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

2 Answers
Apr 24, 2015

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

Apr 24, 2015

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