Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 796

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

\[= (2 + 3x)^{- 2}\]

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

\[= - 6(2 + 3x)^{- 3} = - \frac{6}{(2 + 3x)^{3}}.\]

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

\[= (3 - 2x)^{- 3}\]

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

\[= 6(3 - 2x)^{- 4} = \frac{6}{(3 - 2x)^{4}}.\]

\[3)\ f(x) = \sqrt[3]{(3x - 2)^{2}} =\]

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

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

\[= 2 \bullet (3x - 2)^{- \frac{1}{3}} = \frac{2}{\sqrt[3]{3x - 2}}.\]

\[4)\ f(x) = \sqrt[7]{(3 - 14x)^{2}} =\]

\[= (3 - 14x)^{\frac{2}{7}}\]

\[f^{'}(x) = \frac{2}{7} \bullet ( - 14) \bullet (3 - 14x)^{\frac{2}{7} - 1} =\]

\[= - 4 \bullet (3 - 14x)^{- \frac{5}{7}} =\]

\[= - \frac{4}{\sqrt[7]{(3 - 14x)^{5}}}.\]

\[5)\ f(x) = \frac{1}{\sqrt[3]{3x - 7}} = (3x - 7)^{- \frac{1}{3}}\]

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

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

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

\[6)\ f(x) = \frac{1}{\sqrt[3]{(1 - 2x)^{2}}} =\]

\[= (1 - 2x)^{- \frac{2}{3}}\]

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

\[= \frac{4}{3} \bullet (1 - 2x)^{- \frac{5}{3}} = \frac{4}{3\sqrt[3]{(1 - 2x)^{5}}}.\]