Страница 238 номер 793, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Найти \(f'(x_0)\), если: 1) \(f(x) = x^6\), \(x_0 = \frac{1}{2}\); 2) \(f(x) = x^{-2}\), \(x_0 = 3\); 3) \(f(x) = \sqrt{x}\), \(x_0 = 4\); 4) \(f(x) = \sqrt[3]{x}\), \(x_0 = 8\); 5) \(f(x) = \sqrt{5 - 4x}\), \(x_0 = 1\); 6) \(f(x) = \frac{1}{\sqrt{3x + 1}}\), \(x_0 = 1\).
1) \(f(x) = x^{6}, x_{0} = \frac{1}{2}\)
\(f'(x) = (x^{6})' = 6x^{5}\)
\(f'\left(\dfrac{1}{2}\right) = 6\left(\dfrac{1}{2}\right)^{5} = \dfrac{6}{32} = \dfrac{3}{16};\)
2) \(f(x) = x^{-2}, x_{0} = 3\)
\(f'(x) = (x^{-2})' = -2x^{-3}\)
\(f'(3) = -2 \cdot 3^{-3} = -\dfrac{2}{27};\)
3) \(f(x) = \sqrt{x}, x_{0} = 4\)
\(f'(x) = \left(x^{\frac{1}{2}}\right)' = \dfrac{1}{2}x^{-\frac{1}{2}} = \dfrac{1}{2\sqrt{x}}\)
\(f'(4) = \dfrac{1}{2\sqrt{4}} = \dfrac{1}{4};\)
4) \(f(x) = \sqrt[3]{x}, x_{0} = 8\)
\(f'(x) = \left(x^{\frac{1}{3}}\right)' = \dfrac{1}{3}x^{-\frac{2}{3}} = \dfrac{1}{3\sqrt[3]{x^{2}}}\)
\(f'(8) = \dfrac{1}{3\sqrt[3]{8^{2}}} = \dfrac{1}{3 \cdot 2^{2}} = \dfrac{1}{12};\)
5) \(f(x) = \sqrt{5 - 4x}, x_{0} = 1\)
\(f'(x) = \left((5 - 4x)^{\frac{1}{2}}\right)' = \dfrac{1}{2}(5 - 4x)^{-\frac{1}{2}} \cdot (-4) = \dfrac{-2}{\sqrt{5 - 4x}}\)
\(f'(1) = \dfrac{-2}{\sqrt{5 - 4 \cdot 1}} = -2\)
6) \(f(x) = \frac{1}{\sqrt{3x + 1}}, x_{0} = 1\)
\(f'(x) = \left((3x + 1)^{-\frac{1}{2}}\right)' = -\dfrac{1}{2}(3x + 1)^{-\frac{3}{2}} \cdot 3 = \dfrac{-3}{2\sqrt{(3x + 1)^{3}}},\)
\(f'(1) = \dfrac{-3}{2\sqrt{(3 \cdot 1 + 1)^{3}}} = \dfrac{-3}{2 \cdot 2^{3}} = -\dfrac{3}{16}.\)
