Talk:Chain rule

message: I don't see how Ex 1 is a "This calculation is a typical chain rule application." The calculation doesn't seem to involve differentiation which I would argue is the typical chain rule application. 
--flatfish89 (talk) 18:24, 24 April 2010 (UTC)
thread_title: Example 1

message: I think the basic idea behind the chain rule is getting swamped in a sea of details and special cases. The basic idea of the chain rule is pretty easy to explain in words:
The best linear approximation of the composition is the composition of the best linear approximations.
Every special case of the chain rule, more or less expresses this fact in various situations, with a wide variety of abstraction or concreteness. But, this is the basic idea, and it would be good if it were made a bit more prominent. Revolver 01:43, 9 October 2005 (UTC)
thread_title: Basic idea

message: If I'm not completely mistaken, the definition mixes up f and f \circ g. Also, x isn't some kind of magic symbol, so
f \circ g = f(g(x)) doesn't make sense either. It says that the composition of f and g (which is a function) is equal to a certain value of that function, namely that at x. I think the whole thing should read
:In algebraic terms, the chain rule (of one variable) states: Given function f that is differentiable at g(x) and a function g that is differentiable at x, then the composite h(x) = f(g(x)) (or shorter: h = f \circ g) is differentiable at x and
::
 \frac{d}{dx} h(x) = \frac{d}{dx} f(g(x)) = f'(g(x)) \; g'(x).
Or leave out h completely and just write:
::
 \frac{d}{dx} f(g(x)) = f'(g(x)) \; g'(x).
--K. Sperling (talk) 00:37, 12 November 2005 (UTC)
replies: text: I agree with both points, the current definition is both confusing and (to my understanding)informal. Ariel brunner (talk) 10:18, 14 July 2013 (UTC)
thread_title:  Mix-up?? 

message: I would like to see an extensive expansion of the chain rule for several variables.  Thanks, Silly rabbit 06:00, 15 November 2005 (UTC)
replies: text: What kind of information would you like to see added? I might try to do this. --Monguin61 03:56, 10 December 2005 (UTC)
thread_title:  Chain rule for several variables 

message: Since the "composition of two functions" is technically a "composite function", would its derivative be called a "composite derivative"?  Also, does the secondary (inside) derivative have a special name (something like "harmonic derivative"——though I think that term means something else)?  ~Kaimbridge~20:15, 3 February 2006 (UTC)
replies: text: I did not hear this terminology before. Oleg Alexandrov (talk) 00:27, 4 February 2006 (UTC)
replies: text: What, "composite function"?  I believe that is the legitimate term.  See Thesaurus.Maths, PinkMonkey, ThinkQuest.  I just wanted to know if its derivative had a special name, and if the inside derivative had a special name. ~Kaimbridge~14:48, 4 February 2006 (UTC)
replies: text:  I never heard of composite derivative of harmonic derivative. Oleg Alexandrov (talk) 17:30, 4 February 2006 (UTC)
replies: text:  No, the only term that I know is valid is "composite function".  I'm asking if the derivative of a composite function would be referred to as a "composite derivative" (and, if not, does it have a special, unique name——other than just "derivative of a composite function").  Likewise, does the inside derivative (e.g., for \frac{d}{dx} f(g(x)),\; g'(x)) have a special name——"harmonic derivative" is just some reasonable sounding possibility that I used as an example, not that I'm in any way implying that that is what it is actually called. P=) ~Kaimbridge~18:12, 4 February 2006 (UTC)
replies: text:  That's what I am trying to say. As far as I know, the answer to your questions is "no". I never heard of "composite derivative" or "harmonic derivative". They don't have any special name, either that or any other one, and I don't know why would any body ever want those things to have any special name. Oleg Alexandrov (talk) 19:41, 4 February 2006 (UTC)
thread_title:  Derivative of composite function 

message: This discussion is outdated. Suggesting deletion. AurelienLourot (talk) 19:33, 30 December 2014 (UTC)
thread_title:  Role of eta in proof 

message: I think that the exceptions to the rule should be mentioned. For instance, \frac{d}{dx}\sqrt{25-x^2} \neq \frac{1}{2\sqrt{x}}, as would be suggested by the power rule. The problem is that sqrt{x+a} is just sqrt{x} shifted back, but x+25 still differentiates to 1. This means that according to the power rule, that constant that's added to x has no affect on it, even thought the derivitive should be \frac{-x}{\sqrt{25-x^2}}. Which doesn't follow directly from the Chain Rule. He Who Is 23:33, 3 June 2006 (UTC)
replies: text: Umm... you are missing the whole point of the chain rule? If y=\sqrt{u} and u=25-x^2, then \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} = \frac{1}{2\sqrt{u}} (-2x) = \frac{-x}{\sqrt{25-x^2}} --Spoon! 11:15, 31 August 2006 (UTC)
thread_title:  Exceptions 

message: All that calculus gives me a headache! I have no idea if and/or how they relate, but perhaps the chain rule pertaining to probability theory deserves a place somewhere on this page? You know, the P(A1 n A2 ... n An) = P(A1)P(A2|A1)P(A3|A1 n A2)...P(An|n n-1/i=1 Ai) thing, sorry about the crummy representation.
218.165.75.221 10:04, 30 September 2006 (UTC)
replies: text: I am adding this page. ---- CharlesGillingham (talk) 21:41, 17 September 2009 (UTC)
thread_title:  Chain rule in Probability 

message: The statement (f o g)'(x) = (f(g(x)))' is incorrect and fundamentally misunderstands the prime (f') notation. The prime is a transformation from functions to functions; as such it should be applied before the variable x is evaluated, as on the LHS but not as on the RHS. 69.215.17.209 14:44, 22 April 2007 (UTC)
replies: text: Since the chain rule is a fairly basic concept in calculus and most people at that level haven't taken analysis would it be appropriate to add a note explaining the meaning of the composition operator?  —Preceding unsigned comment added by 76.199.5.236 (talk) 06:47, 13 January 2008 (UTC)
replies: text: Note:  The objection to f(g(x))' refers to an earlier version of the article, and is no longer current.  Silly rabbit (talk) 13:51, 13 January 2008 (UTC)

text: I agree that, in an article about such an elementary concept in calculus, it is inappropriate to use composition operator. ---- CharlesGillingham (talk) 00:21, 16 September 2009 (UTC)

text: Seconded, no need for this operator that's only encountered in 3rd/4th year undergraduate degree courses, this page should be targeting a much less advanced audience. ---- Mirams (talk) 16:27, 14 February 2020 (UTC)
thread_title:  (f(g(x)))' is incorrect and confusing 

message: Perhaps resulting from corrections above, f prime is now invisible when inline (at end of first proof). g prime is visible inline due to different glyph for g not overlapping the prime in the way that it overlaps for f. f prime works ok with display notation instead of inline. Should be a simple fix but I don't know how. 58.175.211.1 (talk)  —Preceding undated comment added 15:26, 7 February 2013 (UTC)
thread_title:  f prime is invisble inline 

message: The examples could use some more description, depending on if we're shooting for "definition" or "instructional detail." Substituting U for X^2+1 makes the plug 'n' chug easier, but it's not strictly necessary. Any objections to expanding current objections to include various applications of chain and detailed description of how and why subsitutions are valid?--Legomancer (talk) 22:45, 8 September 2009 (UTC)
thread_title:  Examples 

message: For any coordinate (real valued function) y on a line (e.g. the real line) and any point p, denote by dyp the equivalence class of y-y(p)1 (where 1 is the constant function, with value 1) modulo functions vanishing at p to higher order. If y=f(x) (i.e. y = f ⚬ x for some other coordinate x on the same line and some f:R→ R) then the definition of differentiability of f at x(p) ensures that dyp = f'(x(p)) dxp because f(x)-f(x(p))1 differs from f'(x(p))(x-x(p)1) by a function vanishing at p to higher order.
The chain rule is an immediate consequence. If u = g(y) then, omitting evaluations/subscripts at p, du = g'(y)dy= g'(f(x))f'(x) dx.
Most arguments formalize this basic idea without discussing the conceptual meaning. Geometry guy 00:57, 2 November 2010 (UTC)
replies: text: The original proof that was in the article is close to being a direct formalization of this.  Unfortunately, it looks like it had gotten a bit over-edited, since it included a version of the proof for several variables that was not really consistent with this philosophy.  I've put back in what I believe is an easier to read version of the old proof.  I wasn't sure whether and how to address the case of several variables: really the article should have a proper statement of the chain rule of a function between two (finite dimensional) Euclidean spaces.  Sławomir Biały (talk) 11:22, 3 November 2010 (UTC)
thread_title: Comment from WPM

message: In the example "Suppose that a skydiver ..." the formula g(t) = 4000 − 9.8t2 should be replaced with g(t) = 4000 − ½9.8t2 isn't it? 2.36.204.64 (talk) 22:21, 19 January 2011 (UTC)
thread_title:  Error in 'First example' 

message: 1) clarify 2nd bullet from "...rate of change in atmospheric pressure at height..." to ...rate of change in atmospheric pressure w.r.t. h, at height...
2) clarify 4th bullet from "...rate of change in atmospheric pressure t seconds after..." to ...rate of change in atmospheric pressure w.r.t. t, t seconds after...
3) the bottom paragraph that starts "It is not true..." is misleading and includes an error.  I would end it with the sentence "This need not have anything to do with the buoyant force ten seconds after the skydiver's jump." and start a new paragraph just below that states the following:
   
It is true that (f o g)'(t) = f'(h) * g'(t).  To find the buoyant force w.r.t. t ten seconds after his jump, we must evaluate g(10), his height ten seconds after he jumps, and substitute the result into f'(h).  g(10) is 3510 meters above sea level, so the true buoyant force w.r.t. t ten seconds after the jump is (proportional to) f'(3510) * g'(10) = 7.133 * -98 = -699.
This example demonstrates the Chain Rule as the product of two rates.  The last sentence that states "g(10) is 3020 meters above sea level, so the true buoyant force ten seconds after the jump is (proportional to) f'(3020)." is erroneous.  To use the Chain Rule you need to multiply by f'(g(t)) by g'(t).  —Preceding unsigned comment added by 69.117.93.37 (talk) 04:41, 31 January 2011 (UTC)
replies: text: I've changed the article. Ozob (talk) 12:34, 31 January 2011 (UTC)
thread_title:  Suggested changes to 'First example'  

message: I might misunderstand the notations, but why is the multivariate chain rule written as:
:D_{\mathbf{a}}(f \circ g) = D_{g(\mathbf{a})}f \circ D_{\mathbf{a}}g
Why is there a composition on the RHS, and not a product of derivatives, as in the univariate case ? Is it to be understood as a matrix operation, in which case composition corresponds to a product, and it that case, shouldn't this be explicitely signaled ? Donvinzk (talk) 11:34, 4 June 2011 (UTC)
replies: text: They are meant to be understood as linear transformations.  The composite is the usual composite of functions.  By choosing a basis you can construct Jacobian matrices, and the composite of linear transformations corresponds to the product of Jacobian matrices.  I've revised the article so that it explains this in more detail.  Is it clearer now? Ozob (talk) 21:02, 4 June 2011 (UTC)
thread_title:  Multivariate chain rule 

message: I have doubts in the comment that the second proof does not need a theorem about products of limits. In the intermediate step, we need to consider the product of \varepsilon(h) and \eta(k_h), which is equivalent to [Q-f'(g(a))]*{[g(x)-g(a)]/(x-a) - g'(a)}. My thought is that both proofs rely on the same theorem. However, I would like to get comments from more experienced editors before changing anything in the article.202.130.125.147 (talk) 09:04, 14 September 2013 (UTC)
replies: text: Nice catch!  You're entirely correct.  I've updated the article.  Ozob (talk) 14:14, 14 September 2013 (UTC)
thread_title:  Second Proof 

message: The second proof contains the following line:
:The above definition imposes no constraints on η(0), even though it is assumed that η(k) tends to zero as k tends to zero.  If we set , then η is continuous at 0.
Is this line really necessary? Wouldn't it be enough to write that η behaves like ε, i.e. it simply tends to zero as its argument tends to zero? 
The second proof terminates finally with the following line:
:The need to define Q at g(a) is analogous to the need to define η at zero.
But I don't see the need to define η at zero in that second proof. Am I missing something? Don't we just need to know that η tends to zero as its argument tends to zero?
AurelienLourot (talk) 19:51, 30 December 2014 (UTC)
replies: text: I see, that's because k_h not only tends to zero but can also be exactly zero. \eta(k_h) would then be undefined otherwise. Feel free to delete this discussion. AurelienLourot (talk) 20:31, 30 December 2014 (UTC)
thread_title:  Continuity of η in second proof 

message: Would it be pertinent/beneficial to add a small section exposing the non-standard approach to the chain rule, i.e. using hyperreal numbers and standard parts?  — Preceding unsigned comment added by Gio97 (talk • contribs) 08:30, 10 April 2015 (UTC)
replies: text: What is the non standard approach of the chain rule? Chain rule is a formula which is the same in standard and non-standard analysis. Only the proof differs slightly, as the definition of the derivative is not the same. A new subsection, called "Proof in non-standard analysis", of the section "Proofs" could be added. Are you willing to write it? D.Lazard (talk) 09:29, 10 April 2015 (UTC)
thread_title:  Non-standard analysis 

message: I tend to agree with he point being made in the edit summary in this revert. Placing the prime after an expression is pushing an abuse of notation too far.  The correct notation would be (f\circ g)'(x) = f'(g(x)) g'(x). The prime just does not have the necessary flexibility (expressive power). —Quondum 02:24, 20 May 2015 (UTC)
replies: text: So the notation under dispute was introduced in  because it was supposed to be easier to understand.  That editor's point was that some students won't understand the notation for function composition.  I think that's both true and a concern for the target audience of article.  Given that assumption, I'm not sure what other notation to use besides f(g(x))' (and I do think that the meaning of that expression is unambiguous and not an abuse).  Ozob (talk) 13:45, 20 May 2015 (UTC)
replies: text: I understand the motivation for trying to use the prime notation for the derivative, but that notation has serious shortcomings, and using it in this way creates problems.  I doubt whether you'll find a reputable textbook using it in such a sloppy fashion, and for the lead it can confuse.  Its *intended* meaning is not even clear.  For example, consider the following invalid substitution: y = g(x), so f(y)′ = f(g(x))′.  Do we really want the student of differential calculus to learn that substitution cannot be done when there is a prime applied to the expression?  —Quondum 14:20, 20 May 2015 (UTC)
replies: text: Good point; I think of the prime mark as implicitly having an x (in which case the substitution you make is valid essentially by definition), but a student isn't likely to have that subtlety down pat.  What do you think we should do?  Insist on function composition notation?  Ozob (talk) 14:34, 20 May 2015 (UTC)
replies: text:  I agree that  writing f(g(x))' is an abuse of notation. However, it is widely used by mathematicians and it is legitimate in some cases. Why it is an abuse of notation? Because, formally, in modern mathematics, the prime is an operator that apply only to functions; therefore, it must normally be written just after the name of a function. But what if the the function has not been named, such as g\circ f or x \mapsto x^2+1? Note that, when Lagrange has introduced his notation, the operator \circ was not yet invented (I believe, but I have not checked), and people did say "let f(x) be a function" rather than "let f be a function of x". At that time, the distinction between an expression containing x and the function that it defines was unclear (it is still unclear for many people). Therefore, it is very common and very often useful to adopt the convention that a prime placed after an expression containing x, indicates the derivative of the function defined by the expression. With this convention, the formula (x^2+1)'=2x becomes correct, and f(g(x))' becomes well defined. For these reasons I'll reinstall the disputed paragraph, with a note explaining the convention. D.Lazard (talk) 15:04, 20 May 2015 (UTC)
replies: text: I understand f(x, y)′ and f(y, x)′ to be functions that are both differentiated with respect to their x-parameter (though actually, it depends on how y is related to x)), and f′(z) entirely depends on how z is a function of x. I can see where the subscript notation came from: fx(x), and here (f(x))x could make some sense.  I would feel more comfortable if the explanatory note was more cautionary about the notation and written in the past tense: this is not a use that should be treated as current or without confusion.  At least Newton's overdot notation is unambiguously with respect to a universal parameter: time. —Quondum 15:45, 20 May 2015 (UTC)
replies: text: Hopefully 's edit is a compromise acceptable to all.  It neatly sidesteps the problem of which of differentiation with respect to the sole parameter or with respect to a specific variable x the prime denotes, and also allows it to be regarded more formally as an operator.  —Quondum 16:43, 20 May 2015 (UTC)
replies: text: The current compromise seems fine to me.  Glad this has been sorted out.50.163.87.238 (talk) 21:04, 20 May 2015 (UTC)
thread_title:  Abuse of prime notation 

message: For the life of me, I can't make heads or tails of that recently-added animation.  I'm not even sure what it's supposed to be showing.  Is it just me?  Perhaps with a bit of clarification it could be a useful addition to the article, but in its present state I'm not sure it'll contribute to anyone's understanding. 96.231.153.5 (talk) 06:07, 26 January 2016 (UTC)
replies: text: I agree. I have removed the animation. It is not even clear that the animation correctly represent the chain rule, and, in any case, it is to fast for allowing understanding. D.Lazard (talk) 09:02, 26 January 2016 (UTC)
thread_title:  Animation 

message: Regarding the example . We strictly have \frac{\partial u(x,y)}{\partial x}=2x, \frac{\partial u(x,y)}{\partial y}=2 and \frac{\partial u(x,y)}{\partial r}=0 since the function does not take r as a function parameter. The article makes use of non standard notation if you apply the chain rule here.
You shall not apply the chain rule here. See https://www.icp.uni-stuttgart.de/~icp/mediawiki/images/7/74/Remark_on_partial_derivatives.pdf please correct this!--94.217.251.2 (talk) 10:24, 26 June 2018 (UTC)
replies: text: The x and y variables are given to be functions of r and t.  It seems like a standard application of the chain rule to me.  Sławomir Biały (talk) 11:27, 26 June 2018 (UTC)
thread_title:  Example is wrong [[Chain_rule#Example]] 

message: I don't think it's wise, didactically, to use a capital F in F(x) = f(g(x)) since capitals are very often used to designate the primitive, as in F'(x) = f(x). Why not just use h or something? — Preceding unsigned comment added by 77.61.180.106 (talk • contribs) 13:00, 4 December 2020 (UTC)
replies: text: However, the main reason for such a change is that  suggests wrongly that the composition is more related to  than to . D.Lazard (talk) 14:30, 4 December 2020 (UTC)
thread_title:  Please don't use capital F in the lead 

message: I have a question about the example function g(x) in the First Proof. Currently the article states: ‘For example, this happens for g(x) = x^2*sin(1 / x) near the point a = 0.’ I think the example function should be a split function as follows: “g(x) = x^2*sin(1 / x)  for x doesn’t equal 0, and g(x) = 0 for x = 0”. As it stands the function is undefined at the point where a = 0, and so not differentiable there. I think the point of the article would be valid if it used the split function I have suggested.  — Preceding unsigned comment added by Matthew.howey (talk • contribs) 10:38, 29 March 2021 (UTC (UTC)
:Please put new talk page messages at the bottom of talk pages and sign your messages with four tildes (~~~~) — See Help:Using talk pages. Thanks.
: Yes, the second part of the Q function mentions g(a), so that should probably be taken into account. - DVdm (talk) 11:36, 29 March 2021 (UTC)
replies: text: Hi – many thanks for your quick response and edit  . Could I suggest it would be clearer if the value of g(x) at x = 0 is explicitly built into the definition of the function itself – perhaps: “this happens for the split function: g(x) = x2sin(1/x) if x ≠ 0,  g(x) = 0 if x = 0”. Happy to make the edit but don’t want to do anything without agreement. Thanks.Matthew.howey (talk) 15:27, 29 March 2021 (UTC)
thread_title:  Correctness of example function in first proof?