Just to elaborate on James Cook's observation in the comments, here is how to obtain these formulas using logarithmic differentiation. I have made the post CW since it does not really answer the question.

$ \begin{align*} P(x) &= f(x)g(x)\\ \log(P(x)) &= \log(f(x))+\log(g(x))\\ \frac{d}{dx} \log(P(x)) &= \frac{d}{dx}\log(f(x))+\frac{d}{dx}\log(g(x))\\ \frac{P'(x)}{P(x)} &= \frac{f'(x)}{f(x)}+\frac{g'(x)}{g(x)}\\ P'(x) &= P(x)(\frac{f'(x)}{f(x)}+\frac{g'(x)}{g(x)})\\ \frac{dP}{dx} &= f(x)g(x)(\frac{f'(x)}{f(x)}+\frac{g'(x)}{g(x)})\\ \end{align*} $

The quotient rule is derived similarly.

I have not thought about logarithmic differentiation in a while. I am a little worried about domain issues. I guess just wrapping the initial equation in absolute value signs resolve these, but this still does not cover zeros of $f$ and $g$.

Actually, zeros are one reason to avoid this form. For example, the usual product rule applies just as well to $f(x)=0$, $g(x) = \sin(x)$, while the modified form does not.

Just to elaborate on James Cook's observation in the comments, here is how to obtain these formulas using logarithmic differentiation. I have made the post CW since it does not really answer the question.

$ \begin{align*} P(x) &= f(x)g(x)\\ \log(P(x)) &= \log(f(x))+\log(g(x))\\ \frac{d}{dx} \log(P(x)) &= \frac{d}{dx}\log(f(x))+\frac{d}{dx}\log(g(x))\\ \frac{P'(x)}{P(x)} &= \frac{f'(x)}{f(x)}+\frac{g'(x)}{g(x)}\\ P'(x) &= P(x)\left(\frac{f'(x)}{f(x)}+\frac{g'(x)}{g(x)}\right)\\ \frac{dP}{dx} &= f(x)g(x)\left(\frac{f'(x)}{f(x)}+\frac{g'(x)}{g(x)}\right)\\ \end{align*} $

The quotient rule is derived similarly.

I have not thought about logarithmic differentiation in a while. I am a little worried about domain issues. I guess just wrapping the initial equation in absolute value signs resolve these, but this still does not cover zeros of $f$ and $g$.

Actually, zeros are one reason to avoid this form. For example, the usual product rule applies just as well to $f(x)=0$, $g(x) = \sin(x)$, while the modified form does not.