If $f(x)$ is differentiable and invertible, and $g(x)=f^{-1}(x)$. Prove $g'(f(x)) = \frac{1}{f'(x)}$
Actually this was a multiple choice question asking which of the following statement is true if $y=3x-1$ is an equation for the tangent line to the graph $y=f(x)$ at $x=1$.
The correct answer is D, where $g'(2)=\frac{1}{3}$, reason given was because $g'(f(x)) = \frac{1}{f'(x)}$
I tried infer this with the inverse function theorem $(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}$
Since $g(x) = f^{-1}(x)$, I got $g'(x) = \frac{1}{f'(g(x))}$
I substituted $f(x)$ into the equation, I got $g'(f(x))=\frac{1}{f'(g(f(x)))}$, I am a bit lost, how do I proceed from here?
First note that $g$ is just defined as an alternative notation for $f^{-1}$ here.
Then note that $g(f(x)) = f(g(x)) = x$, by simple composition.
We'll use both parts of that relation.
Start with $f(g(x)) = x$
Differentiate both sides with chain rule:
$g'(x)\cdot f'(g(x)) = 1$
Now note that that relationship holds identically for any $x$ value in the domain for $g(x)$, which includes $f(x)$ (as can be trivially observed by noting that $g(f(x)) = x$, so the argument $f(x)$ obviously has an image under $g$). So it is permitted to replace $x$ with $f(x)$ to give:
$g'(f(x)) \cdot f'(g(f(x))) = 1$
And now apply the first part of the previous relation to simplify to:
$g'(f(x))\cdot f'(x) = 1$
$g'(f(x)) = \frac 1{f'(x)}$, as required.
Note that you could just as easily have started with the version of the inverse function theorem you already know:
$(f^{-1})'(x) = \frac 1{f'(f^{-1}(x))}$
And replace $f^{-1}$ with $g$:
$g'(x) = \frac 1{f'(g(x))}$
Finally, as before, replace $x$ with $f(x)$ to yield:
$g'(f(x)) = \frac 1{f'(g(f(x)))}$
And note in the RHS that $g(f(x)) = x$, giving:
$g'(f(x)) = \frac 1{f'(x)}$, as required.