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differentiate y = sin^(-1)x

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    i tried using chain rule, with u = sin x

    I get -cosx/sin x

    but answer is

    1/(1-x²)^1/2
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    You must not confuse sin^{-1}x as \frac{1}{\sin x}. The first means the inverse sin function not the recipricol.

    To differentiate the inverse sin function you do the following:

    little hint:
    Spoiler:
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    y=\sin^{-1}x
    \sin y = x and use implicit differentiation

    full working:
    Spoiler:
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    \cos y \frac{dy}{dx} =1
    \frac{dy}{dx}=\frac{1}{cos y}
    But you know that  \sin^2y + \cos^2 y = 1
    therefore \cos y = \sqrt{1 - \sin^2y}
    and you also know that  sin^2y = x^2
    therefore
    Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
    \frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}

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    y=sin-1x

    sin y = x
    cos y = dx/dy
    dy/dx=1/cos y

    and....

    sin y=x
    sin^2y=x^2
    1-cos^2y=x^2
    (1-x^2)^1/2= cos y sub that into 1/cos y

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Updated: October 3, 2006
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