1. ## inverse derivative

Thanks for the Help!

2. Originally Posted by qbkr21
Thanks for the Help!
umm, you use the same formula and method i told you earlier.

$\displaystyle \left( f^{-1} \right)'(5) = \frac {1}{f' \left( f^{-1}(5) \right)}$

you were given all the pieces, now plug them in where they should go

3. ## Re:

Originally Posted by Jhevon
umm, you use the same formula and method i told you earlier.

$\displaystyle \left( f^{-1} \right)'(5) = \frac {1}{f' \left( f^{-1}(5) \right)}$

you were given all the pieces, now plug them in where they should go
But one of the problems I am having with this particular question is that there isn't an f(x) function to work with. I had that exact setup on my paper, but I am wondering about how I would get f'.

Thanks,

-qbkr21

4. Originally Posted by qbkr21
But one of the problems I am having with this particular question is that there isn't an f(x) function to work with. I had that exact setup on my paper, but I am wondering about how I would get f'.

Thanks,

-qbkr21
$\displaystyle f(4) = 5 \implies f^{-1}(5) = 4$

we are also told that $\displaystyle f'(4) = f' \left( f^{-1}(5) \right) = \frac {2}{3}$

that's all the information we need for our formula, right?

5. ## Re:

Thanks!

6. Originally Posted by qbkr21
Thanks!
that's it!