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:
- $f(x) = 2x + \cos{x}$
- ${f}^{\prime}(x) = 2 – \sin{x}$
- Given: ${f}^{-1}(\pi) = \frac{\pi}{2}$
- $\frac{d}{dx}\left[{f}^{-1}(\pi)\right] = \frac{1}{{f}^{\prime}\left({f}^{-1}(\pi)\right)}$
- $\frac{d}{dx}\left[{f}^{-1}(\pi)\right] = \frac{1}{2 – \sin{\left({f}^{-1}(\pi)\right)}}$
- $\frac{d}{dx}\left[{f}^{-1}(\pi)\right] = \frac{1}{2 – \sin{\left(\frac{\pi}{2}\right)}}$
- $\frac{d}{dx}\left[{f}^{-1}(\pi)\right] = \frac{1}{2 – 1}$
- $\frac{d}{dx}\left[{f}^{-1}(\pi)\right] = 1$
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:
- $f(x) = 2x + \cos{x}$
- $y = 2x + \cos{x}$
- $x = 2y + \cos{y}$
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.
Best Answer
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.