a) y=2x^3 - 3 корень х + 2x
y`=6x²-3/2√x +2
b) y=2sin^3x - 3 tg4x - 4
y`=6sin²xcosx-12/cos²4x
а) f(x) = (x^2 - 1) / (x - 2) - 1/3 x^3, x0 = -1
f`(x)=[2x(x-2)-(x²-1)]/(x-2)² -x²=(2x²-4x-x²+1)/(x-2)² - x²=(x²-4x+1)/(x-2)² - x²
f`(-1)=(1+4+1)/(-1-2)² -(-1)²=6/9 -1=2/3-1=-1/3
b) f(x) = 4cos2x - ctg x/2, x0 = pi/3
f`(x)=-8sin2x+1/2sin²(x/2)
f`(π/3)=-8sin(2π/3)+1/2sin²(π/6)=-8*√3/2+1/(2*1/4)=4√3+2