Question

Помогите решить, пожалуйста)

Answer 1

а)

y=2sinx - 4x

y' = 2cosx - 4

2*cos(π/3) - 4 = 2*1/2-4 = 1-4 = -3

б)

y = 1/3(tgx)

y' = 1/3(1/cos²x)

1/3*(1/cos²(-π/3)) = 1/3*(1/cos²(π/3)) = 1/3*4 = 4/3 = 1(1/3)

в)

y = -3cosx + x

y' = 3sinx + 1

3*sin(-π/6) + 1 = -3*sin(π/6) + 1 = -1,5 + 1 = -0,5