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:


*

*$f(x) = 2x + \cos{x}$

*${f}^{\prime}(x) = 2 - \sin{x}$

*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.
 A: 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.
A: 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).
A: I got a similar question:
How to calculate inverse function of a “weird” function like $y=x+cosx$

