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Derivative of the inverse of sin(1/x)

  1. Jan 17, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the derivative of the compositional inverse of [itex] f(x) = sin(1/x) [/itex] restricted to (1,∞). You may use without proof that sin(x) is differentiable with derivative cos(x).

    2. Relevant equations

    [itex] (f^{-1})'(y_0) = \frac{1}{f'(f^{-1}(y_0))}[/itex]

    3. The attempt at a solution

    The compositional inverse of [itex] f(x) = sin(1/x) [/itex] is [itex] f^{-1}(y_0) = \frac{1}{arcsin(y_0)} [/itex].

    Plugging that in to the equation gives;

    [itex] (f^{-1})'(y_0) = \frac{1}{f'(\frac{1}{arcsin(y_0)})} = \frac{1}{sin'(arcsin(y_0))} = \frac{1}{sin'(1/x)} = \frac{1}{(-1/x^2)cos(1/x)} = \frac{-x^2}{cos(1/x)} [/itex]

    And by putting back in

    [itex] x = \frac{1}{arcsin(y_0)} [/itex]

    [itex] (f^{-1})'(y_0) = \frac{-(\frac{1}{arcsin(y_0)})^2}{cos(arcsin(y_0))} = - \frac{1}{\sqrt{1-y_0^2}arcsin^2(y_0)} [/itex]

    However, I'm told that the answer is;

    [itex] \frac {1}{cos(\frac{1}{arcsin(y_0)})} [/itex]

    For the life me, I can't see to get the answer given? Even wolfram alpha confirms that what I have it correct.


    Are we both correct? Because I can't see to show they are both equal to each other.
  2. jcsd
  3. Jan 17, 2012 #2


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    I get the same result you get.
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