Talk:Quotient rule

message: f(x) = \frac{|x|}{|x|} is not at all differentiable at x = 0; it's not even defined there.
138.89.101.39 (talk) 00:18, 10 April 2009 (UTC)
thread_title:  Regarding Limitations 

message: Out of curiosity, what exactly is there about the informal proof that makes it informal? Cburnett 23:22, 3 Feb 2005 (UTC)
replies: text: I guess it's considered an informal proof because it's based off the product rule, while the "formal" proof is based off the difference quotient, which is more direct. I changed the headings to be more specific. - Evil saltine 05:11, 4 Feb 2005 (UTC)

text: The proof is informal because it contains a hidden assumption, that the derivative of f exists. David Radcliffe 02:44, 26 May 2007 (UTC)
thread_title:  Informal proof 

message: The total differential proof uses the fact that the derivative of 1/x is −1/x2. But without the quotient rule, one doesn't know the derivative of 1/x, without doing it directly, and once you add that to the proof, it doesn't seem as "elegant" anymore, but without it, it seems circular. Revolver 16:03, 30 September 2005 (UTC)
replies: text: I would vote to remove that proof altogether. Too many proofs in this article, and they are lengthy and using lots of calculations. Oleg Alexandrov 17:44, 2 October 2005 (UTC)

text: I guess I'll go ahead and get rid of it. I'll save the section incase anyone wants it back. (Frazz 18:04, 7 February 2007 (UTC)

text: This total differential proof uses knowledge of what the derivative of 1/x is.  If you know that, you can prove the quotient rule in two lines using the product and chain rules, not having to go through a huge mumbo-jumbo of differentials.  I really don't know why such a proof is not on this page and numerous complicated ones are. 67.149.103.91 (talk) 04:24, 17 June 2010 (UTC)
thread_title: Total differential proof

message: A proof of the quotient rule is not complete. Why h(x)≠ 0 does implies h(x+\Delta x)\ne 0? I have seen a proof of the quotient rule based on a characterization theorem (don't know its proper English name). --Matikkapoika 01:25, 15 November 2006 (UTC)
replies: text: I know I'm no expert and am probably spouting rubbish, but would it be that h(x)\neq 0\Rightarrow h(x+\Delta x)\neq 0 as \Delta x\to 0, i.e. h(x+\Delta x)\to h(x)\neq 0? 118.90.51.62 (talk) 02:13, 22 June 2008 (UTC)
thread_title: Fix needed in a proof

message: The text:
The rest is simple algebra to make f'(x) the only term on the left hand side of the equation and to remove f(x) from the right side of the equation.
on my PC, I can't see the dash after the f
SethMould 07:53, 16 July 2007 (UTC)
thread_title:  From the product rule 

message: There is an extra open brackets in the third line of the first proof.  —Preceding unsigned comment added by 201.95.65.32 (talk) 17:47, 15 December 2009 (UTC)
thread_title:  Extra '(' 

message: The very first definition of quotient rule is wrong. There is mistake of sign.  — Preceding unsigned comment added by 184.17.73.223 (talk) 23:16, 15 January 2012 (UTC)
thread_title:  Mistake 

message: The second derivative was incorrect. Denominator should be to the power of 3 to make numerator consistent with maths and h-prime term in the numerators 3rd term should have been squared. From formating these look like purely clerical errors. Double checked with maple13.  — Preceding unsigned comment added by 94.195.250.134 (talk) 19:58, 25 March 2012 (UTC)
replies: text: I agree.  I added a fact tag for now. Rlendog (talk) 00:18, 21 December 2018 (UTC)
thread_title:  Mistake 

message: In the introducing article, jumps occur between g and j functions every other rendered example (once j then g then j ...); they appear to be the same. I don't have the expertise to affirm it is wrong in the current form.
In the Proof>Algebraic_Proof section, the jump occurs following the introduction of f'(x) as the question that is to be answered. j(x) and h(x) are used as the functions comprising f(x) when they're divided. Taking 1/delta x as a factor and combining the two remaining fractions shouldn't (algebraically) affect j(x) in any way contrary to what's demonstrated. In this case, g(x) is substituted for j(x) for no apparent reason.
My observations may not be accurate so I need either a confirmation or a clarification. In case of the earlier response, I'll rewrite the TeX code.  — Preceding unsigned comment added by Enamex (talk • contribs) 17:28, 24 September 2012 (UTC)
thread_title:  Jumps between g(x) an j(x) or equivalents in successive rendered(TeX) examples. 

message: I couldn't help noticing that the Quotient Rule article is a stub.  Here's a possible proof that doesn't fall into the same pithole that the previous proof fell through.
:Let some function f(x)=\frac{g(x)}{h(x)}.
:We wish to find \frac{d}{dx}f(x), and this is equivalent to \frac{d}{dx}\frac{g(x)}{h(x)}.
:We also know that g(x)=f(x)*h(x).
:By the product rule, we can say that \frac{d}{dx}g(x)=h(x)*\frac{df(x)}{dx}+\frac{dh(x)}{dx}*f(x).
:From the equation above, \frac{d}{dx}f(x)=\frac{\frac{dg(x)}{dx}-\frac{dh(x)}{dx}*f(x)}{h(x)}
:Because f(x)=\frac{g(x)}{h(x)}, the right side simplifies to \frac{\frac{dg(x)}{dx}-\frac{dh(x)}{dx}*\frac{g(x)}{h(x)}}{h(x)}, which simplifies to \frac{d}{dx}f(x)=\frac{\frac{dg(x)}{dx}*h(x)-\frac{dh(x)}{dx}*g(x)}{h^2(x)}, completing the derivation.
There's another proof stemming from the original definition of the derivative that is also quite simple to understand (and rigorous).  The article would benefit by its addition.
Gummygoodness (talk) 04:23, 11 March 2013 (UTC)
replies: text: At a glance it looks good. My only criticism is that "*" is not normally used for multiplication in maths, in my experience. I think just remove the "*" and have the multiplication implied. Adpete (talk) 11:13, 11 March 2013 (UTC)
thread_title: Derivation (The Uncomplicated One)

message: This article should be rolled back to just before the first edit of July 11, 2016 when someone deleted a big chunk without explaining why and also put in an incorrect formula (the formula was correct before this edit). Jrheller1 (talk) 05:13, 19 August 2016 (UTC)
replies: text: I added implicit differentation and higher order formulas section (which was the useful information contained in the version before July 11, 2016).  So I don't think rolling back would be useful anymore. Jrheller1 (talk) 22:07, 19 August 2016 (UTC)
thread_title:  Rollback 

message: Looking back several years in the history, this article used to be much better.  It had a lot more info and was better written.  But somewhere along the line, it seems to have been gutted and remade poorer.  I'm working on fixing it up some, and I'll add a todo list in a moment with a couple things I think it needs.  --Deacon Vorbis (talk) 16:02, 23 April 2017 (UTC)
thread_title:  What happened to this article? 

message: @Deacon Vorbis:
On the examples and proofs:
# I've been trying to keep a long equation separate from long wording, so I have been breaking them up in an effort to improve readability.
# I usually see a fraction to the left of a long equation (such as for a limit, derivative, or integral, as examples) that it is factored out of, so I have been trying to do that.
# (I understand that the last equation in a section, etc, should have a period at the end of it, so I have left periods there after the first time they got reverted.)
Could you please tell me which changes you are not finding helpful, and why? 
Thank you - I appreciate the cooperation. -- Bwefler (talk) 15:58, 1 February 2018 (UTC)
replies: text: Formulas can certainly be set apart from text, but they should usually be by themselves, not with other running text, and on Wikipedia, the standard is to do so by indenting with a : (although there's some talk about this changing, but that doesn't really matter for now).  Throwing in a bunch of scattered s is certainly not the way to do it.  However, a simple statement of the product rule isn't an especially long or complex formula, and setting too many apart tends to break up the flow of the text.  As for the fraction, it was originally separated on the right hand side, so it's probably good to leave it there.  It could have just as easily been taken out on the left also, but multiplication is commutative, and it's fine either way – there's no particular good reason to change it.  –Deacon Vorbis (carbon • videos) 16:21, 1 February 2018 (UTC)
replies: text: Acknowledged. -- Bwefler (talk) 16:33, 1 February 2018 (UTC)
thread_title:  Which changes are not helpful? 

message: The higher order formula seems incorrect.  At a minimum, the denominator should be h^3, and even with the g/h terms in the numerator that would only get to h^2.  I am not quite getting the same numerator either, but I may have made an error or there may be another algebraic step needed.  For now I added a citation needed tag, but if this can't be supported by a reliable source or by showing more intermediate steps it needs to be removed. Rlendog (talk) 00:21, 21 December 2018 (UTC)
replies: text: What's there is correct.  If you expand out the f' = (g/h)' term in the rightmost expression, it will be as you're expecting.  I'm inclined to remove the  tag per WP:CALC.  –Deacon Vorbis (carbon • videos) 00:49, 21 December 2018 (UTC)
replies: text: I don't think that's obvious from the formulas shown.  I am ok with removing the tag without an explicit source, but then the intermediate formulas need to be shown. Rlendog (talk) 17:36, 21 December 2018 (UTC)
replies: text: Well, I added in one extra step (I think it was there at some point in the past but got removed somewhere along the way).  I'm not sure if that's what you're referring to though.  Earlier I also removed the plain f = g/h substitution at the end as not really necessary.  As it stands now, it gives an expression for f partially in terms of its lower derivatives, as advertised.  Someone could always sub in more, but I'm not sure how illuminating that would be.  –Deacon Vorbis (carbon • videos) 19:03, 21 December 2018 (UTC)
thread_title: Higher Order Formulas

message: Please stop.  I thought this was settled, and yet you've reignited the edit war.  There was even a third opinion generously offered.  And I realize it's not binding, but it's the best we've got.  The version you've reverted to was not the "stable" one.  Please, please, please just let this one go.  –Deacon Vorbis (carbon • videos) 19:27, 24 March 2019 (UTC)
thread_title:  This is nuts 

message: I kept on changing f(x)=g(x)/h(x) to h(x)=f(x)/g(x) even though it’s just notation and not an improvement to the article because the latter is commonly used in textbooks and online resources, but it turns out the former has been long established on this page. Sorry for the unnecessary changes. Also, I replaced k with \Delta x in the limits since normally \Delta x or h but not k is used to represent the change that approaches zero in the derivative, but since h was taken for a function, the remaining common choice was \Delta x. Please take note though that my other edits also got reverted when only the changes in notation were unnecessary. 49.147.83.13 (talk) 16:49, 21 May 2020 (UTC)
replies: text: Your other changes were mostly not improvements either.  There were already a couple reasonably representative examples in place, we don't need more (similar) ones.  Having too many runs afoul of WP:NOTTEXTBOOK.  Likewise, additional proofs are pretty unneeded at this point, and probably more importantly, your "proof" using logarithmic differentiation is only even valid when both functions are positive.  Also, the section headers don't have to state every bit that goes into them; they just have to serve as a means to help tell themselves apart.  On a side note, I find the "reciprocal rule" a silly extra rule that serves no purpose.  Most Calculus texts that I've seen don't even bother to name or prove it separately.  –Deacon Vorbis (carbon • videos) 17:22, 21 May 2020 (UTC)
replies: text: Thank you for explaining that you disagree with the logarithmic differentiation proof and find the reciprocal rule quite trivial. 49.147.83.13 (talk) 17:27, 21 May 2020 (UTC)
thread_title:  Changes in notation that got reverted