# Simple Calculus Questions Having to Do With Inverses

jsewell94

## Homework Statement

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

## 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!

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

Homework Helper
2. Stuff to know..
(f-1)'(x) = 1/f(f-1(x))

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

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. :)

Staff Emeritus
Homework Helper
Gold Member
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. :)

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.

Homework Helper
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. :)

Yes, it looks good.

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

jsewell94
Yes, it looks good.

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

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