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Derivatives of Inverse Functions

  1. Jan 18, 2009 #1
    1. The problem statement, all variables and given/known data
    f(x) = x3+ x. Note that f(2) = 10. Find (f-1)`(10).


    2. Relevant equations



    3. 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-1)`

    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!
     
  2. jcsd
  3. Jan 18, 2009 #2
    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].
     
    Last edited: Jan 18, 2009
  4. Jan 18, 2009 #3
    Okay, and we want x to be 10.

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

    Okay, I think that makes sense now. Thank you!
     
  5. Jan 19, 2009 #4

    HallsofIvy

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    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.
     
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