Найти f'(0) и f'(2), если: 1) f(x) = x^2 - 2x + 1... Найти f'(3) и f'(1), если: 1) f(x) = 1/x + 1/x^2...

### №806 **Ответ:** 1) $f'(0)=-2, f'(2)=2$; 2) $f'(0)=-2, f'(2)=10$; 3) $f'(0)=0, f'(2)=-8$; 4) $f'(0)=1, f'(2)=5$. **Решение:** Используем правила дифференцирования: $(x^n)' = n x^{n-1}$, $(C)' = 0$, $(u ± v)' = u' ± v'$. 1) $f(x) = x^2 - 2x + 1$ $f'(x) = 2x - 2$ $f'(0) = 2 \cdot 0 - 2 = -2$ $f'(2) = 2 \cdot 2 - 2 = 2$ 2) $f(x) = x^3 - 2x$ $f'(x) = 3x^2 - 2$ $f'(0) = 3 \cdot 0^2 - 2 = -2$ $f'(2) = 3 \cdot 2^2 - 2 = 12 - 2 = 10$ 3) $f(x) = -x^3 + x^2$ $f'(x) = -3x^2 + 2x$ $f'(0) = -3 \cdot 0^2 + 2 \cdot 0 = 0$ $f'(2) = -3 \cdot 2^2 + 2 \cdot 2 = -12 + 4 = -8$ 4) $f(x) = x^2 + x + 1$ $f'(x) = 2x + 1$ $f'(0) = 2 \cdot 0 + 1 = 1$ $f'(2) = 2 \cdot 2 + 1 = 5$ --- ### №807 **Ответ:** 1) $f'(3)=-\frac{5}{27}, f'(1)=-3$; 2) $f'(3)=\frac{\sqrt{3}}{6}-\frac{1}{9}, f'(1)=-0,5$; 3) $f'(3)=\frac{2}{27}-\frac{1}{2\sqrt{27}}, f'(1)=4,5$; 4) $f'(3)=\frac{3\sqrt{3}}{2}+\frac{1}{2\sqrt{27}}, f'(1)=3$. **Решение:** 1) $f(x) = x^{-1} + x^{-2}$ $f'(x) = -x^{-2} - 2x^{-3} = -\frac{1}{x^2} - \frac{2}{x^3}$ $f'(3) = -\frac{1}{9} - \frac{2}{27} = -\frac{3+2}{27} = -\frac{5}{27}$ $f'(1) = -1 - 2 = -3$ 2) $f(x) = x^{1/2} + x^{-1} + 1$ $f'(x) = \frac{1}{2\sqrt{x}} - \frac{1}{x^2}$ $f'(3) = \frac{1}{2\sqrt{3}} - \frac{1}{9} = \frac{\sqrt{3}}{6} - \frac{1}{9}$ $f'(1) = \frac{1}{2} - 1 = -0,5$ 3) $f(x) = 3x^{-1/2} - 2x^{-3}$ $f'(x) = 3 \cdot (-\frac{1}{2})x^{-3/2} - 2 \cdot (-3)x^{-4} = -\frac{3}{2x\sqrt{x}} + \frac{6}{x^4}$ $f'(3) = -\frac{3}{2 \cdot 3\sqrt{3}} + \frac{6}{81} = -\frac{1}{2\sqrt{3}} + \frac{2}{27}$ $f'(1) = -1,5 + 6 = 4,5$ 4) $f(x) = x^{3/2} - x^{-3/2}$ $f'(x) = \frac{3}{2}x^{1/2} - (-\frac{3}{2})x^{-5/2} = \frac{3\sqrt{x}}{2} + \frac{3}{2x^2\sqrt{x}}$ $f'(3) = \frac{3\sqrt{3}}{2} + \frac{3}{2 \cdot 9\sqrt{3}} = \frac{3\sqrt{3}}{2} + \frac{1}{6\sqrt{3}}$ $f'(1) = 1,5 + 1,5 = 3$