# Thread: Derivitives of Inverse Functions

1. ## Derivitives of Inverse Functions

Hello out there.

I am fighting my way through Calc, hoping that some of the great help I've seen on here will save me gas going back and forth to the math help lab at school.

I need to find:
(f-1)’(a) for the function f and real number a

f(x)= x^3 - (4/x) with a being 6.

I'm obviously having an issue with the concept as it shouldnt be a hard problem, but I don't know how (other than guess/check) how to determine that "2" is the number I need to put into the inverse in order to get the answer, (1/13)

Thanks!

2. $(f^{-1})'(y_0)=\frac{1}{f'(x_0)}$, where $f(x_0)=y_0$

In this case $y_0=6$. To find $x_0$, we have to solve the equation $f(x)=6$

$x^3-\frac{4}{x}=6\Leftrightarrow x^4-6x-4=0\Rightarrow x=2$

Then $(f^{-1})'(6)=\frac{1}{f'(2)}$

3. Originally Posted by red_dog
$(f^{-1})'(y_0)=\frac{1}{f'(x_0)}$, where $f(x_0)=y_0$

In this case $y_0=6$. To find $x_0$, we have to solve the equation $f(x)=6$

$x^3-\frac{4}{x}=6\Leftrightarrow x^4-6x-4=0\Rightarrow x=2$

Then $(f^{-1})'(6)=\frac{1}{f'(2)}$
This might go back to algebra, but how did you get x=2 from $x^4-6x-4=0$?

4. such a dumb question nobody wants to answer it?