ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 805

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

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

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

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

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

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

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

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

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

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

\[4)\ f(x) = 3\sqrt[6]{x} + 7\sqrt[14]{x}\]

\[f^{'}(x) = 3 \bullet \left( x^{\frac{1}{6}} \right)^{'} + 7 \bullet \left( x^{\frac{1}{14}} \right)^{'} =\]

\[= 3 \bullet \frac{1}{6} \bullet x^{- \frac{5}{6}} + 7 \bullet \frac{1}{14} \bullet x^{- \frac{13}{14}} =\]

\[= \frac{1}{2\sqrt[6]{x^{5}}} + \frac{1}{2\sqrt[14]{x^{13}}} =\]

\[= \frac{1}{2\left( \sqrt[6]{x^{5}} + \sqrt[14]{x^{13}} \right)}.\]