НБ
Наталья Бакалова
a) y` = 1/tg(x) * 1/cos^2(x) = cos(x)/sin(x) * 1/cos^2(x) = 1/ (sinx * cosx) = 2/sin2x
y`` = -2 /sin^2(2x) * cos (2x) * 2 = -4 cos(2x) / sin^2(2x) или - 4ctg(2x) /sin(2x)
б) y` = 3t^2 + 8
y`` = 6t
в) y` = 5t^4 + 2
y`` = 20t^3