Question

Chain rule proof?

Answer 1

We'll do really simple version of chain rule. Single variable scalar function.

Answer 2

\[
\frac{dz}{dx} = \frac{dz}{dy}\frac{dy}{dx}
\]

Answer 3

http://lmgtfy.com/?q=chain+rule+proof

Answer 4

@kc_kennylau Saw those, didn't like them.

Answer 5

What do you mean by "single variable scalar version"?

Answer 6

I'm thinking if we look at definitions, it could help.

We gotta show \[
\lim_{x\to a}\frac{z(x)-z(a)}{x-a} =
\left(\lim_{y\to b}\frac{z(y)-z(b)}{y-b}\right)\left(\lim_{x\to a}\frac{y(x)-y(a)}{x-a}\right)
\]

Answer 7

In this case \(b=y(a)\).

Answer 8

You can't use y as a variable as in z(y) and a function as in y(x) at the same time...

Answer 9

Actually, let me rewrite that...
\[
\frac{d}{dx}f(g(x)) = f'(g(x))g'(x)
\]
We gotta show \[
\lim_{x\to a}\frac{f(g(x))-f(g(a))}{x-a} =
\left(\lim_{g(x)\to g(a)}\frac{f(g(x))-f(g(a))}{g(x)-g(a)}\right)\left(\lim_{x\to a}\frac{g(x)-g(a)}{x-a}\right)
\]

Answer 10

It looks like chain rule could be proved by simple multiplication if you could show that \(g(x)\to g(a)\) and \(x\to a\) are the same in this context.

Answer 11

And if you assume that the functions are differentiable. (hence the limits here exist)

Answer 12

I would write \(g(x)\rightarrow b\) instead

Answer 13

This?\[
\lim_{x\to a}\frac{f(g(x))-f(g(a))}{x-a} =
\left(\lim_{g(x)\to b}\frac{f(g(x))-f(b)}{g(x)-b}\right)\left(\lim_{x\to a}\frac{g(x)-g(a)}{x-a}\right)
\]

Answer 14

Yep but I'm not sure

Answer 15

To be honest, I find the \(g(x)\) part in there the most dubious part.

Answer 16

That is, having a function approach something.

Answer 17

like saying \(x^2 \to -1\) for example.

Answer 18

The only reason why I'm writing this is because all the proofs I've seen are very complex or hand wavy, and there is something about these definitions that seem very intuitive.

Answer 19

But what would be the definition of \(\Large\dfrac d{d\sin x}\cos x\)?

Answer 20

Well \[
\frac{d\cos x}{d\sin x} = \frac{\frac d{dx}\cos x}{\frac d{dx} \sin x}
\]

Answer 21

I mean using the first principle

Answer 22

I'm not sure

Answer 23

I'm also exploring this topic

Answer 24

\[
\lim_{\sin x\to a} \frac{\cos(\arcsin(\sin(x)))-\cos(\arcsin(a))}{\sin(x) - a}
\]

Answer 25

Does this make sense to you?

Answer 26

\(\arcsin(\sin(x))\) -_-

Answer 27

Basically \(x = \arcsin(\sin(x))\) so \(\cos(x) = \cos(\arcsin(\sin(x))\)

Answer 28

In other words... \[
\frac{d}{d\sin x} f(x) = \frac{d}{d\sin x}f(\arcsin(\sin(x)) = \frac{d}{du}f(\arcsin(u))
\]

Answer 29

But how do you make \(\sin x\) approach \(a\)?

Answer 30

In some cases, you can't do it all. In other cases, there is more than one way to do it.

Answer 31

However, we are working in a context where there is some \(a'\) such that if \(x\to a'\) then \(g(x) \to a\)

Answer 32

That is, we are working in a context where \(\lim_{x\to a'}g(x) = a\)

Answer 33

Since we are doing the chain rule, we get to assume this limit exists since chain rule only works if the functions in question are differentiable (hence continuous, hence limit exists).

Answer 34

But what's a'?

Answer 35

All I can tell you is that it exists, and the properties that I've already said about it.

Answer 36

But looking at what was written above.... \(b\) would correspond to \(a\) and \(a\) would correspond to \(a'\). There was a change in notation.

Answer 37

So basically, suppose we are differentiating \(
f(g(x))
\) at the point \(a'\). Then \(g(a') = a\). Does that make sense?

Answer 38

okay...

Answer 39

Look at https://en.wikipedia.org/wiki/Chain_rule#First_proof

Answer 40

Looks nice to me

Answer 41

What did you mean by "Single variable scalar function" anyway?

Answer 42

It's not a vector and doesn't have multiple inputs.

Answer 43

Actually looking at that proof, I have realized something interesting...

Answer 44

Even if \(g(x)\) is continuous at \(a\) and \(g(a) = b\), we still can't say that: \[
\lim_{x\to a}f(g(x)) = \lim_{g\to b}f(g)
\]Since there many be many cases where \(g(x)=b\) and yet \(x\neq a\).

Answer 45

may be^

Answer 46

But that first proof is exactly what I'm looking for, I just need to understand how to thoroughly understand the caveat.

Answer 47

Well you may want to look at the \(\href{ https://en.wikipedia.org/wiki/(%CE%B5,_%CE%B4)-definition_of_limit }{\mbox{Epsilon-Delta definition of limit}}\).

Answer 48

okay some technical problem here but you know what I mean

Answer 49

No, I understand limits and the technical definition quite well.

Answer 50

I meant "technical" as in the LaTeX cannot be displayed

Answer 51

At no point did I actually assume \[
\lim_{x\to a}g(x) = b = g(a) \implies \lim_{x\to a}f(g(x)) = \lim_{g\to b}f(g)
\] was true, I was just curious about it.

Answer 52

I assume you mean \(g(x)\) by \(g\)

Answer 53

No, in this case I actually mean \(g\) as an independent variable.

Answer 54

But g is a function?

Answer 55

Well, I can use another dummy variable to make it more clear to you: \[
\lim_{x\to a}(f\circ g)(x) = \lim_{x\to b}f(x)
\]

Answer 56

Maybe it was wrong to use \(g\) there? I did it to show the pattern.

Answer 57

Anyway I have to go now

Answer 58

Good night.

Answer 59

17:52 here :)