Derivatives of Inverse Functions

[jumbogala]

Homework Statement

f(x) = x³+ x. Note that f(2) = 10. Find (f^-1)`(10).

Homework Equations

The Attempt at a Solution

Note that where I have written ` it denotes prime (as in the derivative of).

- Switch the x and y variables. x= y³ + y

- Differentiate implicitly 1= (3y² + 1)y`

- Solve for y`. y`= 1 / (3y² + 1). Since y = f^-1, then y` = (f<sup>-1</sup>)`

Therefore (f^-1)` = 1 / (3y² + 1)

Since f(2) = 10, f^-1(10)=2. But my equation is(f^-1)`, not just
f^-1! I don't know what to do after this point!

 

[Unco]

- Solve for y`. y`= 1 / (3y² + 1). Since y = f^-1, then y` = (f<sup>-1</sup>)`

Therefore (f^-1)` = 1 / (3y² + 1)

Since f(2) = 10, f^-1(10)=2. But my equation is(f^-1)`, not just
f^-1! I don't know what to do after this point!

Since you let [tex]y=f^{-1}(x)[/tex], what you actually have is [tex](f^{-1})'(x) = \frac{1}{3(f^{-1}(x))^2 + 1}[/tex].

 

[jumbogala]

Okay, and we want x to be 10.

So (f-1)`(10)= 1/13

Okay, I think that makes sense now. Thank you!

 

[HallsofIvy]

Another way: f(x)= [itex]x^3+ x[/itex] so f'(x)= [itex]3x^2+ 1[/itex] and f'(2)= 13. Since f(2)= 10, f^-1(10)= 2 and f'(10)= 1/13.