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February 9, 2016
Posted by **Anonymous ** on Saturday, November 17, 2012 at 12:10am.

- Calculus -
**Steve**, Saturday, November 17, 2012 at 12:43amf' = 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)