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Show that the function $f: ({-\pi \over 2}, {\pi \over 2}) \to \mathbb R$, $f(x) = \tan (x) + 3x +1$, has an inverse function that's differentiable everywhere. Also solve $(f^{-1})'(1)$.

I showed that it's strictly increasing and differentiable and that $f'(x) = 4 + \tan^2(x)$, tried to show that on point $y=\tan(x)+3x+1$ $(f^{-1})'(y) = {1 \over 4+ \tan^2(x)}$ but I'm stuck now.

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    You're almost there. What value of $x$ corresponds to $y=f(x)=1$? You know it is unique if it exists, so take a good guess. Once you have it, plug it into your formula and you're done! – 2017-01-27
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    So I dont have to make the last part such that there's no $x$ at all, just $y$? – 2017-01-27

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If $f$ is increasing, then $f$ is injective. Thus, $f$ is bijective in $Im(f)$, so you can define $f^{-1}$.

To evaluate $(f^{-1})'(1)$, try using the fact that if $y = f(x)$, then

$$(f^{-1})'(y) = \frac{1}{f'(x)}$$

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    But first, you have to show that $f^{-1}$ is indeed differentiable – 2017-01-27