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!

 

[micromass]

Please show an attempt or this will be deleted.

 

[Curious3141]

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. :)

 

[SammyS]

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.

Your method & answer look correct.


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

 

[Curious3141]

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

 

[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?

 

[Bacle2]

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

possibility of your denominator being zero when you answer your question.