Simple Calculus Questions Having to Do With Inverses

1. The problem statement, all variables and given/known data
Suppose [itex]f^{-1}(x)[/itex] is the inverse function of a differentiable function [itex]f[/itex] Let [itex]G(x) = \frac{1}{f^{-1}(x)}[/itex] If [itex]f(3) = 2 [/itex] and [itex]f'(3) = 1/9[/itex] , find [itex]G'(2) [/itex].

2. Stuff to know..
[itex](f^{-1}(x))' = \frac{1}{f'(f^{-1}(x))} [/itex]

3. The attempt at a solution
http://i.imgur.com/X5Q9A.jpg?1?7662
I just have no idea if I'm doing this right or not..or if I've made a stupid calculation error. Thanks for the help!

 

Okay, sorry about the sucky post. I was in a rush to get it up, and I didn't realize the image was screwed up. I also didn't take the time to format it neatly. Hope this is more readable. :)

 

Yeah, you can assume this. We went over it in class, and I managed to prove it (maybe?).

According to the chain rule

[itex](f(g(x))' = f'(g(x))g'(x) [/itex]

So, consider: [itex] f(f^{-1}(x)) = x [/itex]

According to the chain rule,[itex] f'(f^{-1}(x))(f^{-1})'(x) = 1 [/itex]

So, [itex](f^{-1})'(x) = \frac{1}{f'(f^{-1})} [/itex]