Prove Chain Rule with Newton's Approximation

Question

Synopsis

Note that there is an identical question that has no answers based off the same exercise. I hope that I can be a bit more lucky if I provide some more definitions.

The exercise is from Tao's Analysis I and asks simply to prove the chain rule, which he gives as

Let $X,Y$ be subsets of $\mathbb{R}$, let $x_0 \in X$ be a limit point of $X$, and let $y_0 \in Y$ be a limit point of $Y$. Let $f: X \to Y$ be a function such that $f(x_0) = y_0$ and such that $f$ is differentiable at $x_0$. Suppose that $g: Y \to \mathbb{R}$ is a function which is differentiable at $y_0$. The nthe function $g \circ f:X \to \mathbb{R}$ is differentiable at $x_0$, and $$(g \circ f)'(x_0) = g'(y_0)f'(x_0).$$

He recommends utilizing a solution involving a rigorous version of Newton's Approximation we proved earlier in order to avoid the proof which must take into account the case of $f'(x_0) = 0$. He defines Newton's Approximation as such:

Let $X$ be a subset of $\mathbb{R}$, let $x_0 \in X$ be a limit point of $X$, let $f:X \to \mathbb{R}$ be a function, and let $L$ be a real number. Then the following statements are equivalent

1. $f$ is differentiable at $x_0$ on $X$ with derivative $L$.
2. For every $\epsilon > 0$, there exists a $\delta > 0$ such that whenever $x \in X$ and $|x-x_0| \leq \delta$, we have $$|f(x)-(f(x_0)+L(x-x_0))| \leq \epsilon|x-x_0|.$$

I am always sure to add these definitions because Tao defines things in non-standard ways and also uses notation that is a little different from other texts.

Attempt

Let $\epsilon > 0$. Note that our domain is $X$, so $g$ is restricted to $f(X)$. Hence, there exists a $\delta > 0$ such that if $|f(x) - f(x_0)| < \delta$ then $$|g(f(x)) - (g(f(x_0))+g'(y_0)(f(x)-f(x_0)))| \leq \epsilon |f(x)-f(x_0)|.$$ But I have no idea where to go from here. Note that I must show $$|g(f(x)) - (g(f(x_0))+g'(y_0)f'(x_0)(x-x_0)))| \leq \epsilon |x-x_0|$$ and I'm evidently somehow close, but how do I get $f'(x_0)$ into this? It seems like all I'm missing is one piece of the puzzle but I don't know where to find it. I can fix the right hand side pretty easily through the fact that $f$ must be continuous, but all together I'm a bit stuck on how to get that pesky $f'(x_0)$ term in this. Note that I've already tried somehow connecting Newton's approximation for $f$ into it, but I couldn't find a way. I'd appreciate any hints. Thank you.

Answer 1

This is just to follow through with the suggestion of Daniel Fischer:

Let $\varphi:=|f'(x_0)|+|g'(f(x_0))|\geq{0}$, and let $\varepsilon>0$ be arbitrary. We have that $\varphi^2<\varphi^2+4\varepsilon$, so by Proposition $6.7.3$ in the book, $\varphi<\sqrt{\varphi^2+4\varepsilon}$, thus $0<\frac{-\varphi+\sqrt{\varphi^2+4\varepsilon}}{2}:=\psi$. Since $\psi>0$ and $f$ is differentiable at $x_0$, there exists a $\delta>0$ such that for all $x\in{X}:|x-x_0|\leq{\delta}$, $|f(x)-f(x_0)-f'(x_0)(x-x_0)|\leq{\psi{|x-x_0|}}\implies{|f(x)-f(x_0)|\leq{(\psi+|f'(x_0)|)|x-x_0|}}$. Since $g$ is differentiable at $f(x_0)$, it follows that there exists a $\delta'>0$ such that for all $y\in{Y}:|y-f(x_0)|\leq{\delta'}$, $|g(y)-g(f(x_0))-g'(f(x_0))(y-f(x_0))|\leq{\psi{|y-f(x_0)|}}$. Finally, since $f$ is continuous at $x_0$, there exists a $\delta''>0$ such that for all $x\in{X}:|x-x_0|\leq{\delta''}$, $|f(x)-f(x_0)|\leq{\delta'}$.

Thus, let $x\in{X}:|x-x_0|\leq{\min(\delta,\delta'')}$. Then we have that: $$|g(f(x))-g(f(x_0))-g'(f(x_0))(f(x)-f(x_0))|\leq{\psi{|f(x)-f(x_0)|}}\leq{\psi(\psi+|f'(x_0)|)|x-x_0|}$$

Now observe: $$|g(f(x))-g(f(x_0))-g'(f(x_0))f'(x_0)(x-x_0)|=|g(f(x))-g(f(x_0))-g'(f(x_0))(f(x)-f(x_0))+g'(f(x_0))(f(x)-f(x_0)-f'(x_0)(x-x_0))|\leq{|g(f(x))-g(f(x_0))-g'(f(x_0))(f(x)-f(x_0))|+|g'(f(x_0))(f(x)-f(x_0)-f'(x_0)(x-x_0))|}\leq{\psi(\psi+|f'(x_0)|)|x-x_0|+|g'(f(x_0))|\psi{|x-x_0|}}=\psi(\psi+|f'(x_0)|+|g'(f(x_0))|)|x-x_0|$$ But it is a straightforward matter to check that $\psi(\psi+|f'(x_0)|+|g'(f(x_0))|)=\varepsilon$, as desired.