Posted by Anonymous on Saturday, November 17, 2012 at 12:10am.
f' = 6/√(1-x^8) * 4x^3 = 24x^3/√(1+x^8)
recall that f=6arcsin(x^4), so
sin(f/6) = x^4
1/6 cos(f/6) f' = 4x^3
f' = 24x^3/cos(f/6)
but cos(f/6) = √(1-sin^2(f/6)) = √(1-x^8)
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