Proof of Chain Rule for Differentiation - Stoll, Theorem 5.1.6

[Math Amateur]

I need help in understanding the proof of the Chain Rule for differentiation, as presented in Theorem 5.1.6 in Manfred Stoll's book: Introduction to Real Analysis.

Theorem 5.6.1 in Stoll (page 173) reads as follows:View attachment 3925In the above proof we read the following:

" ... ... By identity (3) and then (2),

\(\displaystyle h(t) - h(x) = g(f(t)) - g(f(x))\)

\(\displaystyle = [ f(t) - f(x)][g'(y) + \nu (s) ]
\)
... ... "
I cannot see how (formally) the following equation is true:\(\displaystyle g(f(t)) - g(f(x)) = [ f(t) - f(x)][g'(y) + \nu (s) ]\)Can someone please demonstrate how/why this is the case?

I would really appreciate some help ... ...

Peter

 

[Prove It]

I need help in understanding the proof of the Chain Rule for differentiation, as presented in Theorem 5.1.6 in Manfred Stoll's book: Introduction to Real Analysis.

Theorem 5.6.1 in Stoll (page 173) reads as follows:View attachment 3925In the above proof we read the following:

" ... ... By identity (3) and then (2),

\(\displaystyle h(t) - h(x) = g(f(t)) - g(f(x))\)

\(\displaystyle = [ f(t) - f(x)][g'(y) + \nu (s) ]
\)
... ... "
I cannot see how (formally) the following equation is true:\(\displaystyle g(f(t)) - g(f(x)) = [ f(t) - f(x)][g'(y) + \nu (s) ]\)Can someone please demonstrate how/why this is the case?

I would really appreciate some help ... ...

Peter

It's because in the composition of the functions, $\displaystyle \begin{align*} h(x) = g \circ f(x) \end{align*}$, that would mean to have $\displaystyle \begin{align*} h(t) - h(x) = g \left[ f(t) \right] - g \left[ f(x) \right] \end{align*}$.

But we ALREADY defined a difference in g functions as $\displaystyle \begin{align*} g(s) - g(y) \end{align*}$, and thus that means to work out $\displaystyle \begin{align*} g \left[ f(t) \right] - g \left[ f(x) \right] \end{align*}$, we must have $\displaystyle \begin{align*} s = f(t) \end{align*}$ and $\displaystyle \begin{align*} y = f(x) \end{align*}$. Therefore

$\displaystyle \begin{align*} h(t) - h(x) &= g \left[ f(t) \right] - g \left[ f(x) \right] \\ &= g(s) - g(y) | _{s = f(t), y = f(x) } \\ &= \left( s - y \right) \left[ g'(y) - v(s) \right] |_{s = f(t) , y = f(x) } \\ &= \left[ f(t) - f(x) \right] \left[ g'(y) - v(s) \right] \end{align*}$