<span>f(x)= -sin 2x - cos 2x -2cos2x+2sin2x</span>
<span>f(x)= x^3 - tg4x</span> 3x^2-4/cos^2 (4x)
0,1b+(0,3c+0,4d)
0,1*60=6-b
60+(60*1/2*0,3)+(60*1/4*0,4))=60+(9+6)=60+15=75
B18=b1×q^17
b21=b1×q^20
b18/b21= (b1×q^17)/(b1×q^20) =q^(-3)=27
q^3=1/27
q=1/3
Ответ. 1/3
а) f(x) = 4√x + x² - 2x
f'(x) = 4 * 1/(2√x) + 2x - 2 = 2/√x + 2x - 2
f'(x₀) = 2/√4 + 2 * 4 - 2 = 2/2 + 8 - 2 = 7
б) f(x) = (x³+x)(x²-1) = x⁵ - x³ + x³ - x
f'(x) = 5x⁴ - 1
f'(x₀) = 5 * (-1)⁴ - 1 = 5 * 1 - 1 = 4