Talk:Chain rule

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This page needs a proof and more rigorous maths.

situations for multivariable should be added: if u is a function of x, and y, and both x, and y are functions of t, then,

${\frac {du}{dt}}={\frac {\partial u}{\partial x}}{\frac {dx}{dt}}+{\frac {\partial u}{\partial y}}{\frac {dy}{dt}}$

New proof

I've replaced the current proof with one I saw somewhere. It relies on nothing but the definition of a derivative and the concept of limits. I believe it is quite rigourous and more formal than the previous one, but if there are any flaws with it feel free to point them out or even restore the old one.

Also, if you think it is too long or verbose then correct it.

Someone42 13:43, 25 May 2005 (UTC)[reply]

Isn't that really just the same proof, though? Written out longer. I don't think it is different from the point of view of rigour. Charles Matthews 14:06, 25 May 2005 (UTC)[reply]

I just reverted the thing. Someone42, this is not a math paper, not a math exam, not a math book. This is a general purpose encyclopedia.

I appreciate the many hours you put in that proof, I appreciate your knowledge of mathematics, and your desire for rigor.

However, nobody will read that proof, even mathematicians will gloss over it.

According to Wikipedia:WikiProject Mathematics/Proofs, proofs are discouraged on Wikipedia to start with, and long formal proofs especially. This has been discussed over and over again, and this seems to be the general view over here.

If anything, Wikipedia needs less rigor, not more. This is an encyclopedia for the general public and among its audience mathematicians are less than 10% if not less. Oleg Alexandrov 15:25, 25 May 2005 (UTC)[reply]

Oh well, in retrospect I think I did go a little overboard. Someone42 09:25, 26 May 2005 (UTC)[reply]

I can't really agree about 'rigour'. Anything less than a rigorous proof is just an argument/derivation/bootstrapping. However that is not really the issue here. And I agree with Oleg that the need for proofs is not so high. Charles Matthews 15:29, 25 May 2005 (UTC)[reply]

OK, I did not mean that rigurous proofs are bad overall. I am mathematician too, and I hope I know the value of proofs. I was trying to say that on Wikipedia we can get away with something less than full-blown proofs. So I agree that a proof is not a proof unless it is rigurous. Oleg Alexandrov 22:37, 25 May 2005 (UTC)[reply]

On the other hand, who but a mathmatician (or at least a maths student) would want to know about something as particular as the chain rule? I think Wikipedia is a good maths textbook and dumbing it down so the general public can understand won't add any value... 203.97.255.167 08:37, 22 May 2006 (UTC) (edit: having looked at that proof though, I have to agree that it was a bit overboard and didn't really add any rigour) 203.97.255.167 08:39, 22 May 2006 (UTC)[reply]

In addition, this proof relies on f(g(x+deltax))-f(g(x))=f(g;(x), which it does not. f(g'(x))=f(g(x+deltax)-g(x)). Additionally, the chain rule is not f'(g(x))=f'(g(x))g'(x), as dividing by f'(g(x)) would give you 1=g'(x), which isn't always true. The chain rule is f'(g(x))=f(g'(x))g'(x)

Basic idea

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)[reply]

Mix-up??

If I'm not completely mistaken, the defition 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)[reply]

Chain rule for several variables

I would like to see an extensive expansion of the chain rule for several variables. Thanks, Silly rabbit 06:00, 15 November 2005 (UTC)[reply]

What kind of information would you like to see added? I might try to do this. --Monguin61 03:56, 10 December 2005 (UTC)[reply]

Derivative of composite function

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)[reply]

I did not hear this terminology before. Oleg Alexandrov (talk) 00:27, 4 February 2006 (UTC)[reply]

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)[reply]

I never heard of composite derivative of harmonic derivative. Oleg Alexandrov (talk) 17:30, 4 February 2006 (UTC)[reply]

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)[reply]

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)[reply]

Role of eta in proof

I have no formal training in maths so forgive me if I sound naiive. Near the bottom of the proof it says "Observe that as $\delta \to 0,$ ${\frac {\alpha _{\delta }}{\delta }}\to g'(x)$ and ${\frac {\eta (\alpha _{\delta })}{\delta }}\to 0$ ." Would I be correct to think that " ${\frac {\eta (\alpha _{\delta })}{\delta }}\to 0$ " shows that the "error" (right word?) involved goes to zero as delta goes to zero? 202.180.83.6 03:52, 16 February 2006 (UTC)[reply]

Examples 1 and 2

the primes in example 1 and 2 (where it says f'(x) = ) are very difficult to see. They look exactly like f(x). It may cause a lot of confusion, is there any way to make the primes in f'(x) stand out?

Exceptions

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)[reply]

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)[reply]

Chain rule in Probability

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) M.H.[reply]

Uh, how about no. 69.215.17.209 14:41, 22 April 2007 (UTC)[reply]

Personally, I would like to see some detail about the Chain Rule for probability theory. I have been looking around the internet, and have not been able to find a discussion of it (ideally a step-by-step example or a detailed proof). So, it would be a good thing if wikipedia included something on it. Should the Chain Rule for probability theory be included on this page, the page for probability theory, or it's own page [ex. Chain Rule (Probability Theory)]?? SteelSoul (talk) 17:50, 2 February 2009 (UTC)[reply]

I am adding this page. ---- CharlesGillingham (talk) 21:41, 17 September 2009 (UTC)[reply]

(f(g(x)))' is incorrect and confusing

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)[reply]

I see that this edit was reverted, with the argument that this notation is common enough to be included. I understand that people may sometimes use it (I've never seen it myself, and I challenge anyone to produce examples from a common calculus text), but it is not standard and pedagogically very confusing. It is ambiguous whether the constant f(x) or the function f is being differentiated. I do not think that such poor notation should be perpetuated in an encyclopedia, without evidence that it is at least commonly used. --69.212.231.101 03:52, 26 July 2007 (UTC)[reply]

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)[reply]

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)[reply]

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)[reply]

Examples

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)[reply]