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September 3, 2014

September 3, 2014

Posted by **Mooch** on Thursday, March 15, 2012 at 11:35am.

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**, Thursday, March 15, 2012 at 2:18pmyour 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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