Question

Finding the compound function f'(f^-1(x))?

Answer 1

\[The function is f(x)=\sqrt{x^2+3}, x \epsilon [-2,-1]\]

Answer 2

and im wanting to find\[f'(f^-1(x))\]

Answer 3

Hmmmmm ok let's start with the inner function and work our way outward.

Answer 4

ok thats what i did, must of made an error

Answer 5

\[\Large f(x)=\sqrt{x^2+3}\]\[\Large f^{-1}(x)=?\]

Answer 6

i got -sqrt(x^2-3) for f^-1(x)

Answer 7

then i wanted to differentiate that

Answer 8

Mmmmmmmmm yah I think you're on the right track so far.

Answer 9

so i differentiated and got \[\frac{ -x }{ \sqrt{x^2-3} }\]

Answer 10

And that was wrong? :o

Answer 11

supposedly its not right

Answer 12

Try it without the minus sign.
If that's correct, I can try to explain why it is such :d

Answer 13

its not right, cause for \[\frac{ d }{ dx }f^-1(x)= \frac{ -x }{ \sqrt{x^2-3} }\]

Answer 14

but im wanting to find \[f'(f^-1(x))\]

Answer 15

the compound function

Answer 16

Oh oh oh i see :D

Answer 17

Oh ok ok ok ok.
So here's what we'll do...

Answer 18

We found the inner portion, good good good.
Our next step, find f'(x).
AFTERRRR we do that, we can evaluate the derivative function at f^{-1}(x)

Answer 19

are we putting f^-1(x) into the derivative of f(x)

Answer 20

yes, but only AFTERRRRRR you've taken the derivative of f(x) :)

Answer 21

ok so derivative of f(x)=x/sqrt(x^2+3)

Answer 22

So I guess we didn't need to take the derivative of the inverse :P that was silly hehe

Answer 23

kk

Answer 24

\[\Large f'(\color{orangered}{f^{-1}(x)})=\frac{\color{orangered}{f^{-1}(x)}}{\sqrt{\color{orangered}{f^{-1}(x)}^2+3}}\]

Answer 25

A bit tricky to plug all that in, hmm

Answer 26

ok and then simplify?

Answer 27

Hmm ya i guess so +_+

Answer 28

\[\Large f'(\color{orangered}{f^{-1}(x)})\quad=\quad \frac{\color{orangered}{\dfrac{-x}{\sqrt{x^2-3}}}}{\sqrt{\left(\color{orangered}{\dfrac{-x}{\sqrt{x^2-3}}}\right)^2}+3}\]

Answer 29

woops i didn't put the 3 under the root -_- oh well lol
work it out! :D

Answer 30

ok thanks i understand this now