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Calculus (Derivatives of Inverse Functions)

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Suppose f(x) = sin(pi cos(x)). On any interval where the inverse function y = f^-1(x) exists, the derivative of f^-1(x) with respect to x is:
I've come as far as y = arccos ((arcsin(x))/pi), but I am not certain this is right.

  • Calculus (Derivatives of Inverse Functions) - ,

    your expression for y is right. so,

    Let u = 1/pi arcsinx
    (arccos u)' = -1/sqrt(1-u^2) u'
    = -1/sqrt(1-u^2) * 1/pi * 1/sqrt(1-x^2)
    = - 1/sqrt(1- 1/pi^2 (arcsinx)^2) * 1/pi * 1/sqrt(1-x^2)
    = - 1/[sqrt(pi^2 - (arcsinx)^2)*sqrt(1-x^2)]

    yummm! gotta love it!

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