Did I do this problem correct? Finding slope of a function at a specific x axis.?

f(x) = x cos²(2x)

slope of the function at x=𝜋/3

f'(x) = (x)’ cos²(2x) + (x) (cos²(2x)’)

= 1 (cos²(2x)) + x (2 cos(2x)) (-sin2x (2))

f(𝜋/3) = cos²(2(𝜋/3)) + sin4x

= cos²(2x) – x sin4x

= [(1+ cos4x) – 2x sin4x]/2

= cos4(𝜋/3) - 4(𝜋/3) sin4(𝜋/3)

= 1 + cos(180 + 60) - 4(𝜋/3) sin(180 + 60)

= 1 – cos 60 + 4(𝜋/3) sin(60)

= 1 – √(3)/2 + 4(𝜋/3) (1/2)

= 1 – √(3)/2 + 2𝜋/3.

　
Derivative of cos²(2x) = 2 cos(2x) * −sin(2x) * 2

f(x) = x cos²(2x)

f'(x) = x [2 cos(2x) * −sin(2x) * 2] + [1] * cos²(2x)
f'(x) = cos(2x) (cos(2x) − 4x sin(2x))

f'(π/3) = cos(2π/3) (cos(2π/3) − 4π/3 sin(2π/3))
f'(π/3) = −1/2 (−1/2 − 4π/3 (√3/2))
f'(π/3) = 1/4 + π/√3

The derivative of cos^2 is 2cos(cos)'

f'(x) = cos^2(2x)-4xsin(2x)cos(2x)
f'(pi/3) = 1/4 - (4pi/3)(sqrt3/2)(-1/2)=1/4 + (sqrt3)pi/3