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

Решение:

1. \( f(x) = \frac{1}{x} + \frac{1}{x^2} = x^{-1} + x^{-2} \)

   Производная: \( f'(x) = -x^{-2} - 2x^{-3} = -\frac{1}{x^2} - \frac{2}{x^3} \)

   Значения:

   • \( f'(3) = -\frac{1}{3^2} - \frac{2}{3^3} = -\frac{1}{9} - \frac{2}{27} = -\frac{5}{27} \)
   • \( f'(1) = -\frac{1}{1^2} - \frac{2}{1^3} = -1 - 2 = -3 \)
2. \( f(x) = \sqrt{x+1} + 1 = (x+1)^{\frac{1}{2}} + 1 \)

   Производная: \( f'(x) = \frac{1}{2}(x+1)^{-\frac{1}{2}} = \frac{1}{2\sqrt{x+1}} \)

   Значения:

   • \( f'(3) = \frac{1}{2\sqrt{3+1}} = \frac{1}{2\sqrt{4}} = \frac{1}{4} \)
   • \( f'(1) = \frac{1}{2\sqrt{1+1}} = \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4} \)
3. \( f(x) = \frac{3}{\sqrt{x}} - \frac{2}{\sqrt{x^3}} = 3x^{-\frac{1}{2}} - 2x^{-\frac{3}{2}} \)

   Производная: \( f'(x) = -\frac{3}{2}x^{-\frac{3}{2}} + 3x^{-\frac{5}{2}} = -\frac{3}{2\sqrt{x^3}} + \frac{3}{\sqrt{x^5}} \)

   Значения:

   • \( f'(3) = -\frac{3}{2\sqrt{3^3}} + \frac{3}{\sqrt{3^5}} = -\frac{3}{2 \cdot 3\sqrt{3}} + \frac{3}{9\sqrt{3}} = -\frac{1}{2\sqrt{3}} + \frac{1}{3\sqrt{3}} = -\frac{1}{6\sqrt{3}} = -\frac{\sqrt{3}}{18} \)
   • \( f'(1) = -\frac{3}{2\sqrt{1^3}} + \frac{3}{\sqrt{1^5}} = -\frac{3}{2} + 3 = \frac{3}{2} \)
4. \( f(x) = x^{\frac{3}{2}} - x^{\frac{3}{2}} = 0 \)

   Производная: \( f'(x) = 0 \)

   Значения:

   • \( f'(3) = 0 \)
   • \( f'(1) = 0 \)

Ответ:

• 1) \( f'(3) = -\frac{5}{27}, f'(1) = -3 \)
• 2) \( f'(3) = \frac{1}{4}, f'(1) = \frac{\sqrt{2}}{4} \)
• 3) \( f'(3) = -\frac{\sqrt{3}}{18}, f'(1) = \frac{3}{2} \)
• 4) \( f'(3) = 0, f'(1) = 0 \)