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Math Help - Derivative of Inverse Trig Function

  1. #1
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    Derivative of Inverse Trig Function

    So the problem indicates:
    Suppose y= sin x , for x in [n/2, 3n/2] *note let n=pi
    Then we define P to be the inverse of this function.
    Ex. P(0) = n
    a. Obtain the derivative of P. The derivative is not P'(x) = 1/[(1-x^2)^(1/2)].

    So I'm not entirely sure how to go about this.
    We can write P(x) =y for x in [-1, 1].
    Then I know that on the interval [n/2, 3n/2] the derivative of the sine function will be negative.
    But I'm unsure how to approach it?
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  2. #2
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    Re: Derivative of Inverse Trig Function

    Hi, "turbozz."

    I am puzzled.

    If y = sin(x), and P is the inverse of this function, then P = x = arcsin(y).
    And the range changes: instead of being stated in terms of x, it is stated in terms of y: [1, -1].

    By definition, the derivative of the arcsin function is:

    \frac{\mathrm{d} }{\mathrm{d} y}\arcsin(y) = \frac{1}{\sqrt{(1-y^{2})}}

    Why is the derivative not this standard definition?

    Are you looking for a derivative at one particular point?
    Last edited by DavidB; November 16th 2013 at 11:18 AM.
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  3. #3
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    Re: Derivative of Inverse Trig Function

    I'm puzzled about this to. But I know it has do with fact y=P(x) is defined for x in [-1,1] and y in [n/2, 3n/2], rather than in the conventional way y=arcsinx is defined where y in [-n/2, n/2]. When I draw the graph of P(x) it does vary from the graph of y= arcsinx (for y in
    [-n/2, n/2]), so the slopes must be different => derivatives must be different. But I don't know how to find P'(x).
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  4. #4
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    Re: Derivative of Inverse Trig Function

    The difference is that while "normal" inverse sine returns a value between -\pi/2 and \pi/2, this function is defined to return a value between \pi/2 and 3\pi/2. One way to handle that is to simply "shift" the returned value. sin^{-1}(0)= 0 but, as we are told, P(0)= \pi so we can write P(x)= \pi- sin^{-1}(x). (Notice that we are subtracting sin^{-1}(x), not adding it. That is because the sine function, after x= \pi/2 goes down rather than up as it does after -\pi/4. Of course, the derivative of the constant \pi is 0 so the derivative of P(x) is the negative of the derivative of sin^{-1}(x).
    Thanks from turbozz
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  5. #5
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    Re: Derivative of Inverse Trig Function

    Thank HallsofIvy. It completely agrees with my graph of P(x).
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