Differentiate an Inverse Function, Two Methods?

Question

I would like to take the derivative of this inverse function at $\pi$: $f(x) = 2x + \cos{x}$, given that ${f}^{-1}(\pi) = \frac{\pi}{2}$.

I know that there are two methods of doing it. Let me demonstrate the method that I have down pat, using the fact that $\frac{d}{dx}\left[{f}^{-1}(x)\right] = \frac{1}{{f}^{\prime}\left({f}^{-1}(x)\right)}$.

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Method 1:

1. $f(x) = 2x + \cos{x}$
2. ${f}^{\prime}(x) = 2 - \sin{x}$
3. Given: ${f}^{-1}(\pi) = \frac{\pi}{2}$
4. $\frac{d}{dx}\left[{f}^{-1}(\pi)\right] = \frac{1}{{f}^{\prime}\left({f}^{-1}(\pi)\right)}$
5. $\frac{d}{dx}\left[{f}^{-1}(\pi)\right] = \frac{1}{2 - \sin{\left({f}^{-1}(\pi)\right)}}$
6. $\frac{d}{dx}\left[{f}^{-1}(\pi)\right] = \frac{1}{2 - \sin{\left(\frac{\pi}{2}\right)}}$
7. $\frac{d}{dx}\left[{f}^{-1}(\pi)\right] = \frac{1}{2 - 1}$
8. $\frac{d}{dx}\left[{f}^{-1}(\pi)\right] = 1$

This method make sense. It is this next method that I am a little sketchy on. For the most part it utilizes some algebra for inverse functions...

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Method 2:

1. $f(x) = 2x + \cos{x}$
2. $y = 2x + \cos{x}$
3. $x = 2y + \cos{y}$

The next few steps involve finding the inverse function (can it be done with a function like this?), taking the derivative of that, and plugging in $\pi$ for the answer...

My problem is that I am stuck after this point:

• Am I going about this process correctly?
• Can I find the inverse function of this crazy looking function? It is one-to-one, as shown in the graph below.

Thank you for your time.

Answer 1

It is known that an inverse function $exists$ for any one-to-one function, but in many cases it cannot be expressed in terms of elementary functions. So, your first calculation may be the best you can do without using more machinery.

Answer 2

I got a similar question:

How to calculate inverse function of a “weird” function like $y=x+cosx$