Inverse of the function $f(x) = e^x + \arctan(x) = y$

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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I didn’t change it in the edit, but I think that you mean $f^{-1}(y)=\dots\;$. – Brian M. Scott Feb 25 '12 at 11:39
What is the domain and codomain of your function? – fretty Feb 25 '12 at 11:41
The fact that Wolfram Alpha only give a numerical solution for $f(x)=0$ suggests that there will not be a simple inverse – Henry Feb 25 '12 at 11:51
Thanks Brian, changed it! – Robin Feb 25 '12 at 11:52
Added some more context, hope this helps – Robin Feb 25 '12 at 11:59

1 Answer

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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– anon Feb 25 '12 at 12:05
Ouch, thanks a lot! – Robin Feb 25 '12 at 12:14