Thread: Chain Rule - B&S Theorem 6.1.6 ...

1. I am reading "Introduction to Real Analysis" (Fourth Edition) by Robert G Bartle and Donald R Sherbert ...

I am focused on Chapter 6: Differentiation ...

I need help in fully understanding an aspect of the proof of Theorem 6.1.6 ...

Theorem 6.1.6 and its proof ... ... reads as follows:

In the above text from Bartle and Sherbert we read the following:

"... Since the function $\displaystyle ( \psi \circ f ) \cdot \phi$ is continuous at $\displaystyle c$, and its value at $\displaystyle c$ is $\displaystyle g' (f (c) ) \cdot f'(c)$ , Caratheodory's Theorem gives (11) ...

Could someone please explain exactly how Caratheodory's Theorem gives (11) ...?

Peter

*** NOTE ***

The post above mentions Caratheodory's Theorem ... so I am proving the text of the theorem statement ... as follows:

Originally Posted by Peter
I am reading "Introduction to Real Analysis" (Fourth Edition) by Robert G Bartle and Donald R Sherbert ...

I am focused on Chapter 6: Differentiation ...

I need help in fully understanding an aspect of the proof of Theorem 6.1.6 ...

Theorem 6.1.6 and its proof ... ... reads as follows:

In the above text from Bartle and Sherbert we read the following:

"... Since the function $\displaystyle ( \psi \circ f ) \cdot \phi$ is continuous at $\displaystyle c$, and its value at $\displaystyle c$ is $\displaystyle g' (f (c) ) \cdot f'(c)$ , Caratheodory's Theorem gives (11) ...

Could someone please explain exactly how Caratheodory's Theorem gives (11) ...?

Peter

*** NOTE ***

The post above mentions Caratheodory's Theorem ... so I am proving the text of the theorem statement ... as follows:

After a little reflection I think the answer to my question is along the following lines ...

In the proof of the Chain Rule (Theorem 6.1.6) B&S establish that:

$\displaystyle g(f(x)) - g(f(c)) = [ ( \psi \circ f(x) ) \cdot \phi (x) ] ( x - c )$

Matching this with (10) in Caratheodory's Theorem and noting that (in the notation of that theorem) $\displaystyle \phi (c) = f'(c)$, we then have (in Theorem 6.1.6) that

$\displaystyle [ ( \psi \circ f ) \cdot \phi ] (c) = ( g \cdot f)' (c)$

Is that a correct interpretation ... ?

Peter

3. Yes, it's correct. In my opinion, there is nothing here besides matching a general theorem statement with a concrete situation.

Originally Posted by Evgeny.Makarov
Yes, it's correct. In my opinion, there is nothing here besides matching a general theorem statement with a concrete situation.

Thanks Evgeny ...

Peter