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We have $$ f(x) = \tan(x)-x $$ $$ f: \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \to \Bbb{R}\ $$

Obviously, $$ f'(x) = \tan^{2}(x) $$

We don't know how the inverse function looks like, let's call the inverse function g(x), what we only know is that:

$$ g'(x) = \frac{1}{f'(g(x))} $$

I did this, and I got $$ g'(x) = \frac{1}{\tan^{2}(g(x))} $$

But our task is to come to $$ g'(x) = (g(x)+x)^{-2} $$

And I just don't know how to get there. Can anyone help me please. I tried my best, what I think of is, that you can rewrite $$ \tan(x) $$ as $$\tan(x) = (\tan(x)-x)+x $$

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  • $\begingroup$ Change your misleading title. $\endgroup$
    – user65203
    Jan 8, 2017 at 17:51
  • $\begingroup$ @YvesDaoust I did it, sorry. $\endgroup$
    – Blnpwr
    Jan 8, 2017 at 17:54

1 Answer 1

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You are on the right path, the only remaining thing to do is to rewrite your result as follows: $$g'(x) = \frac{1}{\tan^2(g(x))} = (\tan(g(x))-g(x)+g(x))^{-2} = (f(g(x))+g(x))^{-2} = (x+g(x))^{-2}.$$

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    $\begingroup$ I thank you very much. Obviously, someone voted down your answer, I voted up, therefore you have now 0 votes. Don't get it wrong, please. $\endgroup$
    – Blnpwr
    Jan 8, 2017 at 17:50

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