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I have long wondered why the product rule is taught the way it is. ${ d(UV)=Udv+Vdu}$

Don't get me wrong, I am not a complete NOB when it comes to calc, but the quotient rule states $${d(\frac {U}{V})=\frac {Vdu-Udv}{V^2}}$$ I know this is a matter of semantics, but is just seems to me that (in order to make the quotient rule easier to remember) the the product rule should be taught as ${d(UV)=Vdu+Udv}$ This will allow students to simply change the sign on the product rule and place the difference over $V^2$ when they need to recall the quotient rule so that $${\text{while}\space d(UV)=Vdu+Udv \space \space: d(\frac {U}{V})=\frac {Vdu-Udv}{V^2}}$$

Chris
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  • I'm not sure, but perhaps changing it your way would make students mix the two formulas since they would look more similar. By making them looking more different, it makes it less likely. Don't forget that people are mostly learning these formulas by heart rather that "understanding" them. – Traklon Feb 05 '14 at 14:58
  • my impression of students is that they remember the product rule as $Vdu + Udv$ just as often as they remember it as $Udv + Vdu$. I think telling them to just switch the sign would result in more mistakes than having them memorize it as a second formula. – Jim Feb 05 '14 at 14:59
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    It's wrong to teach $d(uv)=u,dv + v,du$ to begin with. It starts confusing the students' ideas about derivatives and differentials; what's wrong with $(fg)'=f'g+fg'$? – egreg Feb 05 '14 at 15:16
  • @egreg Nothing is wrong with convention but honestly, I had a harder time remembering the order of the quotient rule ( where it does make a difference) as opposed to the product rule where it doesn't – Chris Feb 05 '14 at 15:46
  • @Chris $v$ is at the denominator, so its derivative has a $-$ before it. But, honestly, $(f/g)'=(f'g-fg')/g^2$ is much easier and introduces no extraneous notion. – egreg Feb 05 '14 at 15:54
  • @egreg That would work also. Drop the ${g^2}$ and change the sign, you have the product rule. In fact, disregard my last comment. ${(fg)'=f'g+fg'}$ is equivalent in form to $ {d(UV)= Vdu+Udv}$ and to put it in the form of the quotient rule you change the sign and place the difference over the square of the second function. – Chris Feb 05 '14 at 16:11

2 Answers2

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That's exactly how many of us approach the product rule, consistent with your suggested approach.

I always teach the product rule for $\Big(f(x)g(x)\Big)'$ to be $$f'(x) g(x) + f(x) g'(x),$$ and the quotient rule $$\left(\dfrac{f(x)}{g(x)}\right)' = \dfrac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$$

Either presentation of the product rule is equivalent, thanks to the commutativity of addition.

So use what helps you best remember the product rule and the quotient rule.

amWhy
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  • I always presented it this [your] way also. I seem to recall reading once (perhaps a few years ago in the AP-Calculus listserv, back when I was a regular contributor) that there is a multivariable calculus explanation for the other way, but I don't recall what it is at the moment. – Dave L. Renfro Feb 05 '14 at 15:03
  • https://math.stackexchange.com/a/4261705/945479 – qifeng618 Sep 28 '21 at 01:33
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Probably the answer is just that nobody really cared.

Anyway, it doesn't seem to me so much better that way;

With the product rule, you just have to remember Derivative * Non derivative + Non derivative * Derivative is whatever order you like.

You don't focus on that, so it is easier and faster.

With the quotient rule order matters, and students are gonna learn it.

If we would focus on the order on product rule, students will still have to remeber some ordering, and they would have a bigger chance at getting confused (the minus sign is on the product rule or on the quotient rule? )

Chris
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Ant
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