Symmetric version of product and quotient differentiation rules

Question

The usual way of writing the product rule and the quotient rule in differentiation is $$(fg)'=f'g+fg'$$ $$\left(\frac{f}{g}\right)'=\frac{f'g-fg'}{g^2}\quad\text{where}\quad g\ne 0$$

A few years ago, during a mathematics conference I attended in the Philippines, a foreign speaker presented a "more symmetric" way of writing these rules. Unfortunately, I have forgotten what he wrote and an internet search did not yield it.

I've been playing around and I've found the following "nicer" presentation: $$(fg)'=fg\left(\frac{f'}{f}+\frac{g'}{g}\right)\quad\text{where}\quad f,g\ne 0$$ $$\left(\frac{f}{g}\right)'=\frac{f}{g}\left(\frac{f'}{f}-\frac{g'}{g}\right)\quad\text{where}\quad f,g\ne 0$$

I don't know if this is the version I saw a few years ago, but I'm pretty sure I'm not the first to think of this.

What textbooks or lecture notes do you know that use this form for the product rule and the quotient rule in differentiation?

Edit: As was pointed out in the comments, the "symmetric" version only works when $f$ and $g$ are nonzero, so it is not exactly the same as the "original" version.

Answer 1

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.

Answer 2

I've been going through my papers and have found the source of the idea. Roger Eggleton presented it in a talk at the Severino V. Gervacio Conference on Graph Theory and Combinatorics 2009 held in Manila, Philippines from April 24, 2009 to April 25, 2009.

Aside from the two equations in the original post, he also presented the two equations below: $$\frac{(fg)''}{fg}=\frac{f''}{f}+\frac{g''}{g}+2\frac{f'g'}{fg}$$ $$\frac{(f/g)''}{f/g}=\frac{f''}{f}-\frac{g''}{g}-2\left(\frac{f'}{f}-\frac{g'}{g}\right)\frac{g'}{g}$$

It seems the original reference for this is:

Roger Eggleton and Vladimir Kustov, "The Product and Quotient Rules Revisited," The College Mathematics Journal, Volume 42, Number 4, September 2011, pp. 323-326.