Find the derivative of the inverse function

[ludi_srbin]

So I need to find the derivative of the inverse function. I know that f is one-to-one and is continous. Also I know that f'(x)=1+[f(x)]^2. I found the inverse writing my equation like f(x)=x+(1/3)[f(x)]^3 then I switch the variables and get that my inverse function=(1/3)x^3 - x. Then I just take the derivative and end up with x^2 - 1. Is this correct?

 

[iaskcn]

So I need to find the derivative of the inverse function. I know that f is one-to-one and is continous. Also I know that f'(x)=1+[f(x)]^2. I found the inverse writing my equation like f(x)=x+(1/3)[f(x)]^3 then I switch the variables and get that my inverse function=(1/3)x^3 - x. Then I just take the derivative and end up with x^2 - 1. Is this correct?

First, sorry for my poor english.
I don't think you are right.Because when you switch the variables in the equation you have to pay attention to the ranges of all variables.You can see in your equation f(x)=x^2-1,the range of x is(-∞,∞)，but the range of f(x) is [-1,∞).So what you do is not only switch the variables but also write the range of x at the end of your equation.:tongue2:

 

[ludi_srbin]

Hmmm...But isn't the range of f(x) (-infinity, infinity)?