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The Chain Rule in n Dimensions ... Browder Theorem 8.15 - Another Question ...

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,891
I am reading Andrew Browder's book: "Mathematical Analysis: An Introduction" ... ...

I am currently reading Chapter 8: Differentiable Maps and am specifically focused on Section 8.2 Differentials ... ...

I need some further help in order to fully understand the proof of Theorem 8.15 ...

Theorem 8.15 and its proof read as follows:



Browder - 1 - Theorem 8.15 ... PART 1 ... ....png
Browder - 2 - Theorem 8.15 ... PART 2  ... .....png





In the above proof by Browder we read the following:


" ... ... Now

\(\displaystyle (g \circ f) (p + h) - (g \circ h) (p) = g(q + k) - g(q)\) ... ... "




Can someone please show how/why \(\displaystyle (g \circ f) (p + h) - (g \circ h) (p) = g(q + k) - g(q)\) ... ...




Help will be appreciated ...

Peter
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,891
I am reading Andrew Browder's book: "Mathematical Analysis: An Introduction" ... ...

I am currently reading Chapter 8: Differentiable Maps and am specifically focused on Section 8.2 Differentials ... ...

I need some further help in order to fully understand the proof of Theorem 8.15 ...

Theorem 8.15 and its proof read as follows:










In the above proof by Browder we read the following:


" ... ... Now

\(\displaystyle (g \circ f) (p + h) - (g \circ h) (p) = g(q + k) - g(q)\) ... ... "




Can someone please show how/why \(\displaystyle (g \circ f) (p + h) - (g \circ h) (p) = g(q + k) - g(q)\) ... ...




Help will be appreciated ...

Peter




It now appears to me that the answer to my question is quite straightforward ...

We have ...


\(\displaystyle (g \circ f) (p + h) - (g \circ h) (p) = g(q + k) - g(q)\)

\(\displaystyle = g[ f(p + h) ] - g[f(p)]\) ...

But ... we have that ...

\(\displaystyle k = f(p+h) - f(p)\)

so that

\(\displaystyle f(p+h) = k + f(p) = q + k\)

Therefore

\(\displaystyle g[ f(p + h) ] - g[f(p)] = g(q + k ) - g(q)\) ...


Don't know why I didn't see it ...

Peter