Suppose the $\displaystyle f$ has an inverse and $\displaystyle f(2)=-4$, $\displaystyle f'(2)=\frac{3}{4}$. If $\displaystyle g=\frac{1}{f^-1}$, what is $\displaystyle g'(-4)$?

I have
$\displaystyle f^-1(-4)=2, f(2)=(-4)$
$\displaystyle g(-4)=\frac{1}{2}$
$\displaystyle g'(-4)=\frac{1}{f'(\frac{1}{2})}$

Thanks!

2. Hi

$\displaystyle g=\frac{1}{f^{-1}} = \frac{1}{u}$ where $\displaystyle u = f^{-1}$

$\displaystyle g'=-\frac{u'}{u^2}$

$\displaystyle u' = \frac{1}{f'of^{-1}}$

Therefore

$\displaystyle g'=-\frac{1}{f'of^{-1} \times \left(f^{-1}\right)^2}$

3. ## question on 3rd line

Originally Posted by running-gag
Hi

$\displaystyle g=\frac{1}{f^{-1}} = \frac{1}{u}$ where $\displaystyle u = f^{-1}$

$\displaystyle g'=-\frac{u'}{u^2}$

$\displaystyle u' = \frac{1}{f'of^{-1}}$

Therefore

$\displaystyle g'=-\frac{1}{f'of^{-1} \times \left(f^{-1}\right)^2}$
Can you explain how you got the 3rd line...
$\displaystyle u' = \frac{1}{f'of^{-1}}$?

I think that will clear things up for me.

Thanks!

4. Originally Posted by yvonnehr
Can you explain how you got the 3rd line...
$\displaystyle u' = \frac{1}{f'of^{-1}}$?

I think that will clear things up for me.

Thanks!
Well the derivative of $\displaystyle f^{-1}$ is $\displaystyle \frac{1}{f'of^{-1}}$

Haven't you already seen this relation ?

5. ## thank!

No I don't remember seeing that. I'm going to look for it in my book and see what else I missed in that section.

Now I can try to make sense of your entire response. Thanks again.

6. It comes from the formula $\displaystyle (fog)' = f'og \times g'$

applied to $\displaystyle (fof^{-1})' = f'of^{-1} \times \left(f^{-1}\right)'$

Due to the fact that $\displaystyle fof^{-1} = Id$, $\displaystyle (fof^{-1})' = 1$

Therefore $\displaystyle f'of^{-1} \times \left(f^{-1}\right)' = 1$

which leads to $\displaystyle \left(f^{-1}\right)' = \frac{1}{f'of^{-1}}$