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I'm just making sure, that when I have "the inverse of the derivative of $f(x)$", that it's notated as $$(f')^{-1}(x)$$ and not the opposing "the derivative of the inverse of $f(x)$" which I think would be $$(f^{-1})'(x)$$

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  • $\begingroup$ You are right, although you should ensure your reader doesn't mix up your inverse-notation with $\frac{1}{f'}$ $\endgroup$
    – Gono
    Commented Jun 6, 2017 at 8:09
  • $\begingroup$ Hmm. I would write the former at $f'(x)^{-1}$, possibly adding more parentheses for clarity: $\bigl(f'(x)\bigr)^{-1}$. When you write $(f')^(-1)(x)$, one interpretation could be different: $(f')^{-1}$ would be the inverse of the derivative, itself a map. And then you evaluate it at $x$. $\endgroup$ Commented Jun 6, 2017 at 8:09
  • $\begingroup$ Followup to my previous comment: For example: $f(x)=x^4$, $f'(x)=4x^3$, $(f')^{-1}(y)=(y/4)^{1/3}$. Not what you intended, I think. $\endgroup$ Commented Jun 6, 2017 at 8:14

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You're correct. For example, let $f: (0,\infty)\to\mathbb R$ such that $f(x)=x^2$. Then: $$\begin{align}f'(x) &= 2x\\ f^{-1}(x) &= \sqrt x\\ (f^{-1})'(x) &= \frac{1}{2\sqrt x}\\ (f')^{-1}(x) &= \frac{1}{2x} \end{align}$$ The correct formula is $$(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$$

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  • $\begingroup$ For a possible ambiguity, see my second comment on the question. $\endgroup$ Commented Jun 6, 2017 at 8:16
  • $\begingroup$ Well wait now I feel like something's a little off or ambiguous...that looks like the inverse function theorem, but your last line...it would read as "the derivative of the inverse of f(x)..." "is equal to the reciprocal of the derivative of the inverse of f(x)" $\endgroup$
    – RayOfHope
    Commented Jun 6, 2017 at 8:19
  • $\begingroup$ The RHS would read as "he reciprocal of the derivative of $f$ evaluated at the inverse of $f$ at $x$". $\endgroup$
    – florence
    Commented Jun 6, 2017 at 8:22

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