# 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.

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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.

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Hmm... do I have the right idea, assuming the inverse function can be found? – Oliver Spryn Jan 27 '12 at 4:47
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