Question

Find inverse derivative of f(x) at point a. when f(x) = tan x when a = sqrt 3 /3

Answer 1

Find
\[f^{-1}(x)\]
when \[a = \frac{ \sqrt 3 }{ 3 }\]
\[f(x) = \tan x\]

Answer 2

i'm really confused on this question.

Answer 3

\[y=\tan^{-1}(x)\]

Answer 4

i haven't learned derivative of arctan

Answer 5

yes i will prove to you the derivative ...

Answer 6

huh?
i just need to solve the question. i don't really need a proof -_-

would you take the derivative of sec x?

Answer 7

Hmm I'm not really familiar with this type of problem. I had to look it up on Wikipedia. This is the formula I'm getting for the derivative of the inverse function.\[\large \left[f^{-1}\right]'(a)=\frac{1}{f'\left(f^{-1}(a)\right)}\]
So I think what we need to do issss, let's start by finding \(f^{-1}(a)\).

Answer 8

\[\large f(x)=\tan x \qquad \rightarrow \qquad f^{-1}(x)=\arctan x\]

Answer 9

well you can use the triangle method for this or use implicit differentiation

Answer 10

@mathsmind ...what?

so then x = tan y
so x = pi/6 right?

Answer 11

Our \(a\) value is given to be \(\dfrac{\sqrt3}{3}\).\[\large f^{-1}\left(\dfrac{\sqrt3}{3}\right)=\arctan \left(\frac{\sqrt3}{3}\right)\]

Pi/6? Hmm yes that sounds right! :) good.

Answer 12

yes

Answer 13

but you need to find the derivative

Answer 14

er yeah. i kinda skipped a step.
would you take the derivative of the original equation?
so f'(x) = sec^2 x
and just plug in pi/6
and then flip it??

Answer 15

I think that brings us to this,
\[\large \left[f^{-1}\right]'\left(\frac{\sqrt3}{3}\right)=\frac{1}{f'\left(\dfrac{\pi}{6}\right)}\]

I can stop with the fancy notation if it's confusing you :) lol

Answer 16

Yes that sounds right :)

Answer 17

okay. cuz i have the work for the answers but i was confused on what they did.
thanks :)

Answer 18

cool c: