easy proof of the chain rule?

mikau · 2007-04-29 06:01:24

both calculus books I went through had a fairly lengthy proof of the chain rule. Lying awake in bed the other night, I was wondering why the proof is so dang complicated while it seems fairly obvious. I came up with a fairly simple way to do it. But I'm wondering if its a satisfactory proof or not.

We want to know the limit of [ f(g(u + Δu )) - f(g(u)) ]/Δu as Δu approaches zero.

Now we know Δg = g(u + Δu) - g(u) so g(u + Δu) = g(u) + Δg. Substitute this for the first argument for f()

limit of [ f( g(u) + Δg ) - f(g(u)) ]/Δu as Δu approaches zero.

Δg = g(u + Δu) - g(u) so we can multiply above by g(u + Δu) - g(u) and below by Δg. We then get:

limit of [g(u + Δu) - g(u) ]*[ f( g(u) + Δg ) - f(g(u)) ]/(ΔuΔg) as Δu approaches zero.

Rearranging we get:

limit of [g(u + Δu) - g(u)]/Δu * [f(g(u) + Δg) - f(g(u))]/Δg) as Δu approaches zero.

We know that Δg = g(u + Δu) - g(u) therefore as Δu approache zero, Δg approaches zero. Therefore, by the product rule for limits, the limit is:

limit of [g(u + Δu) - g(Δu)]/Δu as Δu approaches zero * limit of [f(g(u) + Δg) - f(g(u))]/Δg) as Δg approaches zero.

By the definition of derivatives, we get that this is:

g'(u)*f'(g(u)) qed.

That was pretty simple. Why do my textbooks use epsilon proofs that span two pages?
Is this as valid a proof as we could want? Or are there some subtle assumptions being made here?

Last edited by mikau (2007-04-29 06:04:26)

luca-deltodesco · 2007-04-29 07:39:17

not 'completely' off topic, but does the above actualy state the proposistion for proving the chain rule?

actually, using the same argument you used above that Δg will approach 0 as Δu does, and therefore you can multiple them into a single limit, or split into two limits, that would immediately give equality.

note that, im just using his above reasoning backwards basicly here, i've never covered limits formally

Last edited by luca-deltodesco (2007-04-29 07:55:06)

mikau · 2007-04-29 09:39:53

Hmmm. In the second line, something got screwed up. It ends with a "-" sign with nothing after it.

So basically you're saying the reverse, if you multiply g'(u) by f'(g(u)) we end up with the derivative of the composite function, thus supporting my proof. Correct?

luca-deltodesco · 2007-04-29 09:55:54

mikau wrote:

Hmmm. In the second line, something got screwed up. It ends with a "-" sign with nothing after it.
So basically you're saying the reverse, if you multiply g'(u) by f'(g(u)) we end up with the derivative of the composite function, thus supporting my proof. Correct?

im not sure what your screwing up thing is, its fine to me, and yeh, basicly, supposing that product of limits thing holds.

mikau · 2007-04-29 12:11:22

whoops, I meant the third line.
Maybe its just my screen resolution.

from what I remember, the only restriction on limit laws is with the quotient rule, that is, that lim g(x)/lim f(x) as x approaches a, = lim g(x) as x appraoches a, over lim f(x) as x approaches approaches a PROVIDED that lim f(x) as x approaches a is not zero. Otherwise it does not hold.

Here, we do have some zero approaching values in the denominator, but it can be expressed as a product of two finite ratios, provided the derivatives exist. So I don't think it would break the rules to factor out a limit that can be shown to be finite, could it?

Last edited by mikau (2007-04-29 12:12:51)

luca-deltodesco · 2007-04-30 02:28:57

i was thinking about this today, and i think i found a problem in it.

the proof relys on g being continuous does it not? so what happens when g is a function that is not continous?

Zhylliolom · 2007-04-30 10:48:59

When g is not continuous you have to prove it otherwise.

I don't know why everyone thinks the proofs of the chain rule are lengthy or overly complicated. I can think of two or three proofs what are only about a paragraph. There is also a proof using the hyperreal number system that is only a line or two.

mikau · 2007-04-30 12:14:33

but thats my point, Zhylliolom. Why do elementary calculus books present the lengthy proofs instead of ones that are just a paragraph long?

My guess is, those one paragraph proofs are imprecise and not formal enough. Or are they?

Last edited by mikau (2007-04-30 12:15:51)

Zhylliolom · 2007-04-30 12:25:16

One reason they are shorter is because they use another theorem or two previously stated in the same chapter. They're definitely not imprecise or informal. Could you give me an example of a lengthy chain rule proof?

mikau · 2007-04-30 13:31:36

here's a proof one of my books gives. Not terribly long but kind of confusing and much less intuitive.

http://books.google.com/books?id=f5EmxGFKnvoC&pg=RA1-PA227&lpg=RA1-PA227&dq=chain+rule+proof+stewart+calculus&source=web&ots=cEzkpDA7vq&sig=tOgsaP7v7djoZBwlL2aHEM7XPns#PRA1-PA227,M1

mathsyperson · 2007-04-30 23:41:58

Zhylliolom wrote:

One reason they are shorter is because they use another theorem or two previously stated in the same chapter.

Heh, so by the same logic I could give a flawless one-line proof like this:

Proof of the Chain Rule
Refer to mikau's link above. QED

Zhylliolom · 2007-05-01 07:06:33

I suppose it is the same logic, but of course they have to write the proof SOMEWHERE in the book, since the chain rule is such a common thing, and it's not just a direct corollary. Which reminds me that I hate when a book states a less popular theorem and says "see this author for a proof".

mikau · 2007-05-01 09:57:34

So whats the verdict? Is my proof just as good as the proof in the link I listed?

Zhylliolom · 2007-05-01 12:07:44

luca-deltodesco wrote:

the proof relys on g being continuous does it not?

This is the problem with your proof, Mikau. It's somewhat subtle, so I'll point out where:

Mikau wrote:

We know that Δg = g(u + Δu) - g(u) therefore as Δu approache zero, Δg approaches zero

This is the same as saying that if |Δu| = |(u + Δu) - u| < δ, then for any ε > 0, |Δg| = |g(u + Δu) - g(u)| < ε, or in other words that g is continuous. The chain rule does not require continuity, only differentiability.

mikau · 2007-05-01 16:27:02

remind me. what is the difference between continuity and differentiability?
I thought differentiability required continuity, at least at the given location.

Zhylliolom · 2007-05-01 17:56:16

Silly me, if a function f is differentiable at c, then f is continuous at c. Hohoho, good teamwork *high fives*

luca-deltodesco · 2007-05-01 18:27:11

lol

mikau · 2007-05-02 00:10:47

muahaha! Horray!

Incidently, is there a proof of the chain rule on this website? I couldn't find one.

George,Y · 2007-06-03 02:54:59

The biggest flaw of your proof is assuming Δg≠0 always, as long as x around the intevals of x[sub]0[/sub]

Here is a proof covering all points when Δg=0. Elegant.

The Proof

Last edited by George,Y (2007-06-03 02:55:45)

LQ · 2007-06-03 03:48:55

All deriveable functions can be simplified to Ax^a + Bx^b +... + Sx^s

Isn't that so?

Well if so, we all know it applies to them, we can easily see that.

x(x^3 + 3) for instance.

even 1/(x^2 + x) should be possible to simplify to a/x + (b + cx)/x^2 or something of the likes.

Then do we really need to have a seperate proof for the chain rule? I guess it might be good. Because it is so funny?

Ricky · 2007-06-03 12:57:27

All deriveable functions can be simplified to Ax^a + Bx^b +... + Sx^s
Isn't that so?

sin(x)
e^x

There are so many more examples.

George,Y · 2007-06-03 13:56:28

Not so, some cannot. Divergency problem.

LQ · 2007-06-03 23:34:02

Ricky wrote:

All deriveable functions can be simplified to Ax^a + Bx^b +... + Sx^s
Isn't that so?
sin(x)
e^x
There are so many more examples.

All exp, log and trig functions can be simplified like this. Even roots, unfortunately the term diverge. I can only think of ax^bx that doesn't work. How do you derivate that by the way?

LQ · 2007-06-03 23:37:04

George,Y wrote:

Not so, some cannot. Divergency problem.

Aah, you said so, yes. But even though they diverge and you can't solve it, still you can use the numbers to derivate. Right?

George,Y · 2007-06-04 01:18:55

you can search "The radius of convergence" for more details.

Math Is Fun Forum

#1 2007-04-29 06:01:24

easy proof of the chain rule?

#2 2007-04-29 07:39:17

Re: easy proof of the chain rule?

#3 2007-04-29 09:39:53

Re: easy proof of the chain rule?

#4 2007-04-29 09:55:54

Re: easy proof of the chain rule?

#5 2007-04-29 12:11:22

Re: easy proof of the chain rule?

#6 2007-04-30 02:28:57

Re: easy proof of the chain rule?

#7 2007-04-30 10:48:59

Re: easy proof of the chain rule?

#8 2007-04-30 12:14:33

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#9 2007-04-30 12:25:16

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#10 2007-04-30 13:31:36

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#11 2007-04-30 23:41:58

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#12 2007-05-01 07:06:33

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#13 2007-05-01 09:57:34

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#14 2007-05-01 12:07:44

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#15 2007-05-01 16:27:02

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#16 2007-05-01 17:56:16

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#17 2007-05-01 18:27:11

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#18 2007-05-02 00:10:47

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#19 2007-06-03 02:54:59

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