Talk:Product rule

message: 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)
thread_title:  Proof 

message: 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)
thread_title:  Disambiguation  

message: I took away the "informal justification" section, for the following reasons:
#It is virtually identical in essence to the argument using differentials at the beginning of the article. The steps are almost identical, just put "delta-u" instead of "du", or vice versa.
#There is already a rigorous proof given. In other words, there is already a "formal" derivation (using differentials), and a rigorous proof (using difference quotients, so unless this is really a new idea (and it isn't, see #1), why is it necessary?
Revolver 03:21, 20 Feb 2004 (UTC)
thread_title:  "Informal justification" section 

message: 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.
:Is this just silly vandalism, or am I really missing something??? Revolver 16:11, 30 September 2005 (UTC)
replies: text: I couldn't understand this either and have removed it. Eric119 08:34, 16 October 2005 (UTC)
thread_title: ???

message: 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
:\begin{align}
 (f \cdot g)(x+\rho) &{}= f(x+\rho) \cdot g(x+\rho) \\
                     &{}= \left( f(x)+f'(x)\rho+O(\rho^2) \right) \cdot \left( g(x)+g'(x)\rho+O(\rho^2) \right) \\
                     &{}= f(x)g(x) + f'(x)g(x)\rho + f(x)g'(x)\rho + O(\rho^2)
\end{align}
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
: f'(x) = \mathrm{standard}\left( \frac{f(x+\rho) - f(x)}{\rho} \right) , 
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. --KSmrqT 04:26, 28 January 2007 (UTC)
thread_title:  Leibniz and intuition 

message: Formulas like
:(fg)'=f'g+fg' \,
:{d\over dx}(uv)=u{dv\over dx}+v{du\over dx}
are hard to decipher for the beginner. 
Multiplication signs
:(f\cdot g)'=f'\cdot g+f\cdot g' \,
:{d\over dx}(u\cdot v)=u\cdot {dv\over dx}+v\cdot {du\over dx}
make it easier.
Bo Jacoby (talk) 21:43, 11 June 2008 (UTC)
thread_title:  multiplication signs, please 

message: 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 :
(f\times g)'=f'\times g+f\times g' \,
and not 
:(f\times g)'=f'\times g+g'\times f \,
7yl4r (talk) 23:17, 24 February 2009 (UTC)
thread_title: For vector functions

message: 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
:\dfrac{d}{dx}(u\cdot v)=u\cdot \dfrac{dv}{dx}+v\cdot \dfrac{du}{dx}
but only
:\dfrac{d}{dx}(u\cdot v)=u\cdot \dfrac{dv}{dx}+\dfrac{du}{dx} \cdot v
or with brackets, where necessary
:\dfrac{d}{dx}(u\cdot v)=u\cdot \dfrac{d}{dx} v + (\dfrac{d}{dx} u) \cdot v
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)
replies: text: Did you mean commutativity rather than associativity?  Commutativity of multiplication says ab = ba.  Associativity is something else. Michael Hardy (talk) 14:21, 19 January 2010 (UTC)

text: The problem with stating it right is that students will go and forget that they aren't supposed to differentiate the v, very silly I know. Also it makes the two bits look different even though multiplication is commutative for what they're doing. Dmcq (talk) 16:34, 19 January 2010 (UTC)
thread_title:  Associativity ? 

message: 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)
replies: text: that citation you gave said nothing of the sort, perhaps you could point to the version of the book as well or indicate the place better? I'm not doubting Barrow made a geometrical analogue but lots of people did geometrical bits of calculus before it was formalised in an algebraic way. Also I would have thought more should be put into Barrow's article about his geometric lectures before branching out as it were? Dmcq (talk) 16:11, 18 October 2010 (UTC)
replies: text: The 2008 edition is online.  I am not sure what you mean above.  You were not able to find footnote 58 on pages 28-29?  Tkuvho (talk) 17:23, 18 October 2010 (UTC)
replies: text: I see what has happened. Okay I've read it and in it Leibniz is accused of plagiarism as far as I can see. Even though Leibniz quite evidently got it wrong in the first instance. That is so self-evidently silly I don't know why Child wrote it. However I can see the attribution of the product rule to Barrow. I still believe you would be much better off adding a bit about Barrow's geometrical lectures in his article for the reasons I gave above, Child wrote a book about them. Dmcq (talk) 18:22, 18 October 2010 (UTC)
replies: text: I read Child somewhat differently.  My guess is that Barrow had it first, Leibniz discovered it independently, and Newton I don't know.  You seem to know more about Barrow's geometrical lectures than I do; why don't you give it a try?  I can participate if you wish.  Tkuvho (talk) 20:08, 18 October 2010 (UTC)
thread_title:  source for barrow 

message: Isn't the law only valid for constant-mass systems?123Mike456Winston789 (talk) 22:05, 19 January 2011 (UTC)
replies: text: For constant mass you would get F=m(dv/dt), i.e. F=ma.  The formulation F=dp/dt is more general. Tkuvho (talk) 22:16, 19 January 2011 (UTC)
thread_title: Example

message: I'm not versed in non-standard analysis. What does the \text{st} 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)
replies: text: st denotes the standard part function. I have added a few words and links to the section to provide some minimal context. It still is rather bare I'm afraid. — Tobias Bergemann (talk) 08:58, 4 January 2011 (UTC)
replies: text:  Thank you, that's all I wanted. The st page is itself bare, but ah well. Maybe one day I'll have to read a book on non-standard analysis. 67.158.43.41 (talk) 13:17, 9 January 2011 (UTC)
replies: text: What is your background exactly?  To get some motivation it may be helpful to start at Ghosts of departed quantities.  The reason standard part function is "bare" is because it relies on other pages to explain the background.  In an encyclopedia, one can't expect each article to expain every point.  Tkuvho (talk) 01:44, 10 January 2011 (UTC)
thread_title:  <math>\text{st}</math> notation 

message: 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, 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)
replies: text: OK I'm removing it. Even in the unlikely event that it's true, it's not encyclopedic and (to me) is out of place in an otherwise excellent article. Adpete (talk) 06:34, 26 February 2013 (UTC)
thread_title:  "Common Error" - really? 

message: 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)
thread_title:  Formatting error. 

message: 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)
thread_title:  Rigorous to Direct? 

message: Please take a look at Wikipedia_talk:WikiProject_Mathematics#Article_product_rule. Dennui (talk) 06:19, 23 April 2019 (UTC)
thread_title:  [[Wikipedia_talk:WikiProject_Mathematics#Article_product_rule]] 

message: 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)
replies: text: The author of the proof was mechanically distributing the limits over sums and products. Then the limit in question simplifies in the next step. Whether to simplify it a step earlier, during the distribution step, is a matter of taste. For what it's worth, I prefer the current version, over the version that you seem to be proposing. Regards, Mgnbar (talk) 23:21, 27 October 2022 (UTC)
thread_title:  Is this limit superfluous? 

message: 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)
replies: text: It seems that what you call dudv is a differential form of order 2. This has nothing to do with the product rule, and this is described in details in the linked article. Product rule is about differentiation of a product of functions, not about the exterior product of differential forms. D.Lazard (talk) 12:06, 5 November 2023 (UTC)
thread_title:  comparison to it's product rule: cross term 

message: 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)
thread_title:  plausible search not supported