Find the value of the derivative of the function y = sin (-x) + cos2 at x0 = π / 3.

By condition, we are given a function: f (x) = (sіn (-x)) + (cos (2x)).

We will use:

(x ^ n) ‘= n * x ^ (n-1).

(sin (x)) ‘= cos (x).

(cos (x)) ‘= -sin (x).

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

(c) ‘= 0, where c is const.

(c * u) ’= c * u’, where c is const.

(uv) ‘= u’v + uv’.

(u ± v) ‘= u’ ± v ‘.

In this way:

f (x) ‘= ((sіn (-x)) + (cos (2x)))’ = (sіn (-x)) ‘+ (cos (2x))’ = (-x) ‘* (sіn ( -x)) + (-x) * (sіn (-x)) ‘+ (2x)’ * (cos (2x)) + (2x) * (cos (2x)) ‘= -1 * (sіn (- x)) + (-x) * (cos (-x)) + 2 * (cos (2x)) + (2x) * (-sіn (2x)).

Answer: f (x) ‘= -1 * (sіn (-x)) + (-x) * (cos (-x)) + 2 * (cos (2x)) + (2x) * (-sіn (2x )).



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