y = cos(3x) + sqrt(2x - 1)

y' = cos(3x)' + [sqrt(2x-1)]'

cos(3x)' = (3x)' * cos(t), t = 3x

cos(3x)' = 3 * [-sin(3x)] = -3sin(3x)

[sqrt(2x-1)]' = (2x-1)' * [sqrt(t)], t = 2x - 1

[sqrt(2x-1)]' = 2 * [1/(2 * sqrt(2x-1)] = 1/sqrt(2x-1)

y' = 1/sqrt(2x-1) - 3sin(3x)

Ответ: 1/sqrt(2x-1) - 3sin(3x)