Пошаговое объяснение:
1
A) f'(x) = (x⁵+4x³+2x)' = 5x⁴+12x²+2 f'(-1) = 5*(-1)⁴+12(-1)²+2 = 19
Б) f'(x) = cosx f'(п/2) = 0
В) f'(x) = 1/√x f'(2) = 1/√2 = √2/2
Г) f'(x) = (1/x²)' = -2/(x³) f'(1) = -2
2.
A) f(x) = 6x³ +x² -10x x₀= -0.5
Yk = f(x₀)+ f'(x₀)(x-x₀)
f'(x) = 18x₂ +2x -10
f'(-0.5) = -6.5
f(-0.5) = 4.5
Yk = 4.5 -6.5(x+0.5)
или
Yk = -6.5x + 1.25
Б) f(x) = (2x+1) /x = 2 +1/x x₀= -3
f'(x) = -1/x²
f'(-3) = -1/9
f(-3) = 5/3
Yk = 5/3 -1/9(x+3)
или
Yk = -x/9 +4/3
B) f(x) = tgx x₀ = π/4
f'(x) = tg²x +1
f'(π/4) = 2
f(π/4) = 1
Yk = 1+2(x-π/4)
Yk = 2x -π/2 +1
