Question

Найдите производную функции y=sin2x*cos2x в точке x=$frac{pi }{6}$

Answer 1

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