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I know how to prove the product rule for differentiation by using the definition for derivatives, i.e you "add zero" and collect terms. Yesterday a friend showed me another proof that involves the logarithm, implicit differentiation and the chain rule but after looking at it for a while I started to ask myself some questions about the proof. I don't know if it's me thinking about it all wrong or if it's something that actually makes the proof less rigorous. Anyway the proof goes like this (as most of you already know):

Let $$ y = f(x)g(x) $$ and take the logarithm of both sides

$$ \ln y = \ln f(x) + \ln g(x). $$

Now use implicit differentiation with respect to $x$ and solve for $y'$ while using the fact that $y = f(x) g(x)$

$$ \frac{1}{y}y' = \frac{1}{f(x)}f'(x) + \frac{1}{g(x)}g'(x) \quad \Rightarrow \quad y' = f'(x) g(x) + f(x)g'(x). $$

Now the problem I am having with this is that the logarithm is not defined for $f(x) \leq 0$ and $g(x) \leq 0$ so this proof should only apply for product of functions that are positive for all $x$, is not that right?

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    $\begingroup$ You're right that this proof presupposes that $f$ and $g$ are positive functions. Of course the product rule for positive functions immediately implies the product rule for negative functions (use $-f$ and $-g$ instead of $f$ and $g$) and for one positive and one negative functions (use $-f$ and $g$ or use $f$ and $-g$). But there seems to be real trouble with this sort of argument if one or both functions take the value $0$. $\endgroup$
    – Andreas Blass
    Nov 12, 2017 at 15:57
  • $\begingroup$ A statement is true if there exists a proof. It's not wrong just because there exist wrong "proofs". $\endgroup$
    – user436658
    Nov 12, 2017 at 16:03
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    $\begingroup$ It's not a question about whether product rule is true or not, it's a question about a proof for it. I guess what I am asking is: isn't this a proof that is not waterproof since I can pick functions (say for example $f(x) = -x^2$) that I simply cannot plug into the proof? Can this even be considered as a proof since this example fails? $\endgroup$
    – Claessie
    Nov 12, 2017 at 16:07
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    $\begingroup$ You may have a look at this answer math.stackexchange.com/a/2370012/72031 and in case you do have a look at it then don't miss the comments to the answer. $\endgroup$
    – Paramanand Singh ♦
    Nov 12, 2017 at 16:39
  • $\begingroup$ @ParamanandSingh thank you, was helpful! $\endgroup$
    – Claessie
    Nov 12, 2017 at 17:52

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If $f$ and $g$ are differentiable at $x$ then there exists $c$ so that $f+c>0$ and $g+c>0$ in a neighborhood of $x$. So the argument with the logarithm shows that $[(f+c)(g+c)]'=(f+c)'(g+c)+(f+c)(g+c)'$. But $$ [(f+c)(g+c)]'=(fg)'+cf'+cg'$$and $$(f+c)'(g+c)+(f+c)(g+c)'=f'g+fg'+cf'+cg'.$$

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  • $\begingroup$ Can you explain what you just showed because I don't follow? $\endgroup$
    – Claessie
    Nov 12, 2017 at 20:53
  • $\begingroup$ The log trick proves the product rule for $f+c$ and $g+c$, and then with a little algebra that implies the product rule for $f$ and $g$. $\endgroup$
    – David C. Ullrich
    Nov 12, 2017 at 23:05
  • $\begingroup$ Brilliant! Thanks. $\endgroup$
    – Claessie
    Nov 16, 2017 at 11:22

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