to $x$ and solve for $y'$ while using the fact that $y = f(x) g(x)$

$$
\frac{1}{y}y' = \frac{1}{f(x)}f'(x) + \frac{1}{g(x)}g'(x) \quad \Rightarrow \quad y' = f'(x) g(x) + f(x)g'(x).
$$

Now the problem I am having with this is that the logarithm is not defined for $f(x) \leq 0$ and $g(x) \leq 0$ so this proof should only apply for product of functions that are positive for all $x$, is not that right?

Answer

If $f$ and $g$ are differentiable at $x$ then there exists $c$ so that $f+c>0$ and $g+c>0$ in a neighborhood of $x$. So the argument with the logarithm shows that $[(f+c)(g+c)]'=(f+c)'(g+c)+(f+c)(g+c)'$. But $$
[(f+c)(g+c)]'=(fg)'+cf'+cg'$$and $$(f+c)'(g+c)+(f+c)(g+c)'=f'g+fg'+cf'+cg'.$$