Find the general form of the antiderivatives for the function f (x) = 3cos2x.

We need to find our given function: f (x) = 3cos (2x).

Using basic differentiation formulas and differentiation rules:

(x ^ n) ‘= n * x ^ (n-1).

(sin (x)) ‘= cos (x).

(cos (x) ‘= -sin (x).

(c * u) ’= c * u’, where c is const.

(u ± v) ‘= u’ ± v ‘.

y = f (g (x)), y ‘= f’u (u) * g’x (x), where u = g (x).

Thus, the derivative of our given function will look like this:

f (x) ‘= (3sin (2x))’ = (2x) ‘* (3s (2x))’ = 2 * 1 * x ^ 0 * (3) * (-sin (2x)) = 2 * 1 * 3 * sin (2x) = 6sin (2x).

Answer: The derivative of our given function will be equal to f (x) ‘= 6sin (2x).