Calculus (Derivatives of Inverse Functions)
posted by Mooch .
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) 
Steve
your expression for y is right. so,
Let u = 1/pi arcsinx
(arccos u)' = 1/sqrt(1u^2) u'
= 1/sqrt(1u^2) * 1/pi * 1/sqrt(1x^2)
=  1/sqrt(1 1/pi^2 (arcsinx)^2) * 1/pi * 1/sqrt(1x^2)
=  1/[sqrt(pi^2  (arcsinx)^2)*sqrt(1x^2)]
yummm! gotta love it!
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