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

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f(x) = cos^3 x

First we will differentiate f(x):

we know that:

f'(x) = 3cos^2 x * (cosx)'

We know that:

(cosx)' = - sinx

==> f'(x) = 3cos^2 x * -sinx

= -3sinx*cos^2x

But we know that:

cos^2 x = 1-sin^2x

==> f'(x) = -3sinx*(1-sin^2x)

= -3sinx + sin^3 x

= sin^3 x - 3sinx

**==> f'(x) = sin^3 x - 3sinx**

(x) = cos^3x. To find f'(x) in the interval [0 , 2pi].

f(x) = cos^3x.

f'(x) = (cos^3x)'

f'(x) = 3(cos^2x) (cosx)'

f'(x) = 3 cos^2x*sinx.

In the interval f'(0) = f'(pi/2) = f'(pi) = f'(3pi/2) =0

f'(pi/4) = f(3pi/4) = 1/2sqrt2

f'(7pi/4)= f'(5pi/4) = -1/2sqrt2 are some values of f'(x) in [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)**