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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:

  • 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.

enter image description here

Thank you for your time.

  • 0
    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 $f^{-1}(x)$.2012-01-27

2 Answers 2

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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.

  • 2
    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.2012-01-27
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In the equation $x = 2y + \cos(y)$, $y$ is a function of $x$, so the equation may be better written as $x = 2y(x) + \cos(y(x)) = f(y(x))$ but some find this cumbersome/confusing, so the input $x$'s are often omitted. However, we now see that $x$ and $f(y(x))$ agree everywhere (in the domain of $y$), so the derivatives of $x$ and $f(y(x))$ must agree, and the derivative of the latter is found via the chain rule, resulting in $1 = f'(y(x)) y'(x) = (2 - \sin(y)) y'$ or $y' = 1/(2 - \sin(y))$. It remains to evaluate the derivative at the desired point $(x, y) = (\pi, \pi/2)$.

TL;DR: Your starting point is good, but just because you don't know what a function looks like doesn't mean you can't work with it (just as not knowing what the value of a variable is doesn't mean you can't do arithmetic with it).