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

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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Just because a function is 1-1 doesn't mean you can solve for the inverse. Why do you think you can do it here? Did you make up the problem yourself? – Graphth Oct 4 '12 at 17:54
@Graphth: There is an exercise where we have been told that there is an inverse. There is even a b) part where we should find the derivative at a given point, so therefore it seems logical that it should be possible to solve for x. – Curtain Oct 4 '12 at 17:55
@JulianAssange There is an inverse - it just can't be expressed in terms of elementary functions – user39572 Oct 4 '12 at 17:56
@JulienGodawatta: Okay. But is there any way I can get the derivative of the inverse at a given point? – Curtain Oct 4 '12 at 17:57
@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!) – user39572 Oct 4 '12 at 18:05

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