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A while back I saw someone claim that you could prove the product rule in calculus with the single variable chain rule. He provided a proof, but it was utterly incomprehensible. It is easy to prove from the multi variable chain rule, or from logarithmic differentiation, or even from first principles. Is there an actual proof using just the single variable chain rule?

Aaron
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  • Just out of curiosity, is this where you saw the claim? – SCappella Aug 03 '18 at 07:43
  • @user Yes, although when I had last looked at that thread, nobody had responded to the comment. – Aaron Aug 03 '18 at 12:13
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    related: https://math.stackexchange.com/a/208014/13618 https://mathoverflow.net/questions/44774/do-these-properties-characterize-differentiation The product rule and chain rule are not logically independent. Given one, you can prove the other/ –  Aug 03 '18 at 15:36

3 Answers3

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Credit is due to this video by Mathsaurus, but also requires the sum and power rules for derivatives.

Let $u$, $v$ be appropriate functions. Consider $f = (u+v)^2$. By the chain rule, $$f^{\prime} = 2(u+v)(u^{\prime}+v^{\prime})=2(uu^{\prime}+uv^{\prime}+vu^{\prime}+vv^{\prime})\text{.}$$ Now, expand $f$ to obtain $f = u^2+2uv+v^2$, and then we have $$f^{\prime}=2uu^{\prime}+2(uv)^{\prime}+2vv^{\prime}=2[uu^{\prime}+(uv)^{\prime}+vv^{\prime}]\text{.}$$ It follows immediately that $$(uv)^{\prime}=uv^{\prime}+vu^{\prime}\text{.}$$

Clarinetist
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    Very nice. Exactly the kind of thing I was hoping for (since it only relies on the derivative for $x^2$ and the chain rule. – Aaron Aug 02 '18 at 20:01
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Here’s a relatively simple one that only uses the single variable chain rule and some elementary algebra.

Consider two functions $f,g$ and the expression $\frac{d}{dx} (f+g)^2$. Directly applying the chain rule, this becomes $2(f+g)(f’+g’)$.

Instead of directly using the chain rule, one can also multiply out $(f+g)^2$, and the expression becomes $\frac{d}{dx}(f^2+2fg+g^2)$, and we can differentiate term by term to obtain $2ff’+2(fg)’+2gg’$.

Since these two expressions are equal, we have $$ 2(f+g)(f’+g’)=2ff’+2(fg)’+2gg’$$ Using some algebra, $$2ff’+2fg’+2f’g+2g’g=2ff’+2(fg)’+2gg’$$ $$2fg’+2f’g=2(fg)’$$ $$(fg)’=fg’+f’g$$

Tyler6
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It all depends on what other principles you have available to you, but since you mention logarithmic differentiation, note that $\log(fg) = \log(f) + \log(g)$. Take the derivative of both sides and multiply by $fg$, to get $(fg)' = f'g + fg'$.

Sridhar Ramesh
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    Yes, that is the easy proof using log differentiation that I said I already knew. – Aaron Aug 02 '18 at 19:31
  • Ah, sorry, I read too fast and misinterpreted. But note that the proof using logarithmic differentiation is using the single-variable chain rule, in its differentiations of log(whatever). – Sridhar Ramesh Aug 02 '18 at 19:35
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    That's perfectly fine. Log differentiation is technically an application of the chain rule, so you aren't wrong. – Aaron Aug 02 '18 at 19:36
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    @Aaron: This is not perfectly fine. You cannot take logarithm of anything unless you know it is nonzero. Furthermore, it is false that $\ln(xy) = \ln(x)+\ln(y)$ when $x=y=-1$, for any reasonable definition of $\ln$. – user21820 Aug 03 '18 at 08:46
  • @user21820 I meant the fact that he didn't notice that I knew the proof was fine, not that the proof didn't have its limits. However, the issue with signs is easy to deal with for any functions with isolated zeros by (locally) replacing one or both functions with their negatives. – Aaron Aug 03 '18 at 12:23
  • @Aaron: I don't agree; you're still not dealing with the derivative at the zero itself. Furthermore, I challenge you to actually write out a full rigorous proof of the general product rule using your claimed approach. Handwaving is not the way to go here. And there is also a risk of circularity; did you prove the derivative of $\ln$ without proving anything of the same core nature as the product rule? – user21820 Aug 03 '18 at 12:32
  • @user21820 To get to log, we need integration, but showing $\log(ab)=\log(a)+\log(b)$ then only requires $u$-substitution, which is a manifestation of the chain rule. So the logic isn't circular, but it still requires more than I would like. In particular, the proof wouldn't work in a calculus course, where integration is introduced well after the product rule. – Aaron Aug 03 '18 at 13:10
  • @Aaron: Honestly, I'm not even convinced it isn't 'circular'. Just because the theorems used don't have the name "product rule" doesn't mean their proof doesn't do something essentially of the same nature as the proof of the product rule. As an analogy, I could prove the quotient rule, where my proof internally establishes the product rule and chain rule and then uses them to get the quotient rule. Then I could prove the product rule from the quotient rule. Technically it is non-circular, but obviously it is. This is vague, but I think you get what I mean. – user21820 Aug 03 '18 at 13:55
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    @user21820 I've taught calculus enough times to know exactly what goes into the proofs here. You don't need the product rule (though I think you need linearity of the derivative). Yes, circularity can be subtle. It just doesn't happen here. You don't need the product rule to establish the definition of integration, the fundamental theorem of calculus, u-substitution, the definition of log as an integral, or the relevant property of logs. If you don't believe me, sit down and do it. Your doubt is meaningless in the face of something this easy to verify. – Aaron Aug 03 '18 at 15:15
  • @Aaron: Obviously, you don't understand my last comment, so it's pointless to continue. – user21820 Aug 04 '18 at 09:09