Найти y'(0), если y(x)=(x+3)cos2x
Y' (x) = (x + 3)' · cos2x + (x + 3) · (cos2x)' = = 1 · cos2x + (x + 3) · (- 2sin2x) = cos2x - 2sin2x · (x + 3) y' (0) = cos 0 - 2sin 0 · 3 = 1 - 0 = 1