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Given the function

$ f(x) = e^x + \arctan(x) = y\;, $

what is the inverse $f^{-1}(y)=\dots\;$, and how can I find it? I’m looking for solutions including all steps and possible explanations along with each.

To give some wider context, I bumped into this problem as part of a bigger question asking me to prove that $f(x)$ is bijective, (f^{-1})'(y) exists for all $y>-\pi/2$ and to calculate (f^{-1})'(y).

I have proven that $f$ is bijective and the rest of the properties follow from the applicability of the inversion theorem for derivatives, but I have a hard time calculating (f^{-1})'(y) now because I can't find $(f^{-1})(y)$

Thanks for your help!

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    There is no elementary inverse function. Why do you need to find it? – 2012-02-25

1 Answers 1

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To find (f^{-1})'(y) you don't need the inverse function. What you need is to use the inverse function theorem: if $f$ is injective and $f(x)=y$ then (f^{-1})'(y)=\frac{1}{f'(x)}.

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    Ouch, thanks a lot! – 2012-02-25