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.

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    $\begingroup$ I don't know, but I like this form! $\endgroup$ – mweiss Jun 19 '15 at 2:43
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    $\begingroup$ Would this not be a better fit for HSM? $\endgroup$ – Benjamin Dickman Jun 19 '15 at 2:58
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    $\begingroup$ I don't know where to find this, but, I see logarithmic differentiation in those formulas. $\endgroup$ – James S. Cook Jun 19 '15 at 3:14
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    $\begingroup$ The symmetries are appealing, but these formulations don't have the same domain as the originals if $f$ or $g$ vanish. $\endgroup$ – Adam Jun 19 '15 at 19:17
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    $\begingroup$ Personally I just teach my students to rewrite $f/g$ as $fg^{-1}$ and then use the product rule. When students memorize a complicated formula of any kind, there is very little chance that they'll retain it a month after the final exam. $\endgroup$ – Ben Crowell Jun 20 '15 at 14:48

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.

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  • $\begingroup$ I would go further and say: the rules cited by Joel are logarithmic differentiation. And they hold whether the factors are positive, negative, or complex. (Just not zero.) $\endgroup$ – Gerald Edgar Jun 19 '15 at 16:03
  • $\begingroup$ I recall that in Feynman's problem solving tips he recommended using logarithmic differentiation (though he didn't call it like that) when differentiating products of multiple functions. I guess having a more symmetric form helps, and the result is a bit more simplified. $\endgroup$ – Javier Jun 19 '15 at 19:07
  • $\begingroup$ I've never seen this before +1 $\endgroup$ – Karl Jun 20 '15 at 4:16
  • $\begingroup$ Thank you for the detailed explanation. $\endgroup$ – Joel Reyes Noche Jun 20 '15 at 5:06

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.

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  • $\begingroup$ The abstract of the article I cite is "Mathematical elegance is illustrated by strikingly parallel versions of the product and quotient rules of basic calculus, with some applications. Corresponding rules for second derivatives are given: the product rule is familiar, but the quotient rule is less so." $\endgroup$ – Joel Reyes Noche Aug 7 '16 at 1:06
  • $\begingroup$ I note that Eggleton and Kustov's result seems to have been published more than two years after they presented it, which seems pretty long in this case. $\endgroup$ – Joel Reyes Noche Aug 7 '16 at 1:08
  • $\begingroup$ These formulas are natural modifications of the higher-derivative product rule, see en.wikipedia.org/wiki/General_Leibniz_rule $\endgroup$ – James S. Cook Aug 7 '16 at 5:17

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