807 Найти f' (3) и f' (1), если: 1) f (x) = 1/x + 1/x²; 2) f (x) = √x + 1/x + 1; 3) f (x) = 3/√x - 2/x³; 4) f (x) = x² - x 3/2.

Ответ:

1. Дано: f(x) = 1/x + 1/x² = x⁻¹ + x⁻²

   Решение:

   f'(x) = -x⁻² - 2x⁻³ = -1/x² - 2/x³

   f'(3) = -1/3² - 2/3³ = -1/9 - 2/27 = -5/27

   f'(1) = -1/1² - 2/1³ = -1 - 2 = -3

   Ответ: f'(3) = -5/27, f'(1) = -3

2. Дано: f(x) = √x + 1/x + 1 = x^(1/2) + x⁻¹ + 1

   Решение:

   f'(x) = (1/2)x^(-1/2) - x⁻² = 1/(2√x) - 1/x²

   f'(3) = 1/(2√3) - 1/3² = 1/(2√3) - 1/9 = √3/6 - 1/9 = (3√3 - 2)/18

   f'(1) = 1/(2√1) - 1/1² = 1/2 - 1 = -1/2

   Ответ: f'(3) = (3√3 - 2)/18, f'(1) = -1/2

3. Дано: f(x) = 3/√x - 2/x³ = 3x^(-1/2) - 2x⁻³

   Решение:

   f'(x) = 3 * (-1/2)x^(-3/2) - 2 * (-3)x⁻⁴ = -3/(2x^(3/2)) + 6/x⁴ = -3/(2x√x) + 6/x⁴

   f'(3) = -3/(2 * 3√3) + 6/3⁴ = -1/(2√3) + 6/81 = -√3/6 + 2/27 = (-9√3 + 4)/54

   f'(1) = -3/(2 * 1√1) + 6/1⁴ = -3/2 + 6 = 9/2

   Ответ: f'(3) = (-9√3 + 4)/54, f'(1) = 9/2

4. Дано: f(x) = x² - x^(3/2)

   Решение:

   f'(x) = 2x - (3/2)x^(1/2) = 2x - (3/2)√x

   f'(3) = 2*3 - (3/2)√3 = 6 - (3√3)/2 = (12 - 3√3)/2

   f'(1) = 2*1 - (3/2)√1 = 2 - 3/2 = 1/2

   Ответ: f'(3) = (12 - 3√3)/2, f'(1) = 1/2

Ответ: