Question

Let `f(x) = x^3-1`. What is the value of `d/dx f^-1(7)`?

Answer 1

Start by recalling the formula for the derivative of inverse functions:
`d/dx f^-1(x) = 1/(f'(f^-1(x)))`

So `d/dx f^-1(7) = 1/(f'(f^-1(7)))`

To find `f^-1(7)`, we must ask ourselves: "what is x when y=7":
`7 = x^3 - 1`
`8 = x^3`
`x=2`
`f(2) = 7 <=> f^-1(7) = 2`

`d/dx f^-1(7) = 1/(f'(f^-1(7))) = 1/(f'(2)`

Now we need to find `f'(2)`:
`f'(x) = 3x^2`
`f'(2) = 3(2)^2 = 3(4) = 12`

So `d/dx f^-1(7) = 1/(f'(2)) = 1/12`

Answer 2

`1/12`

First get `f^-1 x`. Let y= `x^3` -1 . Interchange x and y and solve for y. Thus x= `y^3` -1, so that y= `(x+1)^(1/3)`. This is `f^-1 x`.

Now `d/dx f^-1 x`= `1/3 (x+1)^(-2/3)`

`d/dxf^-1 (7)`= `1/3 (8)^(-2/3)`

= `1/3(2)^(-2)`= `1/12`