How Do You Calculate the Derivative of the Inverse Function at a Given Point?

[computerex]

Homework Statement

I have been given:
h(x) is the inverse for f(x) = x^3+x
They want to know h'(2).

The Attempt at a Solution

I know that since h(x) and f(x) are inverses:
f(h(x)) = x

differentiating with respect to x gives
f'(h(x))h'(x) = 1

So h'(x) = 1/f'(h(x))

Therefore h'(2) = 1/f'(h(2))

to find h(2)
x^3+x = 2

since f(h(x)) = x
x^3+x-2 = 0

so h(2) = 1

f(x) = x^3+x
f'(x) = 3x^2+1

h'(2) = 1/f'(h(2))
h'(2) = 1/3(1)^2+1
h'(2) = 1/4

My question is: How do you find the inverse of y = x^3+x ?
You interchange x and y then solve for y: y^3+y=x But I don't know how to solve that for y. Any help would be appreciated.

 

[jgens]

We're not supposed to give complete solutions, so here's a hint:
$$y = w - \frac{1}{3w}$$

 

[computerex]

I am sorry but the above doesn't help me solve: y^3+y=x for y. If you will see above, I have already got the solution to the problem using some algebraic gymnastics, I am just curious as to how to find the inverse for y=x^3+x.

 

[jgens]

I know! You want to solve the cubic equation $y^3 + y - x = 0$ and the substitution $y = w + \frac{1}{3w}$ allows you to solve that cubic equation quite nicely.