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.
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!