Найти производную 2sin3x-3cosx/sin2x.

(2 * sin(3 * x) - 3 * cos(x) / sin(2 * x))' = (2 * sin(3 * x))' - (3 * cos(x) / sin(2 * x))'
= 2 * (sin(3 * x))' - 3 * (cos(x) / sin(2 * x))' = 2 * cos(3 * x) * (3 * x)' - 3 *
((cos(x))' * sin(2 * x) - cos(x) * (sin(2 * x))') / (sin(2 * x) ^ 2) = 2 * cos(3 * x) * 3 -
3 * (- sin(x) * sin(2 * x) - cos(x) * cos(2 * x) * (2 * x)') / (sin(2 * x) ^ 2) = 2 *
cos(3 * x) * 3 - 3 * (- sin(x) * sin(2 * x) - cos(x) * cos(2 * x) * 2) / (sin(2 * x) ^ 2).