Решение
1. (u*v)` = u` * v + v` * u
2.
1) f(x) = 2x⁵ - 4/x²
f1(x) = 10x⁴ - 2/x³
2) (√x + 1)*x³ = (1/2√x) * x³ + 3x² * (√x + 1)
3) f(x) = (5x³ + x) / (3x³ + 5) f`(1) - ?
f`(x) = [(15x² + 1)*(3x² + 5) - (6x)*(5x³ + x)]/(3x² + 5)²
f`(1) = [(15 + 1)*(3 + 5) - 6*(5 + 1)] / (3 + 5)² =
= (16*8 - 6*6) / (8²) = 92/64 = 1,4375
3.
f(x) = (x² - 3)/(x + 2)
f`(x) = [2x*(x + 2) - (x² - 3)] / (x + 2)² = (2x² + 4x - x² + 3) / (x + 2)² =
= (x² + 4x + 3)/(x + 2)²
f`(x) = 0
(x² + 4x + 3)/(x + 2)² = 0
x² + 4x + 3 = 0
(x + 2)² ≠ 0, x ≠ - 2
x² + 4x + 3 = 0
x₁ = - 3
x₂ = - 1
4.
f(x) = 8x - x² - x³/3
f`(x) = 8 - 2x - x²
8 - 2x - x² > 0
x² + 2x - 8 > 0
x² + 2x - 8 = 0
x₁ = - 4
x₂ = 2
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-∞ - 4 2 +∞ x
x∈ (- ∞; - 4) (2; + ∞)
Ответ: x∈ (- ∞; - 4) (2; + ∞)