Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 869

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

\[f^{'}(x) =\]

\[= 2 \bullet \left( x^{4} \right)^{'} - \left( x^{3} \right)^{'} + (3x + 4)^{'} =\]

\[= 2 \bullet 4x^{3} - 3x^{2} + 3 =\]

\[= 8x^{3} - 3x^{2} + 3\]

\[2)\ f(x) = - x^{5} + 2x^{3} - 3x^{2} - 1\]

\[f^{'}(x) =\]

\[= - \left( x^{5} \right)^{'} + 2 \bullet \left( x^{3} \right)^{'} - 3 \bullet \left( x^{2} \right)^{'} - (1)^{'} =\]

\[= - 5x^{4} + 2 \bullet 3x^{2} - 3 \bullet 2x - 0 =\]

\[= - 5x^{4} + 6x^{2} - 6x\]

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

\[f^{'}(x) = 6 \bullet \left( x^{\frac{1}{3}} \right)^{'} + \left( x^{- 2} \right)^{'} =\]

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

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

\[4)\ f(x) = \frac{2}{x^{3}} - 8\sqrt[4]{x}\]

\[f^{'}(x) = 2 \bullet \left( x^{- 3} \right)^{'} - 8 \bullet \left( x^{\frac{1}{4}} \right)^{'} =\]

\[= 2 \bullet ( - 3) \bullet x^{- 4} - 8 \bullet \frac{1}{4} \bullet x^{- \frac{3}{4}} =\]

\[= - \frac{6}{x^{4}} - \frac{2}{\sqrt[4]{x^{3}}}\]

\[5)\ f(x) = (2x + 3)^{8}\]

\[f^{'}(x) = {(2x + 3)^{8}}^{'} =\]

\[= 8 \bullet 2 \bullet (2x + 3)^{7} =\]

\[= 16(2x + 3)^{7}\]

\[6)\ f(x) = (4 - 3x)^{7}\]

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

\[= 7 \bullet ( - 3) \bullet (4 - 3x)^{6} =\]

\[= - 21(4 - 3x)^{6}\]

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

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

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

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

\[8)\ f(x) = \frac{1}{\sqrt{1 - 4x}}\]

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

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

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