# Inverse Derivative Help

• Jan 24th 2009, 08:57 PM
Krooger
Inverse Derivative Help
I cannot figure out what to do!!
Given the function $\displaystyle f(x)=5 x^3+2 x+5$ Let g be the inverse function of f. i.e. $\displaystyle g(x)=f^{-1}(x).$
$\displaystyle g^{\prime}(12)=$

I can't manage to find g(x) let alone g'(x)
Thank you for any help.
• Jan 24th 2009, 09:16 PM
Jhevon
Quote:

Originally Posted by Krooger
I cannot figure out what to do!!
Given the function $\displaystyle f(x)=5 x^3+2 x+5$ Let g be the inverse function of f. i.e. $\displaystyle g(x)=f^{-1}(x).$
$\displaystyle g^{\prime}(12)=$

I can't manage to find g(x) let alone g'(x)
Thank you for any help.

note that $\displaystyle g'(x) = \frac 1{f'(g(x))}$ .....not hard to derive. your textbook should do this

now, by definition, $\displaystyle g(a) = x \implies f(x) = a$

thus, $\displaystyle g(12)$ is the x-value so that $\displaystyle f(x) = 12$

by inspection, this is 1, thus $\displaystyle g(12) = 1$

thus, $\displaystyle g'(12) = \frac 1{f'(1)}$

i leave the rest to you
• Jan 25th 2009, 02:27 AM
Prove It
Quote:

Originally Posted by Jhevon
note that $\displaystyle g'(x) = \frac 1{f'(g(x))}$ .....not hard to derive. your textbook should do this

now, by definition, $\displaystyle g(a) = x \implies f(x) = a$

thus, $\displaystyle g(12)$ is the x-value so that $\displaystyle f(x) = 12$

by inspection, this is 1, thus $\displaystyle g(12) = 1$

thus, $\displaystyle g'(12) = \frac 1{f'(1)}$

i leave the rest to you

$\displaystyle f^{-1}(x) \neq \frac{1}{f(x)}$.

Edit: Never mind, I figured out what you were doing. I worked it out using implicit differentiation, but your way works too.
• Jan 25th 2009, 03:24 AM
Moo
Quote:

Originally Posted by Prove It
$\displaystyle f^{-1}(x) \neq \frac{1}{f(x)}$.

Edit: Never mind, I figured out what you were doing. I worked it out using implicit differentiation, but your way works too.

The first formula Jhevon gives comes from implicit differentiation of $\displaystyle f(g(x))=x$

(Thinking)
• Jan 25th 2009, 01:14 PM
Jhevon
Quote:

Originally Posted by Prove It
$\displaystyle f^{-1}(x) \neq \frac{1}{f(x)}$.

that's not the formula i have :p

Quote:

Edit: Never mind, I figured out what you were doing. I worked it out using implicit differentiation, but your way works too.
yes, i worked it out that way as well...er, like Moo said
• Jan 25th 2009, 03:35 PM
Krooger
I was unaware that:
$\displaystyle g'(x) = \frac 1{f'(g(x))}$

...and to be honest upon first glace it still dosen't make sense to me haha. When I get a minute I will have to look it up. Thank You all for the help.
• Jan 25th 2009, 03:37 PM
Jhevon
Quote:

Originally Posted by Krooger
I was unaware that:
$\displaystyle g'(x) = \frac 1{f'(g(x))}$

...and to be honest upon first glace it still dosen't make sense to me haha. When I get a minute I will have to look it up. Thank You all for the help.

since $\displaystyle f(x)$ and $\displaystyle g(x)$ are inverse functions,

$\displaystyle f(g(x)) = x$

now differentiate both sides with respect to x (note you will need the chain rule to differentiate the left side) and solve for $\displaystyle g'(x)$