# Differentiate an Inverse Function, Two Methods?

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 $\frac{d}{dx}\left[{f}^{-1}(x)\right] = \frac{1}{{f}^{\prime}\left({f}^{-1}(x)\right)}$.

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...

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:

• 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.

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The problem is that method 2 is very inefficient. As you already notice, isolating $y$ from the equation $x=2y+cosy$ is really hard (it might be actually impossible in some cases). – azarel Jan 27 '12 at 4:35
Yes... but for my Calc II class I'm required to know the second method. Hope they won't ask to method 2 on a function like this. Yikes!!! – Oliver Spryn Jan 27 '12 at 4:47
Some of the lines in your first solution are technically wrong, though you arrive at the right answer. In lines $4$ to the end, it looks as if you are differentiating a constant. The derivative of a constant is $0$. The simplest notational trick is to say let $g(x)=f^{-1}(x)$. Then $g'(x)=\dots$. Therefore $g'(\pi)=\dots$. The reason is that it is awkward to put a "prime" on $f^{-1}(x)$. – André Nicolas Jan 27 '12 at 6:33

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.
Yes, that would work, IF you can find the inverse, but thats a big IF. I am pretty sure that for your $f$, no "nice" inverse exists. – Ravi Jan 27 '12 at 4:53