Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 4 down vote accepted

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.

share|cite|improve this answer
Thanks. I assumed I should use that theorem, but I was not sure. That theorem has our lecturer been going through. Thank you! – Curtain Oct 4 '12 at 18:03
@JulianAssange You should edit your question, so that this will actually be an answer. – David Mitra Oct 4 '12 at 18:09
@DavidMitra: Yeah, of course. :) – Curtain Oct 4 '12 at 18:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.