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Math Help - Where am I going wrong?

  1. #1
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    Where am I going wrong?

    In my textbook i have the problem:

    f(x)=cos^-1(sin(x))

    And the answer in the back of the textbook is -1

    What i did was:

    f'(x) = -1/(sqrt(1-sin(x)^2) * cos(x)

    f'(x) = (-cos(x))/(sqrt(1-sin(x)^2))

    But i can't see where i went wrong, any ideas?
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  2. #2
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    Re: Where am I going wrong?

    \frac{1}{\sqrt{1 - \sin^2(x)}} = \frac{1}{\sqrt{\cos^2(x)}} = \frac{1}{\cos(x)}

    (we can take the positive square root, because we only define an inverse for cosine on the interval [0,pi]).

    note that:

    f(x) = cos-1(sin(x)) = cos-1(cos(π/2 - x)) = π/2 - x.

    it's pretty clear that for f(x) = π/2 - x, f'(x) = -1.
    Last edited by Deveno; December 30th 2012 at 10:33 AM.
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  3. #3
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    Re: Where am I going wrong?

    Hello, mariusg!



    Differentiate: . f(x)\:=\:\cos^{-1}(\sin x)

    And the answer in the back of the textbook is -1.

    What i did was:

    f'(x) \:=\: \frac{-1}{\sqrt{1-\sin^2x}}\cdot\cos x \:=\: \frac{-\cos x}{\sqrt{1-\sin^2x}}

    But i can't see where i went wrong . . . any ideas?

    You did nothing wrong . . . except stop too soon.

    You have: . f'(x) \;=\;\frac{-\cos x}{\sqrt{1-\sin^2x}}

    You know this, don't you?
    . . \sin^2\!x + \cos^2\!x \:=\:1 \quad\Rightarrow\quad \cos^2\!x \:=\:1 - \sin^2\!x \quad\Rightarrow\quad \cos x \:=\:\sqrt{1-\sin^2\!x}

    Got it?
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