Chain Rule

[prasannapakkiam]

All the proofs I have found for the Chain Rule involve limits and the fundamental theorem of Algebra...

So I came up with a PROOF, not a derivation. But my teacher claims that my proof is invalid. Is it? If so, why???


let:
u=z(x)
y=f(u)=f(z(x))

therefore: dy/du = f ' (u)
therefore: dy = (f ' (u)) * du -->

therefore: du/dx = z ' (x)
therefore: 1/dx = (z ' (x))/du -->


therefore dy/dx = dy*(1/dx)
substituting...
therefore: dy/dx = ((f ' (u))*du)*(z ' (x))/du
which simplifies to:
dy/dx=(f ' (u))*(z ' (x))=(f ' (z(x)))*(z ' (x)) ==>
or alternatively substituting...
dy/dx=dy/du*du/dx ==>

[Office_Shredder]

therefore: dy = (f ' (u)) * du -->

Uhh... is this rigorous? You're just manipulating notation. While extremely convenient, the use in a true proof is questionable.

Of course a proof of the chain rule will involve limits.... because the definition of a derivative is based on a limit, and the chain rule is a proof about a derivative. You can often expect proofs to rely on definitions

[Werg22]

Your proof is valid. Though there is a simpler proof.

du / dx = z'(x)

df / du = f'(u)

If we multiply both, we get df/dx, which is what we are looking for. Hence df / du = f'(u) * z'(x)

[prasannapakkiam]

Well, (to "Office Shredder"), the reason, I used arrows, was because my teacher could not follow my proof. Anyway, thanks. I just wanted to check whether my proof is valid. I insist that it is not a derivation. I see it like induction - this concept involves proving the statement/equation by proving that it will work for all the numbers defined within the set in which it claims that it will work... Although this is a simplistic thought, it underlies; the reason why I think my proof is valid. If I am "manipulating notation", then can you tell me why this is invalid in a 'Proof'???

[Werg22]

Uhh... is this rigorous? You're just manipulating notation. While extremely convenient, the use in a true proof is questionable.

Of course a proof of the chain rule will involve limits.... because the definition of a derivative is based on a limit, and the chain rule is a proof about a derivative. You can often expect proofs to rely on definitions

Manipulation of infinitesimals as such is valid because the meaning of the limit they represent is not lost.

Edit: e.g. With the expression 1/dx = = (z ' (x))/du, the OP substituted such as dy/dx = ((f ' (u))*du)*(z ' (x))/du. This expression, interpreted as a limit, is still valid. Since du is a function of dx such as du = du(dx), if we let dx go to a very small value, so does du and the expression (z ' (x))/du becomes closer and closer to 1/dx. Since it gets closer and closer to 1/dx, ((f ' (u))*du)*(z ' (x))/du gets closer and closer to (f ' (u))*du / dx. Hence, if (f ' (u))*du / dx has a limit, the other expression will have the same limit because the two expressions become increasingly close as dx goes to 0.

[d_leet]

Your proof is valid. Though there is a simpler proof.

du / dx = z'(x)

df / du = f'(u)

If we multiply both, we get df/dx, which is what we are looking for. Hence df / du = f'(u) * z'(x)

How exactly is this a proof? du/dx and df/du are not fractions.

[Werg22]

How exactly is this a proof? du/dx and df/du are not fractions.

As said before, infinitesimals can be treated like numbers because the the ending result always represents the limit we are looking for once we "convert" this ending result to a limit. This said, the simpler proof can be understood in another way: as in limits, dy, df and dx are not 0. The expressions dy/dx and df/dx are thus fractions. Their factor, df/dx, is also a fraction. However, it is more convenient to look at df/dx as a product. The definition of the derivative is dy/dx = f'(x) + k where k is increasingly small for dx going to 0. Hence we would have

df/dx = (z'(x) + k )*(f'(u) + l ) = z'(x)*f'(u) + lk + l(...) + k(...)

You can see that as dx becomes closer to 0, the values l and k become very small, and so do the terms lk + l(...) + k(...). If we introduce a variable m, such as m = lk + l(...) + k(...), we obtain

df/dx = z'(x)*f'(u) + m

Since m can be made as small as we wish, this new expression fits the definition of the derivative and hence z'(x)*f'(u) is the derivative.

[prasannapakkiam]

Thanks. Your Proof to my proof is exactly what I asked for...

[Office_Shredder]

werg, your definition of dy/dx doesn't mean anything as far as I can tell; can you clarify it?

[mathwonk]

my stomach hurts. this is horse****

[Werg22]

werg, your definition of dy/dx doesn't mean anything as far as I can tell; can you clarify it?

I have no definition of dy/dx: I just assigned dy/dx to a fraction for the sake of being practical. dy/dx in my explanation really means f(x+h) - f(x)/h. Mathwonk, please elaborate...

[d_leet]

I have no definition of dy/dx: I just assigned dy/dx to a fraction for the sake of being practical.

It doesn't seem like such a great idea, at least for rigorous proofs, to arbitrarily assign new definitions to things that are already well defined.

dy/dx in my explanation really means f(x+h) - f(x)/h.

But that is not what dy/dx is, it is the limit of this ratio.

[Office_Shredder]

I have no definition of dy/dx: I just assigned dy/dx to a fraction for the sake of being practical. dy/dx in my explanation really means f(x+h) - f(x)/h. Mathwonk, please elaborate...


Then you hit the wall that

lim_{h->0}\frac{f(x+h)-f(x)}{h} =/= \frac{lim_{h->0}f(x+h)-f(x)}{lim_{h->0}h}

[NeoDevin]

my stomach hurts. this is horse****

While I agree with the sentiment... not your most constructive post ever.

(Nor mine :rolleyes: )

[Gib Z]

Werg22, I am all for your treatment of the differentials in such a manner, but as a nice trick that use in calculations, not as a *proof*. If you wish to justify your treatment rigorously, please prove that in every case the differentials can be treated as such, retaining its original definition.

[prasannapakkiam]

Okay, the arguments seem to have swerved. But nobody has explained whether my PROOF (NOT derivation) is valid. If not, nobody has yet stated why...

[Gib Z]

In The strictest sense, it is not valid as you have treated notation which just appears to look like a fraction, as a fraction. We have discussed the pros and cons and if you read our response you would have realised we have already stated why it is not valid.

[Werg22]

It doesn't seem like such a great idea, at least for rigorous proofs, to arbitrarily assign new definitions to things that are already well defined.



But that is not what dy/dx is, it is the limit of this ratio.

I don't think you got the whole principle. Forget dy/dx. Pretend we are talking about f(u+h) - f(u) / i

where h is a small change change in u that is itself a function of the change in x, which we shall denote i.

If we multiply the expression by the change h top and bottom, we get the fraction

f(u+h) - f(u) / h * h/i

Note that this a fraction and no limit has been evaluated. Now as i - > 0, so does h. But however small i and h, the expression simplifies to f(u+h) - f(u)/i.
Hence, the limit as i goes to 0 of THAT expression, is the same as the limit of f(u+h) - f(u)/i as i goes to 0. We have (definition of a limit):

f(u+h) - f(u) / h = f'(u) + k, h/i = z'(x) + l

where k and l are increasingly small for smaller and smaller i and h. This gives us the following expression for f(u+h) - f(u) / i:

f(u+h) - f(u) / i = f(u+h) - f(u) / h * h/i = f'(u)*z'(x) + kl + k(...) + l(...)

Here again, I have only dealt with fractions. Now look at what happens: if we let i go to 0, so does h. Hence, for a very small i, k and l will be very small. Now you can see that the expression kl + k(...) + l(...) approaches 0 and is, from a certain point, constantly approaching 0 without being bounded to a value close to it. Hence we can write

f(u+h) - f(u) / i = f'(u)*z'(x) + m(i)

where m(i) is a function of i and is equal to kl + k(...) + l(...). Now since we have deduced that m(i) is increasingly small (absolute value) from a certain point and approaches 0, this new expression fits exactly what we mean by a limit. Hence f'(u)*z'(x) is defined as the limit. This is exactly what I did with my explanation that made dy/dx a real fraction: it was for the sake of being pragmatic. In the explanation, dy/dx were no longer infinitesimals, but just values such as f(u+h) - f(u), i and u. Now that I didn't use any of this, I hope it's clearer. Anyway, the real use of dy/dx spares us allot of time as it spares us to consider everything with limits. And for being rigorous, the explanation itself is plenty rigorous, just not the presentation - a ridiculous expectation on a forum.

[Werg22]

Then you hit the wall that

lim_{h->0}\frac{f(x+h)-f(x)}{h} =/= \frac{lim_{h->0}f(x+h)-f(x)}{lim_{h->0}h}

No wall at all, h is left to be the same on top and bottom so it's the same expression.

[Office_Shredder]

Except the RHS of that inequality doesn't even exist.

So you've proven no function has a derivative. Congratulations

Notice how, when being rigorous, one can only prove lim(a/b) = lim(a)/lim(b) only if lim(a), lim(b) both exist and lim(b) =/= 0 (assuming lim(a/b) exists here)

[Gib Z]

In general

\lim_{x\to c} \frac{f(x)}{g(x)} \not= \frac{\lim_{x\to c} f(x)}{\lim_{x\to c} g(x)}

[Werg22]

Except the RHS of that inequality doesn't even exist.

So you've proven no function has a derivative. Congratulations

Notice how, when being rigorous, one can only prove lim(a/b) = lim(a)/lim(b) only if lim(a), lim(b) both exist and lim(b) =/= 0 (assuming lim(a/b) exists here)

RHS? Meaning? And the condition lim (b) = 0 is the whole foundation of differential calculus, so you're gonna have to explain what you mean here. And as I calrified before, h on the top and h on the bottom are kept equal, which does not change the limit.

[Werg22]

In general

\lim_{x\to c} \frac{f(x)}{g(x)} \not= \frac{\lim_{x\to c} f(x)}{\lim_{x\to c} g(x)}

If the change h is kept equal on the denominator and numerator, the limit IS the same! And what do you mean in general? This inequality is only true if the limit of the denominator is 0.

[Office_Shredder]

RHS means right hand side (of the equation).

And the condition lim (b) = 0 is the whole foundation of differential calculus

No, the foundation of differential calculus is that limits of the form lim(a/b) where lim(a) and lim(b) go to zero can converge. This doesn't mean it converges to the expression lim(a)/lim(b) (since then it wouldn't, because lim(a)/lim(b) doesn't even exist).

Do you even know what the definition of a limit is? Because this should be pretty obvious

EDIT

If the change h is kept equal on the denominator and numerator, the limit IS the same! And what do you mean in general? This inequality is only true if the limit of the denominator is 0.

That's the point, if the denominator is 0 or if either limit doesn't exist, then it's not true.

For example, if c=0, f(x)=1/x, g(x)=1/x2, then limf(x)/limg(x) doesn't exist, but lim[f(x)/g(x)] = 0

[Werg22]

RHS means right hand side (of the equation).



No, the foundation of differential calculus is that limits of the form lim(a/b) where lim(a) and lim(b) go to zero can converge. This doesn't mean it converges to the expression lim(a)/lim(b) (since then it wouldn't, because lim(a)/lim(b) doesn't even exist).

Do you even know what the definition of a limit is? Because this should be pretty obvious

This is very frustrating. I will repeat it one last time. It converges towards the same value if the value of h in f(x+h) - f(x) and h itself are kept the same!!!!!!!!!! If h = 0.001 on top, it's equal to 0.001 on the bottom!!!!

f(x)=1/x, g(x)=1/x:

f(0+ h) = 1/h , g(0+h) = 1/h^2

now the expression f(h)/g(h) = h. And to what value does that expression converge, provided that the value of h is kept the same in both functions, as h goes to 0? 0.

[Office_Shredder]

Noo... once you split it into two limits,, there's no reason to think the h's go to zero just as fast or anything like that (again, definition of a limit?).

The h is just a dummy variable and could as easily be called k and j, or whatever.

I'm not responding anymore until you post the definition of a limit, because otherwise the conversation is meaningless, because you're not actually arguing from a logical basis

[Werg22]

Noo... once you split it into two limits,, there's no reason to think the h's go to zero just as fast or anything like that (again, definition of a limit?).

The h is just a dummy variable and could as easily be called k and j, or whatever.

I'm not responding anymore until you post the definition of a limit, because otherwise the conversation is meaningless, because you're not actually arguing from a logical basis

The irony!

[mathwonk]

a proof is a logical deduction starting from meaningful definitions.

in your argument you have made assertions without justification, like dy/du times du/dx equals dy/dx.

this assertion is in fact the statement of the chain rule that you are trying to prove.

this is not trivial, at least not with the usual limit definitions of dy/du and du/dx. any other non standard definitions, involing say "infinitesimals" need to be accompanied by and justified in terms of, appropriate defijnitions of infinitesimals.

[mathwonk]

here is a fairly complete proof of the chain rule.

f is differentiable at z0 if and only if the limit of [f(z)-f(z0)]/(z-z0) exists and is finite. if so this limit is denoted f'(z0).

let u(x) be differentiable at x0 and also y(u) be differentiable at u0 = y(x0).

then we claim y(u(x)) is differentiable at xo with derivative y'(u0).u'(x0).

lemma: u is also continuous at x0.
proof: since [u(x)-u(x0)]/(x-x0) -->u'(x0), as x-x0-->0, then

[u(x)-u(x0)] = [u(x)-u(x0)]/(x-x0) . [x-x0]-->0, as x-x0-->0. QED


Now we must show that [y(u(x))-y(u(x0))]/(x-x0)-->y'(uo).u'(x0),

as x-->x0.

case 1) assume u(x) is never equal to u0 on some punctured nbhd of x0.

then the fraction [y(u(x))-y(u(x0))]/(x-x0), equals the product

{[y(u(x))-y(u(x0))]/(u(x)-u0)}.{[u(x)-u(x0)]/(x-x0).

since in the product, the factors approach y'(uo) and u'(x0) respectively, the product rule for limits finishes it.

case 2) u(x) equals u0 for some x on every punctured nbhd of xo. now we cannot consider the denominator u(x)-uo as x-->x0, for such points x but no matter.

as x-->xo through other values, the previous argument still works, and now we know that the limit u'(x0)= 0, since it exists and equals zero on every punctured nbhd of x0.

thus as x-->xo through values where u(x) differs from u0, the limit of

[y(u(x))-y(u(x0))]/(x-x0) is zero.

But also at the other points x where u(x) = u(x0), now
the numerator of [y(u(x))-y(u(x0))]/(x-x0) equals zero, so the limit is also zero as x-->x0 through points where u(x) = u(x0).

[Werg22]

I don't see what's the point of going through all of that. If we have f(z(x + h)) - f(z(x)) / z(x + h) - z(x), certainly the limit of that expression as z(x + h) - z(x) goes to 0 is f'(z(x)). if we multiply the expression f(z(x + h)) - f(z(x)) / h by [z(x + h) - z(x)] / [z(x + h) - z(x)], the limit doesn't change since this is equal to 1. So:

lim h - > 0 f(z(x + h)) - f(z(x)) / h = lim h - > 0 f(z(x + h)) - f(z(x)) / z(x + h) - z(x) * z(x + h) - z(x) / h.

The limit of the last expression is f'(z(x))*z(x), and so is the limit of the original expression. If this is not rigorous, then I might as well just go to a nuts house.

[JonF]

If the change h is kept equal on the denominator and numerator, the limit IS the same! And what do you mean in general? This inequality is only true if the limit of the denominator is 0.

please explain what you mean by "keeping the change equal" in a limit using the formal definition of a limit...

[Werg22]

What I mean by this is: Suppose we have f(x+h) - f(x) /h. Now we chose a very small value of h that is the same on top and bottom. For example h = 0.000001. We get f(x+0.000001) - f(x) /0.000001. If we let h approach 0 whilst respecting that condition, then the expression f(x+h) - f(x) /h tends towards the limit. In the expression of the derivative such as f(z(x+i)) - f(z(x)) /i, the same concept applies. If we multiply top and bottom by h = z(x+i) - z(x), then we can rewrite the expression as the product of two quotients. Now, if we let i go to very small values, so does h. The expression h/i thus approaches the limit z'(x), and the expression f(z(x+i) - f(z(x))/ h approaches f'(z(x)). Expressing the expression in such a way shows us that

f(z(x+i)) - f(z(x)) /i = f'(z(x))*z'(x) + kl + l(..) + k(...)

Where k and l are values that get smaller and smaller in scale as i approaches 0. You can see that kl + l(..) + k(...) doesn't have any lower bound: we make it as small as we wish by conveniently reducing i. This is the very definition of a limit: the limit of a function f(x) at c, is a constant C such as that we have an equality

f(c + h) = C + \epsilon (h)

in which epsilon can be made as small in scale as we wish by making h conveniently smaller. THIS is the definition of a limit, or at least the definition I have always believed in.

[mathwonk]

werg, you do not seem to realize that it is possible the fraction you wrote has a zero denominator for infiniteoy many values of x on every nbhd of x0.

then it is not at all clear that multiplying by 0/0 is multiplying by 1.

i.e. you say "certainly the limit of that expression as z(x + h) - z(x) goes to 0 is f'(z(x))"

but the point is that the denominator has to go to zero through non zero values for that statement to be true.

and when you are just letting x go to x0, that may not be the case.

[Werg22]

I don't really understand the meaning of this... z(x + i) - z(x) is never 0, it's a function of i, which is itself never 0. z(x + i) - z(x) goes through strictly positive or negative values past a certain lower bound on h, this depending on the monotonicity of the function z on on the point at which we are evaluating the derivative. This said z(x + i) - z(x) tends towards 0, meaning it can be made as small as we wish.


but the point is that the denominator has to go to zero through non zero values for that statement to be true.

To repeat myself, z(x + i) - z(x) is never 0, as i is never 0. This is best shown by assigning the value z(x + i) - z(x) to h. We write in simplified terms:

f(z(x+i)) - f(z(x)) / z(x + i) - z(x) = f(z(x) + h) - f(z(x)) / h.

Since h tends towards 0, this expression, here again, tends towards the derivative at point z(x). The same goes with h/i. And since their product of the two ratios is always f(z(x+i)) - f(z(x)) / i, here again reasserting the fact that neither h nor i are 0, we get the expression

f(z(x+i)) - f(z(x)) / i = f'(z(x))*z(x) + m

Where m gets increasingly small as the other two ratios get closer to the actual value of the their respective derivatives. The above expression is exactly what we are looking for: f'(z(x))*z(x) is the limit as i goes to 0, because m is a function of i and can be made as small as we wish.

[prasannapakkiam]

Sorry about this late Reply. But I asked another teacher. He showed that my proof is valid:

BUT TO FULLY PROVE IT, I must take the limit of x-->0 in the END.

This creates:

lim(x-->0) (dy/du*du/dx)
= lim(x-->0) (dy/du) + lim(x-->0) (du/dx)

This proves it. But there is a problem with: lim(x-->0) (dy/du)

But as x-->0, u-->0
so this proves it. Agreed?

[NeoDevin]

so this proves it. Agreed?

No, most of what you wrote was nonsense.

Read mathwonk's post with the proof. All these "proofs" treating derivatives as fractions (if they are valid at all) require first proving that you can use them in such ways (which, often you can't). It is generally the best practice when you want to prove something, to go to the definition, and prove from there, or go to theorems which you've already proven, and go from there.

Up to this point in your math career, I'm pretty sure that you haven't proven that you can separate dy/dx like that (or you wouldn't have asked in the first place). It's often convenient to remember theorems and methods by systems like that. But it is not always true, or even meaningful to write it out.

[Office_Shredder]

To repeat myself, z(x + i) - z(x) is never 0, as i is never 0.

Wow. Just wow. Similiarly, x2-1 is never zero if x is never zero

[Werg22]

Wow. Just wow. Similiarly, x2-1 is never zero if x is never zero

Thank you for having validating my discussion: you just proved you are a total ignorant. Keep your mediocrity for yourself.

[Office_Shredder]

Thank you for having validating my discussion: you just proved you are a total ignorant. Keep your mediocrity for yourself.

I agree, what I posted was pretty stupid

Look at sin(1/x) as it x approaches zero. On any interval (-a,a) , sin(1/x) is zero infinite times

[Werg22]

But we are only interested in a region in which the function is strictly monotonic in one direction! If we chose to derivative sin (1/x) at say 2, we chose the whereabouts of 2 so that the function is monotonic in that neighborhood, thus making h never equal to 0.

[Office_Shredder]

But we are only interested in a region in which the function is strictly monotonic in one direction!

This is a rather brash assumption considering in general you don't even know what function you're dealing with

we chose the whereabouts of 2 so that the function is monotonic in that neighborhood, thus making h never equal to 0.

You don't get to choose where you're taking the derivative at either

EDIT:

For example, f(x) = x2sin(1/x) if x=/= 0, f(0) = 0

What is f'(0)?

[Werg22]

The function you presented is undefined at x = 0, and unless we define it at this point, differientating at that point is meaningless. Now back to the main point: for any regular continuous function that is constructed out of elementary function, for any point \xi, there is a value {\epsilon} such as if |\lambda|<\epsilon, then

f(\xi + \lambda) - f(\xi)

is either strictly negative or strictly positive. This is pretty obvious: around any point, there exist a region in which the function is monotonic. Hence if i is inferior to a certain episilon (the so called lower bound on i I mentionned in previous posts), the value of z(x+i) - z(x) is either strictly negative or strictly positive and is thus assured of never reaching 0. Certainly, if we were to evaluate the ratio with |i| superior to this value, we could get values that osciliate between negative and positive and through 0. However, once i is small enough, this is no longer the case, and the ratio converges steadily.

[Office_Shredder]

The function you presented is undefined at x = 0, and unless we define it at this point, differientating at that point is meaningless.

For example, f(x) = x2sin(1/x) if x=/= 0, f(0) = 0

What is f'(0)?

Your point that you claim is absolutely false, as shown by this example

[Werg22]

You did exactly what I mentionned: you defined the function at that point! 0 is the limit of the value of x^2*sin(1/x) as x goes to 0, not the actual definition! It seems you don't understand much and you're trying to cover that up by constantly contradicting me with examples that aren't very sensible as to what we are talking about here. I am done explaning myself: everything that had to be proven and justified already has.

[Office_Shredder]

No, that was what I wrote in my first post. Note I editted it two hours before you posted, I hope it didn't take you 2.5 hours just to write what you did

[Werg22]

For your information, I was in exam session and answered you as soon as I had access to a computer.

[mathwonk]

werg, take a look at the function z(x) = xsin(1/x), as x goes to zero.

this function is zero on every nbhd of x. i.e. this function is not monotonic on any nbhd of zero.

this little complexity is the traditional difficulty in proofs of the chain rule.

you are missing this case.

most books from hardy on, i.e. for the last century, have said this case makes the proof you are giving invalid. I am pointing out that the proof can be made valid by covering this case separately, as was done in ancient books, and a few more recently.

[Office_Shredder]

For your information, I was in exam session and answered you as soon as I had access to a computer.

That's my point, I didn't just edit it in while you were posting, it was sitting there in my post long before you read it for you to notice, you just happened to miss it.

It's not a huge deal that you didn't read it, but the derivative definitely is defined by the definition of a derivative, and the function definitely does not work how you want it to.

[Werg22]

Mathwonk, this dosen't impose that much of a difficulty: as long as we are assured that the product of the ratios converges. Even if the denominator can be 0, there would always exist a value i small enough to ''go back'' to a non-zero denominator. The point is: the product of the two ratios IS the ratio f(z(x+i)) - f(z(x)) / i for every value at which the other two ratios are defined. Even if we have z(x+i) - z(x) = 0, making the product undefined at that point, for a smaller i, the product becomes defined once again. The only difference here is that value k in f(z(x+i)) - f(z(x))/i = z'(x)*f(z(x)) + k, ossiliates between the negative and positive indefinitly, but it decreases in scale: its absolute value constantly decreases, hence making z'(x)*f(z(x)) the derivative by definition.

Edit: Ok sorry, I did not read what you said about treating it seperatly.

Office_Shredder: I admit that my proof did not extend to the case in which there is no monotnic neighbourhood. The explanation above validates the chain rule for this special case.

[Office_Shredder]

I'm a bit confused now... the statement

f(z(x+i)) - f(z(x))/i = z'(x)*f(z(x)) + k

doesn't really prove anything

[mathwonk]

well you are in good company. as i said, even the great g.h. hardy, and many many other calculus authors since his time, have presented the problem with ratio canceling proof apparently without noticing the special case can be treated easily, but separately, once it is noticed.

but you still have to give a proof that the difference k, in

f(z(x+i)) - f(z(x))/i = z'(x)*f(z(x)) + k, does go to zero even when

z(x+i) = z(x), which is exactly what i was doing in the part where you said you failed to see why i was going to so much trouble. i.e. this is precisely the import of my case 2.


[I have been explaining this proof to people for almost 40 years now, even though as i said, it appears correctly in 100 year old books on analysis.]

[mathwonk]

i just checked edwards and penney, hass weir thomas, grossman, courant, even joseph kitchens great book, and none of them give this proof.

kitchen gives a two case proof but makes the second case look more complicated. the point is that in the second case the LHS of werg's equation

f(z(x+i)) - f(z(x))/i = z'(x)*f(z(x)) + k, is zero, so it suffices to show that

in that case z'(x) is also zero, which is what i remark as my proof of case 2. i.e. in case 2, k is identically zero. and in case 1, the elementary proof shows it goes to zero.

courant does remark in a footnote that the proof can be carried out by a special argument, but that he prefers to use a proof that does not have two separate cases.

courants proof, i.e. hardy's, is also the one given in , grossman, mattuck, etc ........ edwards penney give a proof like that of spivak in edition 4 but the usual hardy proof in edition 6 as i recall.

grossman even declares that a different approach from ours is REQUIRED in case 2, which of course is false. many other books imply the same.

spivaks proof, also given in 4th edition of edwards penney, builds a special function phi, whose definition has two cases, essentially building our two case argument into the definition of this function thus making it look afterwards as if the proof has one case. but the proof there is totally unmotivated, and the two cases are there but hidden.

[Werg22]

Hummm I seee... but there's an ambiguity in my mind... I still do not understand what is to be proven with k for the case z(x+i) = z(x). I see it as such: the product of the two ratios is not defined for z(x+i) = z(x), however it's defined for an infinity of i in the neighbourdhood of 0. We need not the product to be always defined: as long as we have enough information on the behaviour of f(z(x+i)) - f(x) /i, we have enough: we know that for a conviniently small i, the absolute value of k gets smaller. This truth is not constrained by the fact that the value of the product is not always defined...

[Office_Shredder]

You haven't actually proven it, you've just said something around the lines of

'well, from curves that behave relatively well, it appears as if it should become small'

But either way, you're assuming that df(u(x))/dx = df/du*du/dx in the first place just to make that equality so you can say k becomes small

[mathwonk]

i rewrote my post to explain this i hope.

[mathwonk]

now that we all aprpeciate the issues, let me try again to give the proof:

we want to show that if z'(x0) exists and f'(z0)) exists where z0 = z(x0),

then also g'(x0) exists where g = foz, and in fact that g'(xo) = f'(z0)z'(x0).

so we want to show the limit of the quotient [f(z(x))-f(z(x0))]/[x-x0]

exists as x-->x0, and is equal to the product f'(z0)z'(x0).

case 1) z'(x0) is not zero. then there is a punctured interval around x0 where z(x) never equals z0.

then the quotient [f(z(x))-f(z(x0))]/[x-x0] equals the product of two quotients

[f(z(x))-f(z(x0))]/[z(x)-z0] . [z(x)-z(x0)]/[x-x0].

and we are done by definition of the derivative and the product rule for lim its.


case 2) z'(x0) = 0. this time all we have to show is that the quotient

[f(z(x))-f(z(x0))]/[x-x0] has limit zero as x-->x0, and it suffices to show this separately as x-->x0 through values where z(x) = z0, and those where it does not.

at those x where z(x0) = z0, there is nothing to show, as then the quotient is identically zero.

for values where z(x0) = z0, the argument from case 1 applies again. QED.

[Werg22]

You haven't actually proven it, you've just said something around the lines of

'well, from curves that behave relatively well, it appears as if it should become small'

But either way, you're assuming that df(u(x))/dx = df/du*du/dx in the first place just to make that equality so you can say k becomes small

While mathwonk made me realise I did not exactly prove the general case, it's for a totally different reason than whatever you are speaking of.

[JonF]

You don't prove any case, you have yet to state anything formal. I mean go back and look at what you responded to my previous question and note how you ignored the part that said "using the formal definition of a limit"

[Werg22]

No, it's just that you don't understand.

[JonF]

Why don't you try stating the formal definition of a limit and then comparing it to what you said about "keeping the change equal" and then tell me who is failing to understand. But hey, "you don't undertand" is a good fall back position.

[Werg22]

How about you check post 32.

[prasannapakkiam]

That is it.

Arguments seem to be revolving around whether dy/dx or dy/du can be split up into a fraction. How about we take a look at: f'(x)=lim h--> 0 (f(x+h)-f(x))/h. How did we derive it? We used small values of y and x i.e. dy and dx. Remember that all the equal signs in my proof are the "aprox." signs. The limit in the end ties it all up to a differential. But the limit thing is neccessiarily neccessary...

[Werg22]

That is it.

Arguments seem to be revolving around whether dy/dx or dy/du can be split up into a fraction. How about we take a look at: f'(x)=lim h--> 0 (f(x+h)-f(x))/h. How did we derive it? We used small values of y and x i.e. dy and dx. Remember that all the equal signs in my proof are the "aprox." signs. The limit in the end ties it all up to a differential. But the limit thing is neccessiarily neccessary...

This was the argument but it no longer is. The argument now is that the equality of the ratio and the product of the other ratios as well as the equality the limit of the first ratio with the product of the other ratios are doubtful. But as I said, for the case in which there exist a neighborhood in which the functions are monotonic, there's nothing to be doubted.

[JonF]

How about you check post 32.This is not the definition of the limit of a function, and your explanation was far from formal, and didn’t really what exactly it means to take to limits keeping “equal change”. Did you go over the delta epsilon definition of a limit in your class?

[Office_Shredder]

That is it.

Arguments seem to be revolving around whether dy/dx or dy/du can be split up into a fraction. How about we take a look at: f'(x)=lim h--> 0 (f(x+h)-f(x))/h. How did we derive it? We used small values of y and x i.e. dy and dx. Remember that all the equal signs in my proof are the "aprox." signs. The limit in the end ties it all up to a differential. But the limit thing is neccessiarily neccessary...

Yes, someone said'hey, if we use this as a definition, we get the slope for a bunch of nice looking curves. So let's extend this, and define it to be the slope for all curves'

Why? Because for all curves, that includes curves that aren't nice. So yes, you can use this intuitive sense of derivative, and if your curve behaves welll, expect to be right. But the point of the definition is that it's the only way to 100% accurately describe the derivative of a function, no matter which function you're using.

Technically, werg has yet to prove that he's even covered all the cases (and as we can see by his initially missing the case where it hits zero infinite times in any interval, it's not obvious that all the cases are covered).

However, by definition using the definition of a derivative hits all the cases

[Werg22]

Notice how these [insert you know what here] keep on attacking my proof without actually putting anything forth. Office_Shredder, you are so laughable. As soon as Mathwonk stepped in, you changed your tune. It's pretty obvious now that you really don't know what you are talking about.