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I was doing a long exercise when come to this point: calculate the inverse function of $y=2x+\sin x (x \in\mathbb R) $ and its derivative. I know that the derivative of an inverse function is $1/f'(x)$ but it is not enough as $x=f^{-1}(y)$. So I tried to find the inverse function but I'm completely stuck just in this point. Can somebody help me please? Thanks in advance for your help!!

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  • $\begingroup$ Do you mean $\sin x$ $\endgroup$ – Adi Dani Dec 22 '13 at 12:55
  • $\begingroup$ yes, in Italy we used to write senx $\endgroup$ – Dipok Dec 22 '13 at 12:56
  • $\begingroup$ Hi @Dipok! $$\color{blue}{\Large{\text{Welcome to Math.SE!}}}$$ Don't worry about it now but you might like to know that we use MathJax here (e.g. $\theta$ for $\theta$). $\endgroup$ – Shaun Dec 22 '13 at 12:57
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    $\begingroup$ @Shaun thank you very much :-) $\endgroup$ – Dipok Dec 22 '13 at 12:59
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    $\begingroup$ I don't think it's possible to find a closed form of the inverse of your function. $\endgroup$ – David Mitra Dec 22 '13 at 13:01
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Hint: write $x=2y+\sin(y)$, then $1=2y'+\cos(y)\cdot y'$, hence $y'=1/(2+\cos(y))$, that will be the derivative of the inverse. If you can solve that differential equation you'll have the inverse ...

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