Is my proof for the Chain Rule invalid?

[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 doesn'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 realize 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.