# Simple Calculus Questions Having to Do With Inverses

1. Aug 22, 2012

### jsewell94

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

2. Stuff to know..
$(f^{-1}(x))' = \frac{1}{f'(f^{-1}(x))}$

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!

Last edited by a moderator: Aug 22, 2012
2. Aug 22, 2012

### micromass

Staff Emeritus
Please show an attempt or this will be deleted.

3. Aug 22, 2012

### Curious3141

What you posted here makes no sense. Use Latex to format these expressions so they're more legible.

4. Aug 22, 2012

### jsewell94

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. :)

5. Aug 22, 2012

### SammyS

Staff Emeritus
Yes that makes sense now. I see that you also fixed one or more typos.

BTW: You may be warned about an overly large image by one of the moderators.

6. Aug 22, 2012

### Curious3141

Yes, it looks good.

But can you assume $(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$ without proof?

7. Aug 22, 2012

### jsewell94

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

According to the chain rule

$(f(g(x))' = f'(g(x))g'(x)$

So, consider: $f(f^{-1}(x)) = x$

According to the chain rule,$f'(f^{-1}(x))(f^{-1})'(x) = 1$

So, $(f^{-1})'(x) = \frac{1}{f'(f^{-1})}$

8. Aug 22, 2012

### chiro

You can prove it if you want. Assuming an inverse exists, then y = f(x) which implies x = f-1(y) = f(f-1(x)) = f-1(f(x)) (by the properties of the inverse). Now can you differentiate these with respect to x using the differentiability rules to get it in the form of the above?

9. Aug 23, 2012

### Bacle2

A kind-of-obvious point for the OP: you may want to also address the issue of the