Найдите производную функции y=sin2x*cos2x в точке x=

Y ' = (sin2x) ' cos2x + (cos2x) ' sin2x =

= 2 cos2x cos2x - 2sin2x sin2x =

= 2 cos^2 (2x) - 2sin^2 (2x)

x = pi/6

2 cos^2 (2*pi/6) - 2sin^2 (2*pi/6) =

= 2 cos^2 (pi/3) - 2sin^2 (pi/3) =

= 2 * (1/2) ^2 - 2 * (√3/2) ^2 =

= 2*1/4 - 2*3/4 =

= 1/2 - 3/2 =

= - 2/2 = - 1