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)
If f(x)=6arcsin(x4), find f'(x).
1 answer
1 answer