# inverse derivative

• Sep 12th 2007, 06:27 PM
qbkr21
inverse derivative
Thanks for the Help! :)
• Sep 12th 2007, 06:45 PM
Jhevon
Quote:

Originally Posted by qbkr21
Thanks for the Help! :)

umm, you use the same formula and method i told you earlier.

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

you were given all the pieces, now plug them in where they should go
• Sep 12th 2007, 06:49 PM
qbkr21
Re:
Quote:

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

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

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
• Sep 12th 2007, 06:53 PM
Jhevon
Quote:

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

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

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

that's all the information we need for our formula, right?
• Sep 12th 2007, 07:14 PM
qbkr21
Re:
Thanks!:D
• Sep 12th 2007, 07:42 PM
Jhevon
Quote:

Originally Posted by qbkr21
Thanks!:D

that's it! (Rock)