# Find the derivative of the inverse of this real function $f(x) = 2x + \cos(x)$

I don't know how to attack this problem. The last I've tried is using a differential equation, but I don't know how to solve it.

Let $y$ be $f^{-1}(x)$. Knowing that $x=f(y)= 2y + \cos(y)$ and derivating I obtained the following non-linear first order differential equation: $y' \cdot (2-\sin(y))=1$

I would thank you if you can help me.

Edit: I haven't said, but it is trivial to check that the function is injective, so it has an inverse, because $\forall x \in \mathbb{R}$ $f'(x) \neq0$

• There's a typo in the derivative, I think it should be $y' · (2 - \sin(y)) = 1$. You can try solving the differential equation now, it seems more easy as before. Dec 3, 2014 at 11:54
• Yes, you are right. So stupid from my part, sorry. But how could I solve the differential equation, anyway? Dec 3, 2014 at 12:01
• Perhaps the answer that you're expected to provide is that $$(f^{-1})'(x) = \frac{1}{2 - \sin f^{-1}(x)}$$. (That's what I'd be asking for if I had asked the question.) Dec 3, 2014 at 12:06
• @JohnHughes No, I want a explicit form of the funtion. Not in terms of the inverse function. Dec 3, 2014 at 12:11
• Hmm I think John Hughes is right, I think the inverse of this function cannot be written explicitly Dec 3, 2014 at 12:13

Given function $$y= 2 x + \cos x \tag{1}$$ Inverse function $$x = 2y + \cos y \tag{2}$$ Differentiating with respect to $x$ $$1 = 2 y^{'} - \sin y \, y^{'}$$ $$y^{'} = \dfrac{1}{2-\sin y}$$
Cannot be further put in terms of $x$ as (1) and (2) are transcendental.
• $x = 2y + \cos y$ ? Jun 26, 2015 at 10:42