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Hi,
the proof given for the product rule is a little hard to follow. What about rewriteing it, using the x -> x0 notation? And explaining the steps in more depth (I've had a little difficulty proving the tricky part to myself, when g(x)h(x) - g(x0)h(x0) becomes g(x) (h(x) - h(x0)) + h(x) ( g(x) - g(x0)).) PAStheLoD 18:49, 9 June 2006 (UTC)
---
Thanks, Anome. But if we have separate "formal" proofs that:
a) As δx -> 0, δy/δx -> dy/dx (what I have called "differentiation from first principles")
b) The limit of a product is the product of the limits, and vice versa (what I have called the "product rule for limits")
then does that make the proof count as a formal one? Kidburla2002.
The justification is informal, because it uses infinitesimals in an informal way: it works under most reasonable assumptions, but what about unreasonable ones? To do it formally, you can either use limits formally (with δ - ε type stuff), or a formalised system of infinitesimals. The Anome
I changed the formal proof; I agree that it was hard to follow. I think it's much easier to prove this as δx -> 0 rather than as x -> x0. It also fits better with the initial "intuitive" proof, so it makes it clearer how make an intuitive proof rigorous. I tried to be fairly clear (without being pedantic) about definitions, hypotheses, conclusions, and where we use properties of limits. I also was careful to write it in a way that kept f and g in the same order, so that the proof will "work" in the non-commutative case. — Preceding unsigned comment added by Natkuhn ( talk • contribs) 00:36, 26 December 2012 (UTC)
Why was this moved? What is going at Product rule? -- Tarquin
I am not aware of any other product rule, so a disambiguation like product rule (calculus) is unnecessary and I will move back to product rule. AxelBoldt 19:54 18 Jun 2003 (UTC)
The Product Rule is a proper name Pizza Puzzle
There's plenty of other identities which can be called product rules. There must be 15 of them involving curls and divs and grads and what not. But I'm an opponent of pre-emptive disambiguation, so I say leave it unless there's a real need to move it. Certainly you can call this one the product rule, and get away with it. -- Tim Starling 14:24 19 Jun 2003 (UTC)
There is another "product rule" that is often used in combinatorics. (see Rule of product) which (i think) would be better listed as "product rule" than "rule of product". -- Hughitt1 23:09, 3 April 2006 (UTC)
I took away the "informal justification" section, for the following reasons:
Revolver 03:21, 20 Feb 2004 (UTC)
This can also be seen as a barber shop analogy. For example, in the above example, u stands at one side while v takes the haircut.
The argument of Leibniz can be adapted to provide better intuition than more common proofs.
Let ρ be an infinitesimal, and write g(x+ρ) as g(x)+g′(x)ρ+O(ρ2); likewise for f(x+ρ). Then (f·g)(x+ρ) will be
When we take the derivative we subtract off the f(x)g(x) term, divide by ρ, and discard any remaining terms containing ρ — which will be the O(ρ2) bit. In other words, the product rule is merely the distributive law followed by the discarding of higher order terms. This also quickly explains the result for a multiple product; any product term with two or more derivatives will be higher order, thus discarded.
The derivative here is, of course, a non-standard analysis version, of the form
where ρ is an infinitesimal.
Sometimes this idea is conveyed with a picture of areas.
░░░░░░░░░░░░█ ▒▒▒▒▒▒▒▒▒▒▒▒░ ▒▒▒▒▒▒▒▒▒▒▒▒░ ▒▒▒▒▒▒▒▒▒▒▒▒░ ▒▒▒▒▒▒▒▒▒▒▒▒░ ▒▒▒▒▒▒▒▒▒▒▒▒░ ▒▒▒▒▒▒▒▒▒▒▒▒░
Here the top right corner corresponds to the higher order terms in ρ. -- KSmrq T 04:26, 28 January 2007 (UTC)
Formulas like
are hard to decipher for the beginner.
Multiplication signs
make it easier.
Bo Jacoby ( talk) 21:43, 11 June 2008 (UTC).
I'm new (sorry about sloppy/lack of code) and not a math-whiz (sorry if this is irrelevant), but in reading this article I thought it might be worth noting in the section "For vector functions" the difference between dot and cross products in this circumstance. Or perhaps just denoting the equation with the appropriate multiplication sign as noted above would suffice.
... that is :
and not
7yl4r ( talk) 23:17, 24 February 2009 (UTC)
I think it is very unfortunate that associativity is assumed throughout this article without ever mentioning it, then, under Generalisations suddenly the sequence of multiplicands is observed. IMHO the product rule should NEVER be stated
but only
or with brackets, where necessary
This would be appreciated by everyone who works with operators (quantum mechanics), matrices, or just wants to apply the product rule to vector products (as 7yl4r found out already). In this respect, unfortunately, many of the math lines need reworking - anyone? 213.68.42.99 ( talk) 13:15, 19 January 2010 (UTC)
See "The early mathematical manuscripts of Leibniz", Gottfried Wilhelm Leibniz, translated by J. M. Child; page 29, footnote 58. Tkuvho ( talk) 15:44, 18 October 2010 (UTC)
The claim that "Leibniz himself made this error initially" is absurd, though it has been recycled in some publications more interested in amusing the students than historical accuracy. Leibniz specifically says that he would like to determine whether the rule d(xy)=dxdy applies, makes a few calculations, and a few lines later rejects this. This was dealt in detail in Margaret Baron's book (last chapter). Tkuvho ( talk) 15:54, 18 October 2010 (UTC)
Isn't the law only valid for constant-mass systems? 123Mike456Winston789 ( talk) 22:05, 19 January 2011 (UTC)
I'm not versed in non-standard analysis. What does the notation in the "Using non-standard analysis" section mean? (I'd look it up in a non-standard analysis article but somehow I'm not comfortable editing a section about which I know very little.) 67.158.43.41 ( talk) 05:57, 4 January 2011 (UTC)
The calculation in the section "using nonstandard analysis" is incorrect. Tkuvho ( talk) 15:20, 13 February 2011 (UTC)
The article says "It is a common error, when studying calculus, to suppose that the derivative of (uv) equals (u ′)(v ′)". I dispute that. I don't see why it's intuitive (and the article doesn't say), and in my experience most students simply follow whatever recipe they're given. But leaving aside my personal opinion, it's unreferenced. It's been sitting there since the very first version of this article (Nov 2002) when WP was in its infancy, [1] and it's time it was removed. The reference for Leibniz' error can simply be moved to the history section. Adpete ( talk) 11:52, 25 February 2013 (UTC)
I'm getting the error
"Failed to parse(unknown function '\begin'): {\begin{aligned}d(u\cdot v)&{}=(u+du)\cdot (v+dv)-u\cdot v\\&{}=u\cdot dv+v\cdot du+du\cdot dv.\end{aligned}}"
in the first line of the discovery section. I don't know enough about wiki formatting math equations to fix this or if this might just be an error with my browser. Someone with better experience of formatting maths expressions should probably have a look at fixing it. — Preceding unsigned comment added by 110.20.69.222 ( talk) 07:56, 10 February 2014 (UTC)
Two proofs of the Product Rule are provided. As far as I can see they are both rigorous and they are both nicely presented. However, the title of the second proof, `Rigorous Proof', suggests, to me at least, that the proof by factoring is not rigorous. The second proof proceeds directly from the definition of the derivative. I suggest changing the title to `Direct Proof'. Michealefr ( talk) 08:24, 13 September 2015 (UTC)
Please take a look at Wikipedia_talk:WikiProject_Mathematics#Article_product_rule. Dennui ( talk) 06:19, 23 April 2019 (UTC) [2]
/info/en/?search=Wikipedia_talk:WikiProject_Mathematics/Archive/2019/Apr#Article_product_rule Dennui ( talk) 13:53, 5 July 2020 (UTC)
In the final line of the proof it seems to me that the limit applied to f(x) [i.e., "\lim_{\Delta x\to 0} f(x)"] shouldn't be there. After all, f(x)=f(x) by definition. This is a minor thing but if it's superfluous it makes the proof more complicated than it needs to be. Joe in Australia ( talk) 21:39, 27 October 2022 (UTC)
the current articles doesn't discuss dudv and why this is not used In differentiable deterministic calculus but is required in stochastic calculus. Ito's product rule is not cited or referenced in article which is therefore missing discussion of important concepts. Rdsk2014 ( talk) 08:22, 5 November 2023 (UTC)
If I google for "product rule for derivatives", this article should be among the first hits presented. As it is, it doesn't appear at all. So, this expression ought to be incorporated into the article in some way, perhaps as part of a list of keywords.
Big-picture takeaway: I guess we can't just blithely assume that a garrulous / comprehensive article on a topic will automatically contain the most plausible / relevant search strings for it. Maybe a deliberate program of inclusion of keywords for every article needs to be undertaken. Kontribuanto ( talk) 14:58, 5 March 2024 (UTC)
This
level-5 vital article is rated B-class on Wikipedia's
content assessment scale. It is of interest to the following WikiProjects: | |||||||||||
|
Hi,
the proof given for the product rule is a little hard to follow. What about rewriteing it, using the x -> x0 notation? And explaining the steps in more depth (I've had a little difficulty proving the tricky part to myself, when g(x)h(x) - g(x0)h(x0) becomes g(x) (h(x) - h(x0)) + h(x) ( g(x) - g(x0)).) PAStheLoD 18:49, 9 June 2006 (UTC)
---
Thanks, Anome. But if we have separate "formal" proofs that:
a) As δx -> 0, δy/δx -> dy/dx (what I have called "differentiation from first principles")
b) The limit of a product is the product of the limits, and vice versa (what I have called the "product rule for limits")
then does that make the proof count as a formal one? Kidburla2002.
The justification is informal, because it uses infinitesimals in an informal way: it works under most reasonable assumptions, but what about unreasonable ones? To do it formally, you can either use limits formally (with δ - ε type stuff), or a formalised system of infinitesimals. The Anome
I changed the formal proof; I agree that it was hard to follow. I think it's much easier to prove this as δx -> 0 rather than as x -> x0. It also fits better with the initial "intuitive" proof, so it makes it clearer how make an intuitive proof rigorous. I tried to be fairly clear (without being pedantic) about definitions, hypotheses, conclusions, and where we use properties of limits. I also was careful to write it in a way that kept f and g in the same order, so that the proof will "work" in the non-commutative case. — Preceding unsigned comment added by Natkuhn ( talk • contribs) 00:36, 26 December 2012 (UTC)
Why was this moved? What is going at Product rule? -- Tarquin
I am not aware of any other product rule, so a disambiguation like product rule (calculus) is unnecessary and I will move back to product rule. AxelBoldt 19:54 18 Jun 2003 (UTC)
The Product Rule is a proper name Pizza Puzzle
There's plenty of other identities which can be called product rules. There must be 15 of them involving curls and divs and grads and what not. But I'm an opponent of pre-emptive disambiguation, so I say leave it unless there's a real need to move it. Certainly you can call this one the product rule, and get away with it. -- Tim Starling 14:24 19 Jun 2003 (UTC)
There is another "product rule" that is often used in combinatorics. (see Rule of product) which (i think) would be better listed as "product rule" than "rule of product". -- Hughitt1 23:09, 3 April 2006 (UTC)
I took away the "informal justification" section, for the following reasons:
Revolver 03:21, 20 Feb 2004 (UTC)
This can also be seen as a barber shop analogy. For example, in the above example, u stands at one side while v takes the haircut.
The argument of Leibniz can be adapted to provide better intuition than more common proofs.
Let ρ be an infinitesimal, and write g(x+ρ) as g(x)+g′(x)ρ+O(ρ2); likewise for f(x+ρ). Then (f·g)(x+ρ) will be
When we take the derivative we subtract off the f(x)g(x) term, divide by ρ, and discard any remaining terms containing ρ — which will be the O(ρ2) bit. In other words, the product rule is merely the distributive law followed by the discarding of higher order terms. This also quickly explains the result for a multiple product; any product term with two or more derivatives will be higher order, thus discarded.
The derivative here is, of course, a non-standard analysis version, of the form
where ρ is an infinitesimal.
Sometimes this idea is conveyed with a picture of areas.
░░░░░░░░░░░░█ ▒▒▒▒▒▒▒▒▒▒▒▒░ ▒▒▒▒▒▒▒▒▒▒▒▒░ ▒▒▒▒▒▒▒▒▒▒▒▒░ ▒▒▒▒▒▒▒▒▒▒▒▒░ ▒▒▒▒▒▒▒▒▒▒▒▒░ ▒▒▒▒▒▒▒▒▒▒▒▒░
Here the top right corner corresponds to the higher order terms in ρ. -- KSmrq T 04:26, 28 January 2007 (UTC)
Formulas like
are hard to decipher for the beginner.
Multiplication signs
make it easier.
Bo Jacoby ( talk) 21:43, 11 June 2008 (UTC).
I'm new (sorry about sloppy/lack of code) and not a math-whiz (sorry if this is irrelevant), but in reading this article I thought it might be worth noting in the section "For vector functions" the difference between dot and cross products in this circumstance. Or perhaps just denoting the equation with the appropriate multiplication sign as noted above would suffice.
... that is :
and not
7yl4r ( talk) 23:17, 24 February 2009 (UTC)
I think it is very unfortunate that associativity is assumed throughout this article without ever mentioning it, then, under Generalisations suddenly the sequence of multiplicands is observed. IMHO the product rule should NEVER be stated
but only
or with brackets, where necessary
This would be appreciated by everyone who works with operators (quantum mechanics), matrices, or just wants to apply the product rule to vector products (as 7yl4r found out already). In this respect, unfortunately, many of the math lines need reworking - anyone? 213.68.42.99 ( talk) 13:15, 19 January 2010 (UTC)
See "The early mathematical manuscripts of Leibniz", Gottfried Wilhelm Leibniz, translated by J. M. Child; page 29, footnote 58. Tkuvho ( talk) 15:44, 18 October 2010 (UTC)
The claim that "Leibniz himself made this error initially" is absurd, though it has been recycled in some publications more interested in amusing the students than historical accuracy. Leibniz specifically says that he would like to determine whether the rule d(xy)=dxdy applies, makes a few calculations, and a few lines later rejects this. This was dealt in detail in Margaret Baron's book (last chapter). Tkuvho ( talk) 15:54, 18 October 2010 (UTC)
Isn't the law only valid for constant-mass systems? 123Mike456Winston789 ( talk) 22:05, 19 January 2011 (UTC)
I'm not versed in non-standard analysis. What does the notation in the "Using non-standard analysis" section mean? (I'd look it up in a non-standard analysis article but somehow I'm not comfortable editing a section about which I know very little.) 67.158.43.41 ( talk) 05:57, 4 January 2011 (UTC)
The calculation in the section "using nonstandard analysis" is incorrect. Tkuvho ( talk) 15:20, 13 February 2011 (UTC)
The article says "It is a common error, when studying calculus, to suppose that the derivative of (uv) equals (u ′)(v ′)". I dispute that. I don't see why it's intuitive (and the article doesn't say), and in my experience most students simply follow whatever recipe they're given. But leaving aside my personal opinion, it's unreferenced. It's been sitting there since the very first version of this article (Nov 2002) when WP was in its infancy, [1] and it's time it was removed. The reference for Leibniz' error can simply be moved to the history section. Adpete ( talk) 11:52, 25 February 2013 (UTC)
I'm getting the error
"Failed to parse(unknown function '\begin'): {\begin{aligned}d(u\cdot v)&{}=(u+du)\cdot (v+dv)-u\cdot v\\&{}=u\cdot dv+v\cdot du+du\cdot dv.\end{aligned}}"
in the first line of the discovery section. I don't know enough about wiki formatting math equations to fix this or if this might just be an error with my browser. Someone with better experience of formatting maths expressions should probably have a look at fixing it. — Preceding unsigned comment added by 110.20.69.222 ( talk) 07:56, 10 February 2014 (UTC)
Two proofs of the Product Rule are provided. As far as I can see they are both rigorous and they are both nicely presented. However, the title of the second proof, `Rigorous Proof', suggests, to me at least, that the proof by factoring is not rigorous. The second proof proceeds directly from the definition of the derivative. I suggest changing the title to `Direct Proof'. Michealefr ( talk) 08:24, 13 September 2015 (UTC)
Please take a look at Wikipedia_talk:WikiProject_Mathematics#Article_product_rule. Dennui ( talk) 06:19, 23 April 2019 (UTC) [2]
/info/en/?search=Wikipedia_talk:WikiProject_Mathematics/Archive/2019/Apr#Article_product_rule Dennui ( talk) 13:53, 5 July 2020 (UTC)
In the final line of the proof it seems to me that the limit applied to f(x) [i.e., "\lim_{\Delta x\to 0} f(x)"] shouldn't be there. After all, f(x)=f(x) by definition. This is a minor thing but if it's superfluous it makes the proof more complicated than it needs to be. Joe in Australia ( talk) 21:39, 27 October 2022 (UTC)
the current articles doesn't discuss dudv and why this is not used In differentiable deterministic calculus but is required in stochastic calculus. Ito's product rule is not cited or referenced in article which is therefore missing discussion of important concepts. Rdsk2014 ( talk) 08:22, 5 November 2023 (UTC)
If I google for "product rule for derivatives", this article should be among the first hits presented. As it is, it doesn't appear at all. So, this expression ought to be incorporated into the article in some way, perhaps as part of a list of keywords.
Big-picture takeaway: I guess we can't just blithely assume that a garrulous / comprehensive article on a topic will automatically contain the most plausible / relevant search strings for it. Maybe a deliberate program of inclusion of keywords for every article needs to be undertaken. Kontribuanto ( talk) 14:58, 5 March 2024 (UTC)