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?
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$\begingroup$ Just out of curiosity, is this where you saw the claim? $\endgroup$– SCappellaAug 3, 2018 at 7:43
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$\begingroup$ @user Yes, although when I had last looked at that thread, nobody had responded to the comment. $\endgroup$– AaronAug 3, 2018 at 12:13
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2$\begingroup$ related: math.stackexchange.com/a/208014/13618 mathoverflow.net/questions/44774/… The product rule and chain rule are not logically independent. Given one, you can prove the other/ $\endgroup$– user13618Aug 3, 2018 at 15:36
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3 Answers
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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 $$(uv)^{\prime}=uv^{\prime}+vu^{\prime}\text{.}$$
answered Aug 2, 2018 at 19:58
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2$\begingroup$ 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. $\endgroup$– AaronAug 2, 2018 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, $$2ff’+2fg’+2f’g+2g’g=2ff’+2(fg)’+2gg’$$ $$2fg’+2f’g=2(fg)’$$ $$(fg)’=fg’+f’g$$
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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 $(fg)' = f'g + fg'$.
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2$\begingroup$ Yes, that is the easy proof using log differentiation that I said I already knew. $\endgroup$– AaronAug 2, 2018 at 19:31
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$\begingroup$ 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). $\endgroup$ Aug 2, 2018 at 19:35
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1$\begingroup$ That's perfectly fine. Log differentiation is technically an application of the chain rule, so you aren't wrong. $\endgroup$– AaronAug 2, 2018 at 19:36
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1$\begingroup$ @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 $\ln$. $\endgroup$ Aug 3, 2018 at 8:46
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2$\begingroup$ @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. $\endgroup$– AaronAug 3, 2018 at 15:15