if f(x) = cos^3 x . Find f'(x) for the interval [0,2pi]

To calculate the first derivative of the given function,
we'll apply the chain rule:

[u(v(x))]' =
u'(v)*v'(x)*x'

We'll put u(v) = u^3 => u' =
3u^2

v(x) = cos x => v' = - sin
x

(uov)(x) = [u(v(x))] = [u(cos x)]^3 = (cos
x)^3

[u(v(x))]' = [(cos x)^3]'
(1)

[u(v(x))]' = u'(v)*v'(x)*x'
(2)

We'll put (1) =
(2)

u'(v)*v'(x)*x' = [(cos x)^3]' = -3(cos
x)^2*(sinx)

f'(x) =  -3(cos
x)^2*(sinx)