Determine the derivative of

- cosx - cos^3x/3

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We'll note the given expression with the function f(x).

f(x) = - cosx - cos^3x/3

We'll differentiate f(x) with respect to x:

f'(x) = [- cosx - (cosx)^3/3]'

f'(x) = (- cosx)' + [- (cosx)^3/3]'

f'(x) = -(-sin x) - 3(cosx)^2*(-sin x)/3

We'll simplify and we'll get:

f'(x) = sin x + (cosx)^2*(sin x)

We'll substitute (cosx)^2 = 1 - (sin x)^2:

f'(x) = sin x + [1 - (sin x)^2]^2*(sin x)

We'll remove the brackets:

f'(x) = sin x + sin x - (sin x)^3

We'll combine like terms:

f'(x) = 2sin x - (sin x)^3

You need to find the derivative of the function using the derivative of composed function for the member `(cos^3 x)/3` , such that:

`(d(f(x)))/(dx) = - (d(cos x))/(dx) - (d(cos^3 x))/3)'`

`(d(f(x)))/(dx)= sin x + 3(cos^2 x)*(sin x)/3`

Reducing duplicate factors yields:

`(d(f(x)))/(dx) = sin x + cos^2 x*sin x`

Factpring out `sin x` yields:

`(d(f(x)))/(dx) = sin x*(1 + cos^2 x)`

**Hence, evaluating the derivative of the given trigonometric function yields `(d(f(x)))/(dx) = sin x*(1 + cos^2 x).` **

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