Prove that $g'(f(x)) = \frac{1}{f'(x)}$ Explanation behind intuition of the proof. Question: Suppose $f'(x) > 0)$ in $(a,b)$. Prove that $f$ is strictly increasing in $(a,b)$, and let $g$ be its inverse function. Prove that $g$ is differentiable, and that $$ g'(f(x)) = \frac{1}{f'(x)}$$ for $(a< x < b)$.

Relevant concepts needed

1) Definition: Let $f$ be defined and real valued on $[a,b]$. For any $x\in [a,b]$ we form the quotient: $$\phi(t) = \frac{f(t) - f(x)}{t-x}$$ for $(a < t < b, t \neq x)$ and define $$f'(x) = \lim_{t \rightarrow x} \phi(t)$$ Provided the limit exists. This is our defintion of the derivative at $x$.
2) Since $g$ is defined as the inverse function of $f$, then $g(x) = f^{-1}(x)$


I've already done the first and second parts of the question: Showing $f$ is strictly increasing and $g$ is differentiable. My issue is with the third portion, showing $ g'(f(x)) = \frac{1}{f'(x)}$. 
I made some attempts and was not successful and then I found solutions for the question both of which were similar in style. I'm going to reproduce the solution here but with questions about the reason / intuition behind each step. I've reached a stage in my mathematical journey where I feel I'm hitting a glass ceiling preventing me from making the jump to possess the ability to think more "creatively"  about these concepts. So I hope trying to understand how the masters approach these situations would be enlightening. Without further ado the solution:
Solution: 
Fix $\epsilon > 0$ and $x \in (a,b)$. $f'(x) > 0$ means that there exists $\eta > 0$ such that $ 0 < |x - t| < \delta_{1}$ implies $$\Bigg|\frac{f(t) - f(x)}{t-x} \Bigg| > \eta$$
$$\Rightarrow \Bigg| \frac{1}{\frac{f(t) - f(x)}{t-x}} \Bigg| < \frac{1}{\eta}$$
$\textit{$\textbf{Comment/Question}$}$: What is the reason behind this step? The idea behind the existence of an $\eta$ makes sense to me by Archimedian properties of the Reals, but I don't see the intuition behind needing to establish this? I get vibes that there is some sort of argument that is going to involve continuity, but still not sure of its placement.
From the definitiuon of $f'(x)$, we also have for $\eta f'(x) \epsilon > 0$ there exists a $\delta_{2} > 0$ such that $ 0 < |x - t| < \delta_{2}$ implies $$\Bigg|f'(x) - \frac{f(t) - f(x)}{t-x} \Bigg| > \eta f'(x) \epsilon$$
$\textit{$\textbf{Comment/Question}$}: Again I have no idea the purpose of this step, but I again get continuity sort of vibes from this portion.
Let $y = f(x)$ and $\delta = min\{\delta_{1},\delta_{2} \}$. Then for any $u \in (f(x-\delta), f(x + \delta)$, let $g(u) = t$ so $t \in (x-\delta, x + \delta)$ and $$\Bigg| \frac{g(u) - g(y)}{u-y} - \frac{1}{f'(x)} \Bigg|= \\"magical\  manipulations" < \frac{ \eta f'(x) \epsilon}{\eta f'(x)} = \epsilon$$
Which shows $ g'(f(x)) = \frac{1}{f'(x)}$.
$\textit{$\textbf{Comment/Question}$}$: I get how to do the mechanical manipulations to arrive at the result, but what was the purpose of defining $u \in (f(x-\delta), f(x + \delta)$, let $g(u) = t$ so $t \in (x-\delta, x + \delta)$ ? I get the purpose of it in terms of trying to align our argument with the definition of continuity, but it still seems uncomfortable that we can arbitrarily just "let this equal this". This probably falls under my inability to think more creatively about the solution.
 A: Intuition: on an interval small enough, and with the origin shifted appropriately, $f(x)$ is close enough to a linear function with positive slope.  For such a function, to be denoted $f_{L}$, we have:
$$
f_{L}(x) = \alpha x, \quad \alpha > 0.
$$
By direct inspection:
$$
f_{L}'(x) = \alpha > 0, \quad
g_{L}(y) = f_{L}^{-1}(y) = \alpha^{-1} y, \quad
g_{L}'(y) = \alpha^{-1} = {1 \over f_{L}'(x)},
$$
where $y = f_{L}(x)$.
So, the key here is that a smooth monotone function is locally equivalent to a linear function with nonzero slope.  The rigorous statement of this is one of the versions of the Implicit Function Theorem (see, for example, Arnol'd's Ordinary Differential Equations, any edition after the 1st).
For a proof in your second question, it can be done without "epsilonics".
Let's form the function $\phi(x)$ as follows:
$$
\phi(x) = g \circ f (x).
$$
Now, note that
$$
x = \phi(x) = g \circ f (x)
$$
and differentiate using the Chain Rule.
A: For geometric intuition (not necessarily help with the proof) use the fact that the graph of the inverse function of $f$ is the reflection of $f$ over the line $y=x$. Then look at the relation between the slopes of the tangents at reflected points.
For example draw the graphs of $y=x^2$ and $y = \sqrt{x}$ and compare the slopes at $(2,4)$ and $(4,2)$.
A: Let $A=f((a, b)) $ then by the properties of continuous functions $A$ is an open interval. Let $x\in A$ and $y\in(a, b) $ be such that $f(y) =x, y=g(x) $. Let $h\neq 0$ be such that $x+h\in A$. Then by strict monotone nature of $f$ we have a $k\neq 0$ such that $y+k\in(a, b) $ and $f(y+k) =x+h, g(x+h) =y+k$ and $kh>0$.
The key is to observe that as $h\to 0$ the variable $k$ also tends to $0$. This is more commonly stated as "inverse of a continuous function is also continuous" (you should prove this as this is non-trivial). Now the evaluation of the derivative of $g$ is pretty standard. We have $$\frac{g(x+h) - g(x)} {h} =\frac{y+k-y} {f(y+k) - f(y)} $$ and the right side tends to $1/f'(y)$ as $h, k$ tend to $0$. Thus $g'(f(y)) =1/f'(y)$. 
