3*sin^2(x)*cos(x)/(cos(2x)) найти производнуюпомогите, срочно надо

Решите задачу:

$( \dfrac{3sin^2xcosx}{cos2x})'=3 \dfrac{(sin^2xcosx)'cos2x-(cos2x)'sin^2xcosx}{cos^22x} = \\ =3 \dfrac{((sin^2x)'cosx+(cosx)'sin^2x)cos2x+2sin2xsin^2xcosx}{cos^22x}= \\ =3 \dfrac{(2sinxcos^2x-sin^3x)cos2x+4sin^3xcos^2x}{cos^22x}= \\ =3 \dfrac{2sinxcos^2xcos2x-sin^3xcos2x+4sin^3xcos^2x}{cos^22x}= \\ =3 \dfrac{2sinxcos^2x(cos2x+2sin^2x)-sin^3xcos2x}{cos^22x}= \\ =3 \dfrac{2sinxcos^2x-sin^3xcos2x}{cos^22x}= \\= \dfrac{6sinxcos^2x-3sin^3xcos2x}{cos^22x}$