General

There's a very simple solution to this problem.
This is the formula for derivatives of inverses:
g'(x)=1/f'(g(x))

Since f(x)=x^3+x and g(x) is the inverse of f(x), we can use this equation to figure out g'(2).

Going by the formula above, g'(2)=1/f'(g(2))

Because we know that g(x) is the inverse of f(x), g(2)=1. f(1)=2.

The derivative of f is easy to solve, it ends up being f'(x)=3x^2+1.

Finally, put everything together and you get:
g'(2)=1/f'(g(2)) = 1/f'(1) = 1/4

Final answer is 1/4

[Edit] nvm misread.

i got stuck on this question:
f(x) = x^3 + x, g(x) = f^-1(x), find g'(2)

what i did so far:
f^-1(x): x = [f(x)]^3 + f(x)

[b]g(x) = f^-1(x) = x - [f(x)]^3[/b]
I don't get this step. why did positive f(x) ^3 become negative?

If I am right, g(x)= [f(x)]^3 + [f(x)]
so derivative of this is g'(x) = 3[f(x)]^2 + 1
so plug in 2 and you get f(x)= 10 and therefore g'(2) is 301

or is it g(x) = f^-1(x) = [x - [f(x)]^3 = f(x)]
cuz thats that whole thing in the square bracket is f^-1(x) isn't it

Heh I did this a month ago in my math class. I'd help if I wasn't busy :S