Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 792

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\[1)\ f(x) = \sqrt[3]{2x + 7} = (2x + 7)^{\frac{1}{3}}\]

\[f^{'}(x) = \frac{1}{3} \bullet 2 \bullet (2x + 7)^{1 - \frac{1}{3}} =\]

\[= \frac{2}{3} \bullet (2x + 7)^{- \frac{2}{3}} = \frac{2}{3\sqrt[3]{(2x + 7)^{2}}}.\]

\[2)\ f(x) = \sqrt[4]{7 - 3x} = (7 - 3x)^{\frac{1}{4}}\]

\[f^{'}(x) = \frac{1}{4} \bullet ( - 3) \bullet (7 - 3x)^{\frac{1}{4} - 1} =\]

\[= - \frac{3}{4} \bullet (7 - 3x)^{- \frac{3}{4}} =\]

\[= - \frac{3}{4\sqrt[4]{(7 - 3x)^{3}}}.\]

\[3)\ f(x) = \sqrt[4]{3x} = (3x)^{\frac{1}{4}}\]

\[f^{'}(x) = \frac{1}{4} \bullet 3 \bullet (3x)^{\frac{1}{4} - 1} =\]

\[= \frac{3}{4} \bullet (3x)^{- \frac{3}{4}} = \frac{3}{4\sqrt[4]{(3x)^{3}}} =\]

\[= \frac{3}{4\sqrt[4]{27x^{3}}}.\]

\[4)\ f(x) = \sqrt[3]{5x} = (5x)^{\frac{1}{3}}\]

\[f^{'}(x) = \frac{1}{3} \bullet 5 \bullet (5x)^{\frac{1}{3} - 1} =\]

\[= \frac{5}{3} \bullet (5x)^{- \frac{2}{3}} = \frac{5}{3\sqrt[3]{(5x)^{2}}} =\]

\[= \frac{5}{3\sqrt[3]{25x^{2}}} = \frac{\sqrt[3]{5}}{3\sqrt[3]{x^{2}}}.\]