$f'(f^{-1}(x))$. How do I take the derivative of this?

Question

I am told that $f'(f^{-1}(x))$ where $f(x) = \sin(x)$ and $f^{-1}(x) = \arcsin(x)$.

When I see $f'(f^{-1}(x))$, I instantly think of applying the chain rule since we have a composite of functions. However, my textbook does it differently; it does not apply the chain rule:

$f(x) = \sin(x)$

$\implies f'(x) = \cos(x)$

$\therefore f'(f^{-1}(x)) = f'(\arcsin(x)) = \cos(\arcsin(x))$.

I do not understand why the textbook does the above calculations instead of using the chain rule. I would greatly appreciate it if people could please take the time to clarify this and explain the reasoning behind the decision.

Answer 1

Reverse engineering from the solution given in the book, it is clear to me you misread the problem. The problem did not ask you to differentiate the function $f'(f^{-1}(x))$. The problem asked you to find (a formula for) $f'(f^{-1}(x))$.

You are right that the former problem requires the chain rule (and Sonnhard's answer shows you the answer to that question). But that is not the question you were asked, assuming the solution in the book is right.

To answer the question you were given, you just need to evaluate the function $f'(x)=\cos x$ at the inverse of $f$. These are exactly the steps your book follows. (One could go a step further and write $\cos(\arcsin x)$ as an algebraic function, $\sqrt{1-x^2}$, but that is another matter.)

Answer 2

it is just the chain rule $$f''(f^{-1}(x))\cdot (f^{-1}(x))'$$