Proof of the chain rule

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Mondo

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Apr 23, 2021
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107
Hello,

I try to follow the proof of a chain rule for a single variable functions:

1630744571907.png

I don't get how in the world we got the multiplication in the bottom part? I mean how was [imath]\frac{f(g(x)) - f(g(c))}{x-c}[/imath] transfered to [imath]\frac{f(y)) - f(d)}{y-d} \times \frac{g(x) - g(c)}{x-c}[/imath].

Thanks for help.
 
Recall that y=g(x) and d=g(c). This is given.

So f(g(x)) - f(g(c)) = f(y) - f(d)

Also, y - d = g(x) - g(c).

Does that help you see what happened?
 
@Harry_the_cat, no. What you wrote is obvious substitution. What I don't understand is why they multiply by [imath]\frac{g(x) - g(c)}{x-c}[/imath]. It was written out of the blue while this is the essence of the proof.
 
[imath]\frac{F(x)-F(c)}{x-c}=\frac{f(g(x))-f(g(c))}{x-c} = \frac{f(y)-f(d)}{x-c}[/imath] *

Now [imath]g(x)-g(c)=y-d[/imath] (by definition)

So [imath]\frac{g(x)-g(c)}{y-d}=1[/imath] (when [imath]y≠d[/imath])
So the expression in * equals:

[imath]\frac{f(y)-f(d)}{x-c} \times \frac{g(x)-g(c)}{y-d}[/imath]

[imath]=\frac{f(y)-f(d)}{y-d} \times \frac{g(x)-g(c)}{x-c}[/imath]
...
 
lex said:
[imath]\frac{F(x)-F(c)}{x-c}=\frac{f(g(x))-f(g(c))}{x-c} = \frac{f(y)-f(d)}{x-c}[/imath] *

Now [imath]g(x)-g(c)=y-d[/imath] (by definition)

So [imath]\frac{g(x)-g(c)}{y-d}=1[/imath] (when [imath]y≠d[/imath])
So the expression in * equals:

[imath]\frac{f(y)-f(d)}{x-c} \times \frac{g(x)-g(c)}{y-d}[/imath]

[imath]=\frac{f(y)-f(d)}{y-d} \times \frac{g(x)-g(c)}{x-c}[/imath]
...
Click to expand...
That's what I was trying to get Mondo to realise for her/himself.
 
lex said:
[imath]\frac{F(x)-F(c)}{x-c}=\frac{f(g(x))-f(g(c))}{x-c} = \frac{f(y)-f(d)}{x-c}[/imath] *

Now [imath]g(x)-g(c)=y-d[/imath] (by definition)

So [imath]\frac{g(x)-g(c)}{y-d}=1[/imath] (when [imath]y≠d[/imath])
So the expression in * equals:

[imath]\frac{f(y)-f(d)}{x-c} \times \frac{g(x)-g(c)}{y-d}[/imath]

[imath]=\frac{f(y)-f(d)}{y-d} \times \frac{g(x)-g(c)}{x-c}[/imath]
...
Click to expand...

Hmm but if this is really the case and we can write the second term because it is equal to 1 then why do we write it in the first place?
I mean the proof does not show to me why the derivative of composition function is the product of derivatives. In the same way I could say that [imath]2 \times 2 = 4 \times \frac{e^x}{e^x}[/imath]...

I have more doubts regarding this theory:
1630767360421.png

My doubts:
1. Why do we care to have functions f and g defined on an open intervals I and J? Wouldn't it be enough to just provide that the image of g(x) is in the domain of f(x)?
2. Why we need [imath]g(I) \subseteq J[/imath] instead of just [imath]g(I) \subset J[/imath]? The later guarantees all elements of g(I) are in J and both sets don't need to be equal which first relation suggeest.
 
Whenever a factor apparently appears from nowhere, there probably is a factor of 1 involved. It is a standard technique; you first learn it when rationalizing fractions with a radical in the denominator.

To prove the chain rule, we need two factors, one a derivative of f and one a derivative of g. That means two Newton quotients. And obviously

[math]y \equiv g(x) \text { and } d \equiv g(c) \implies y - d = g(x) - g(c) \implies\\ \dfrac{g(x) - g(c)}{y - d} = 1, \text { provided } y \ne d.[/math]
So

[math]x \ne c \implies \dfrac{F(x) - F(c)}{x - c} = \dfrac{f(g(x)) - f(g(c))}{x - c} = \dfrac{f(y) - f(d)}{x - c}.[/math]
That is only one quotient, and furthermore the numerator and denominator are not consistent with a Newton quotient. Thus, we look for a fraction = 1 that can get us the two quotients, one with a denominator of y - d

[math]\therefore x \ne c \text { and } y \ne d \implies \dfrac{f(y) - f(d)}{x - c} * 1 =\\ \dfrac{f(y) - f(d)}{x - c} * \dfrac{g(x) - g(c)}{y - d} = \dfrac{f(y) - f(d)}{y - d} * \dfrac{g(x) - g(c)}{x - c}.[/math]
Now we have two Newton quotients, both with useful denominators. You have to remember where you want to end up.
 
@Mondo
We write this expression simply to help us produce a proof of the chain rule.
Yes, that line is only one step in the proof. The rest of the proof of the chain rule is not here. Presumably you have the rest of the proof?

f and g are defined on open intervals simply for the sake of them being differentiable.

For the composite function to exist you are right we need [imath]g(I)[/imath] to be a subset of J.

You seem to be misinterpreting [imath]g(I) \subseteq J[/imath]
This means [imath]g(I)[/imath] is a proper subset of J OR [imath]g(I)=J[/imath]. It does not mean that it needs to be equal to J.
It means:
in the case when [imath]g(I) \subset J[/imath] then f(g) exists and...
and in the case when [imath]g(I)=J[/imath] then f(g) exists and...
 
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