Question

I'm given that, \[f(x)= e^{1+x}-e^{1-x}\]
and that \[g(x) = f^{-1}(x)\]
I need [help!] to evaluate \[g'(1)\]

Answer 1

So the way I tried to work through it was to change f(x) to \[f(x) = e \times e^x- e \times e^{-x}\]then find the inverse of f(x) by changing x and y...but I couldn't isolate for y so I tried using implicit differentiation and everything went down the toilet.

I have a solution to the problem but wanted to know if there was an easier way to do it

Answer 2

First I think you want to solve \(x\) where\[
f(x) = 1
\]

Answer 3

Ah, why would I do that? ;o

Answer 4

Hmmm, well, we can start with the derivative I guess.

Answer 5

\[
g(f(x)) = x
\]

Answer 6

To make the notation less confusing, let \(y=f(x)\). Then \[
g(y) = x
\]We can differentiate this with implicit differentiation.

Answer 7

It gives me the wrong answer...unless the sol'n I'm given is wrong

Answer 8

differentiating implicitly

Answer 9

Well, what did you for the derivative?

Answer 10

\[y=e \times e^{x} - e \times e^{-x}\]\[y=e(e^{x}-\frac{ 1 }{ e^{x}})\]then I switched x and y so that it's the inverse--is that correct to do or is that my mistake?
But I'll continue, \[\frac{ x }{ e }=e^{y}-e^{-y}\]\[xe^{-1}=e^{y}-e^{-y}\]\[e^{-1}=e^{y} \times \frac{ dy }{ dx }-e^{-y} \times \frac{ -dy }{ dx }\]\[\frac{ dy }{ dx }=\frac{ e^{-1} }{ e^{y}+e^{-y} }\]

Answer 11

sorry it takes me so much time to make eqns ;s

Answer 12

Okay, what did you do after that?

Answer 13

Nothing, I got confused. I have no value for y, right? I don't know what to dooooo

Answer 14

Remember that \(y\) is the inverse function.

Answer 15

You have \(x=1\). And so \(y=g(1)\).

Answer 16

You asked why I said you should find \[
f(x)=1
\]

Answer 17

;o so if I solve for g(1) or the x value for which f is 1 and plug it in I'll get the answer? But the answer's insane

Answer 18

Ok?

Answer 19

Can't you at least try?

Answer 20

And how do you know what sort of answer you should get anyway?

Answer 21

Because someone showed me their solution ^^ I just don't understand it

Answer 22

I can't solve for when f(x)=1, but I did try
taking the ln of both sides isn't helping. Stuck at\[x-1=\ln(e^{2x}-1)\]

Answer 23

\[
1 = e(e^x-e^{-x})\implies e^{-1} = e^x-e^{-x} = 2\sinh(x)
\]

Answer 24

loool what's sinh

Answer 25

\[
\sinh(x) = \frac{e^x-e^{-x}}{2}
\]

Answer 26

I'm not expected to know that lol I give up on this question

Answer 27

\[
\sinh^{-1}(x) = \ln\left(x+\sqrt{x^2+1}\right)
\]

Answer 28

They never talked about \(\sinh\)? Okay, is the topic of the lesson?

Answer 29

no they haven't, and no topic it's just a question