1)y=-6x^6+4x^3; y`=13x^2-36x^5
2)y=3x^2+6x+12+20/(x-2)
y`=6x+6+20*(-1)(x-2)^(-2)=6x+6-20/((x-2)^2)
Решение
2sinx*cosx - 2cos²x = 0
sinx ≠ 0, x ≠ πk, k∈Z
2cosx(sinx - cosx) = 0
1) cosx = 0
x₁ = π/2 + πn, n∈Z
2) sinx - cosx = 0
tgx - 1 = 0
tgx = 1
x₂ = π/4 + πm, m∈Z
48=18+х+(4,6+х)
48=18+х+4,6+х
х+х=18+4,6-48
2х=-25,4
х= -25,4:2
х= -12,7
1 - 10 + 4X = - X - 3
- 9 + 4X = - X - 3
4X + X = - 3 + 9
5X = 6
X = 1.2