This is motivated by looking at the calculus exams of some of my undergraduate students. A recurring mistake is assuming that the derivative of the product of functions is a product of derivatives and the derivative of the quotient of two functions is the quotient of their derivatives.

This might be an ill formed question, but do their exist a pair of functions such that these rules hold. Explicitly, do there exist $f(x),g(x)$ satisfying all of: \begin{align*} f&\neq 0;\\ \frac{d(fg)}{dx}&=\frac{df}{dx}\cdot \frac{dg}{dx}\\ \text{and }\frac{d(f/g)}{dx}&=\frac{df}{dx}/ \frac{dg}{dx}\text{ ?} \end{align*}

EDIT: Just to clarify my question, looking at the responses so far. I am looking for functions satisfying all three conditions above simultaneously.

  • $\begingroup$ @Timothy Wagner: you also want $g\neq 0$, so the first one does not have trivial answers. $\endgroup$ – Arturo Magidin Nov 19 '10 at 4:57
  • $\begingroup$ @Arturo: I was looking for functions satisfying all three conditions. For just the product rule, $f=g=e^{2x}$ is one solution, if I did the algebra right. Anyway, $g\neq 0$ is fair enough. $\endgroup$ – Timothy Wagner Nov 19 '10 at 5:02
  • $\begingroup$ @Timothy Wagner: Oh; sorry. $\endgroup$ – Arturo Magidin Nov 19 '10 at 5:03
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    $\begingroup$ An anecdote. Long ago I put on a calculus exam a random differentiation problem. By coincidence, if the student mistakenly computed the derivative of a quotient as the quotient of the derivatives, the right answer would result. $\endgroup$ – GEdgar Nov 19 '10 at 16:17
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    $\begingroup$ @GEdgar And did anybody get points for getting the right answer in the wrong way? =P $\endgroup$ – Adrián Barquero Nov 19 '10 at 16:35

I don't think there is a solution.

$f'g'=f'g+g'f$ implies $\frac{f'}{f}=\frac{g'}{g'-g}$ where the denominators are nonzero.

$\frac{f'}{g'}=\frac{f'g-g'f}{g^2}$ implies $\frac{f'}{f}=\frac{(g')^2}{g(g'-g)}$ where the denominators are nonzero.

If $f$ and $g$ are nonzero and $g'-g$ is nonzero, this implies that $\frac{g'}{g'-g}=\frac{(g')^2}{g(g'-g)}$, which implies $g'=g$. Thus $g=g'$ if $f$ and $g$ are nonzero, so $g(x)=ce^x$. But now from $f'g'=f'g+g'f$ we have $ce^xf(x)=0$, which implies $f=0$.

For solutions to just the product rule, the equations above suggest taking $g$ such that $\frac{g'}{g'-g}$ is integrable, and $$f(x)=\exp\left(\int_a^x\frac{g'(t)}{g'(t)-g(t)}dt\right),$$ which in particular works for $f(x)=e^{bx}$ and $g(x)=e^{cx}$ with $bc=b+c$ (as seen in a now deleted answer). Similarly, for just the quotient rule, one could try finding $g$ such that $\frac{(g')^2}{g(g'-g)}$ is integrable, and then take $$f(x)=\exp\left(\int_a^x\frac{g'(t)^2}{g(t)(g'(t)-g(t))}dt\right).$$ For example, $f(x)=e^{bx}$ and $g(x)=e^{cx}$ with $bc=b+c^2$.

  • $\begingroup$ Thanks, that's neat. I need to get better at manipulating these equations. $\endgroup$ – Timothy Wagner Nov 19 '10 at 5:34
  • $\begingroup$ I got the first, but you beat me. Good work. $\endgroup$ – Ross Millikan Nov 19 '10 at 6:02
  • $\begingroup$ @Ross: Thanks. I'm afraid I don't understand your comment. You got the first what? $\endgroup$ – Jonas Meyer Nov 19 '10 at 6:08
  • $\begingroup$ The first part of your answer-above the break line. Frustrated that your solution came in while I was typing up mine. But still gave an upvote. $\endgroup$ – Ross Millikan Nov 19 '10 at 6:11

Another answer to your question from a different angle is this: They hold in a different kind of calculus.

If you swap subtraction for division and division for taking roots in the usual definition of the derivative, you get

$$f^*(x) = \lim_{\Delta x \to 0} \left(\frac{f(x+\Delta x)}{f(x)}\right)^{\frac{1}{\Delta x}}.$$

It's not too hard to prove that $(fg)^* = f^* g^*$ and $(f/g)^* = f^*/g^*$ under this definition.

The calculus that results from this definition of the derivative (and the corresponding definition of the integral) goes by various names, such as "multiplicative calculus," "non-Newtonian calculus," and "product calculus." Many of the standard results in the usual calculus (e.g., Fundamental Theorem, Mean Value Theorem, l'Hopital-type rules, Taylor's Theorem) can be redone for this multiplicative calculus. The idea goes back at least to Vito Volterra in the late 1800s.

According to the Wikipedia page, "Opinions differ as to the usefulness of the multiplicative calculi," but there are some applications. For instance, the product derivative measures the multiplicative rate of change of some function and so is useful when you're interested in looking at, say, growth rates of stock prices. The product integral has some corresponding applications; my favorite is that it can be used to calculate geometric means (in the same way that the usual integral can find arithmetic means).

Multiplicative calculus also has a close relationship to the usual calculus. For instance, the derivatives are related via $$f^*(x) = \exp\left(\frac{d}{dx}\ln |f(x)|\right) = e^{f'(x)/f(x)}.$$ The product integral and the usual integral have a similar relationship. In fact, I think part of the reason some folks don't find this multiplicative calculus all that interesting is that product derivatives and integrals can so easily be expressed in terms of the usual ones. The question, I suppose, is whether there is really anything new and different going on here. (Note also that this relationship I just gave means that the product derivative is just $\exp$ of the logarithmic derivative of $f$.)

Anyway, a few years ago I wrote a survey paper and had a student do a summer research project on this before I realized how much was already out there and before the Wikipedia page appeared. There are lots of other references and information on the Wikipedia page, so if you find this interesting you should check it out. The page on the product integral is also a good source.

  • $\begingroup$ Interesting. Thanks and thanks for the link to the paper. $\endgroup$ – Arturo Magidin Nov 19 '10 at 5:50
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    $\begingroup$ Thanks for that answer. This was interesting to know. This is the great thing about posting on math SE, even innocuous questions lead to interesting, unexpected answers. $\endgroup$ – Timothy Wagner Nov 19 '10 at 5:51
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    $\begingroup$ @Timothy: Thanks, and I agree. Sometimes I find myself wanting to post questions I already know the answers to, just to see what new insights the collective minds on this site will provide. $\endgroup$ – Mike Spivey Nov 19 '10 at 5:54
  • $\begingroup$ Yes, beautiful, just what I thought. $\endgroup$ – mathreadler Jun 10 '17 at 10:46

The product rule equality gives $f'g+g'f = f'g'$, or $-fg' = f'g-f'g'$. From this, using the quotient rule equality we get $$ \frac{f'}{g'} = \frac{gf'-fg'}{g^2} = \frac{gf' +f'g - f'g'}{g^2} = f'\left(\frac{2g-g'}{g^2}\right). $$ If $f$ is not constant, then $f'\neq 0$. So we get $\frac{1}{g'} = \frac{2g-g'}{g^2}$, or $g^2 - 2gg'+(g')^2 = 0$, or $(g-g')^2 = 0$; hence $g=g'$.

Therefore, from the product rule again we have $f'g'=f'g+fg'$, or $f'g = f'g+fg$, so $fg=0$. This requires $g=0$, which makes the quotient rule impossible.

If $f$ is constant, then the product rule equality requires $g$ constant, and then the quotient rule equality cannot hold.

  • $\begingroup$ Thanks, Arturo. This is similar to Jonas' solution. $\endgroup$ – Timothy Wagner Nov 19 '10 at 5:35
  • $\begingroup$ @Timothy Wagner: Yes: he posted it while I was double-checking my algebra... I was going the same route, but this way the only assumption you need to do is $f'\neq 0$, and don't have to worry about $g-g'\neq 0$. $\endgroup$ – Arturo Magidin Nov 19 '10 at 5:36
  • $\begingroup$ In any case, it is nice to see your complementary solution. I actually liked worrying about $g-g'\neq0$, because to my surprise the assumption that $g-g'\neq0$ leads to the conclusion that $g-g'=0$. $\endgroup$ – Jonas Meyer Nov 19 '10 at 5:42
  • $\begingroup$ @Jonas Meyer: which of course just means that $g-g'=0$, since $(\neg p\rightarrow p)\rightarrow p$ is a tautology. (-: $\endgroup$ – Arturo Magidin Nov 19 '10 at 5:46
  • $\begingroup$ Precisely. $\endgroup$ – Jonas Meyer Nov 19 '10 at 5:50

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