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Math Help - Inverse functions + derivative

  1. #1
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    Inverse functions + derivative

    Hi, I have a question that is troubling me:

    Given that for f: R -> R such that f(x) = 2 x + cos x, find g' (1) where g is the inverse of f.

    I've tried trying to find g but I get stuck at: x = 2y + cos y.


    If anyone could help, it would be greatly appreciated.

    Cheers.
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  2. #2
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    Quote Originally Posted by shinn View Post
    Hi, I have a question that is troubling me:

    Given that for f: R -> R such that f(x) = 2 x + cos x, find g' (1) where g is the inverse of f.

    I've tried trying to find g but I get stuck at: x = 2y + cos y.


    If anyone could help, it would be greatly appreciated.

    Cheers.
    You're on a hiding to nothing trying to first get the rule for the inverse.

    Clearly the inverse is defined as an implicit function of x by x = 2y + \cos y. Therefore, using implicit differentiation:

    1 = 2 \frac{dy}{dx} - \sin y \frac{dy}{dx}.

    So 1 = 2 g'(x) - g'(x) \sin g(x). You require the value of g'(1): 1 = 2 g'(1) - g'(1) \sin g(1).

    Note that the value of g(1) is given by the solution to 1 = 2y + \cos y. By inspection, y = 0, that is, g(1) = 0.

    Therefore ....
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