# Let $m$ be the inverse function of $h(x) = 3x + \cos(2x)$. Find $m'(\frac{3\pi}{4})$

I wasn't sure how to solve for $x$ and create the inverse function given the $\cos(2x)$ term. Whenever I tried to take $\arccos$ there was an $x$ on the other side of the equality which meant I was achieving nothing. How do I proceed with this question?

Edit: I'm still nto entirely following. When I apply the formula I get stuck with the $m(\frac{3\pi}{4})$ term. I have a note which says that by inspection, $m(\frac{3\pi}{4}) = \frac{\pi}{4}$ which I don't follow.

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Use the formula – The Chaz 2.0 Mar 23 '12 at 0:35
@edit: By definition, $m(y) = x$ if and only if $y = h(x)$. So to show that $m(3\pi/4) = \pi/4$, try calculating $h(\pi/4)$... – TMM Mar 23 '12 at 11:40

You don't need to solve for $x$ (which is a good thing, because it can't be done).
"$m$ is the inverse of $h$" tells you $3m(x)+\cos(2m(x))=x$ (why?). Now differentiate, using the chain rule. You will have to figure out what $m(3\pi/4)$ is, but you should be able to find a number $x$ that makes $h(x)=3\pi/4$.
we have the formula, $(f^{-1})^{\prime}(x)=\frac{1}{f^{\prime}(f^{-1}(x))}$, so we have for that case, $m^{\prime}(\frac{3\pi}{4})=\frac{1}{h^{\prime}(m(\frac{3\pi}{4}))}$
That's great, if you're good at remembering formulas. Also, you still have to figure out what $m(3\pi/4)$ is. – Gerry Myerson Mar 23 '12 at 4:55