Find the value of the derivative of the function y = f (x) at the point x = x0 y = 1nx + 3cosx, x0 = П / 3

Let’s find the derivative of our given function: f (x) = cos ^ 2 (x / 3).

Using the basic formulas and rules of differentiation:

(x ^ n) ‘= n * x ^ (n-1).

(cos (x) ‘= -sin (x).

(c) ‘= 0, where c is const.

(c * u) ’= c * u’, where c is const.

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

Thus, the derivative of our given function will be as follows:

f (x) ‘= (cos ^ 2 (x / 3))’ = (x / 3) ‘* (cos (x / 3))’ * (cos ^ 2 (x / 3)) ‘= (1 / 3) * (-sin (x / 3)) * 2 * (cos (x / 3)) = (2/3) * (-sin (x / 3)) * (cos (x / 3)).

Answer: The derivative of our given function will be equal to f (x) ‘= (2/3) * (-sin (x / 3)) * (cos (x / 3)).