f(x) = (cos^2 x) * ln ( x^2)

We need to determine the first derivative ( f'(x)).

We notice that f(x) is a product of two functions. Then, we will use the product rule to solve.

Let f(x) = u * v such that:

u = ( cos^2 x) ==> u' = -2*cosx*sinx = -sin2x

v= ln x^2 ==> v' = 2x*1/x = 2

Then, we know that:

f'(x) = u'*v + u*v'

= (cos^2 x)* 2 + (-sin2x( ln x^2)

= 2cos^2 x - sin2x * ln x^2

**==> f'(x) = 2cos^ 2x - 2lnx*sin2x**

Since we have to determine the first derivative of a composed function, we'll apply the chain rule and also the product rule.

f'(x) = [(cos x)^2]'* ln (x^2) + (cos x)^2* [ln (x^2)]'

We'll apply the chain rule for the terms:

[(cos x)^2]' = 2 cos x*(cos x)'

[(cos x)^2]' = - 2cos x*sin x

[ln (x^2)]' = (x^2)'/x^2

[ln (x^2)]' = 2x/x^2

We'll simplify and we'll get:

[ln (x^2)]' = 2/x

The result of sifferentiating the given function is:

f'(x) = - 2cos x*sin x*ln (x^2) + 2(cos x)^2/x

**f'(x) = -sin 2x*ln (x^2) + 2(cos x)^2/x**