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I learn about the inverse function theorem which says the derivative of the inverse function is the reciprocal of the derivative of the original function, and I wonder if it is possible that I can find the inverse function of the original function by applying this theorem? e.g. If I have $$ f(x) = \frac{2^x}{2^x+1} $$ then I can find $$f^{-1}(x)$$ by $$f^{-1}(x)=\int\frac{dx}{f'(x)}.$$

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  • $\begingroup$ Please, clarify what you are asking for. My answer might be wrong because you claim to use inverse function theorem which is used for the inverse of the derivative... $\endgroup$ – Edu Oct 10 '16 at 12:01
  • $\begingroup$ No, see my answer. Your argument would yield $(x^2)^{-1}=\int\frac{dx}{2x}=\frac12\log x$, which is wrong. $\endgroup$ – Yves Daoust Oct 10 '16 at 12:20
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The inverse function theorem says $$(f^{-1}(x))'=\frac1{f'(f^{-1}(x))}$$ which can be reworked by integration, $$f^{-1}(x)+C=\int\frac{dx}{f'(f^{-1}(x))}.$$ Unfortunately, $f^{-1}$ appears in both members and this is an integral equation, usually much more difficult than the function inversion itself.

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It's not likely to be helpful in a way that you're looking for. Take a simpler example, $f(x)=\sin(x)$. Then the inverse function theorem says that, there exists a $g_{x_0}(y)$ an inverse of $f(x)$ near $x_0$, and $g_{x_0}'(f(x))=\frac{1}{f'(x)}=\frac{1}{\cos(x)}$. If this were expressed in terms of $f(x)$, we could now integrate to find $g(y)$, but it's not. The inverse function theorem tells you the inverse as a function of $x$, which is probably not enough for you. In order to start integrating, we already have to answer the question: "okay, for a given $y$ value, what value of $x$ do I use to get the inverse from the inverse function theorem?" In other words, you already need to know the inverse.

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For functions of a single variable like yours, the Inverse function theorem states that

If $f$ is a continuously differentiable function with nonzero derivative at a point $x_0$, then $f$ is invertible in a neighborhood of $x_0$, the inverse is continuously differentiable, and is given by $$(f^{-1})'(f(x_0))={1\over f'(x_0)}.$$

Since your function $f$ fulfills the hypothesis at every point in $\mathbb{R}$, yes, you can apply this theorem to find the derivative of the inverse.

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  • $\begingroup$ The question is about finding the inverse function, not its derivative. $\endgroup$ – Yves Daoust Oct 10 '16 at 11:57
  • $\begingroup$ @Yves I just realized that OP is asking for different thigns, yes, because it says inverse function in the title and inverse function theorem in the body. I focused on the body and thought the request was about the inverse of the derivative. $\endgroup$ – Edu Oct 10 '16 at 11:59

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