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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.

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    You are not asked to find the derivative of the function $f'\circ f^{-1}$. You are asked to *determine* that function, given $f$ and $f^{-1}$. – 2017-02-28
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    You also are not "told that $f'(f^{-1}(x))$". Indeed that clause makes no grammatical sense: it is like saying "I am told that 4." You are told to *find* $f'(f^{-1}(x))$, or something like that. Pay closer attention to the English, not just the symbols, which do indeed involve a composition of functions. – 2017-02-28
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    @symplectomorphic thanks for the response. Can you please elaborate on your first answer? I am not entirely understanding what you're saying. – 2017-02-28

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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.)

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    What do you mean when you say, "find (a formula for) $f'(f^{-1}(x))"? I'm really struggling to understand because this seems like a simple derivatives problem. – 2017-02-28
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    @The Pointer: please copy and paste here **verbatim** exactly what the problem says to do. Or take a photo. If you do so I will be able to point out what you are failing to read. – 2017-02-28
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    For some further intuition: if I ask you for $f'(g(x))$ where $f(x)=\sin x$ and $g(x)=x^2$, do you understand that I'm only asking you to *find* the composite $f'\circ g$ rather than *take the derivative of* the composite $f'\circ g$? In this case $f'(x)=\cos x$, so $f'(g(x))$ is just $\cos(x^2)$. You are being asked to *calculate* a composite function, not *differentiate* one. – 2017-02-28
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    The derivative of $f'(g(x))$ is written $(f'(g(x))'$ and by the chain rule becomes $f''(g(x))g'(x)$. That isn't what you want. You want just $f'(g(x))$, not $(f'(g(x))'$. – 2017-02-28
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    It is a proof for $\arcsin'(x) = \dfrac{1}{\sqrt{1 - x^2}}$. It gives $\arcsin'(x) = f^{-1}'(x)$ $= \dfrac{1}{f'(f^{-1}(x))}$ $= \dfrac{1}{\sin'(arcsin(x))}$ $= \dfrac{1}{\cos(arcsin(x))}$ and so on ... – 2017-02-28
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    I apologise, but I can't find the MathJax error for the above comment. – 2017-02-28
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    @The Pointer: It doesn't matter. Read my comments. You are making an extremely elementary error. – 2017-02-28
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    Your additional comments have clarified my misunderstanding. Thank you very much for the assistance; I appreciate it. – 2017-02-28
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it is just the chain rule $$f''(f^{-1}(x))\cdot (f^{-1}(x))'$$