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Find the derivative of the inverse function of: $ f(x)=\frac{1}2\sin(2x) + x $

I already know that this function is one-to-one.

What I've done:

$ y=\frac{1}2\sin(2x) + x $

$ 2y - 2x = \sin(2x) $

$ \frac{\arcsin(2y - 2x)}2 = x $

Is this a suitable way to do it, and how do I eleminate the x that is left inside arcsin?

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    @JulianAssange If you want to know more about difficult inverses, you might be interested in reading the Wiki page on Lagrange inversion (which allows you to sometimes express the inverse when it would otherwise be impossible - it isn't pretty though!)2012-10-04

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From Calculus by Varberg, Purcell, and Rigdon:

Theorem: Let $f$ be differentiable and strictly monotonic on an interval $I$. If $f'(x) \neq 0$ at a certain $x$ in $I$, then $f^{-1}$ is differentiable at the corresponding point $y = f(x)$ in the range of $f$ and

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

It is a common exercise in a calculus class to find the derivative of the inverse at the point even if it is very hard, or even impossible, to find the inverse function itself. This theorem is how you do it.

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    @DavidMitra: Yeah, of course. :)2012-10-04