ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 985

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\[1)\ f^{'}(x) = x^{4} = \frac{1}{5} \bullet 5x^{4} + 0 =\]

\[= \frac{1}{5} \bullet \left( x^{5} \right)^{'} + (C)^{'} = \left( \frac{x^{5}}{5} + C \right)^{'};\]

\[Ответ:\ \ f(x) = \frac{x^{5}}{5} + C.\]

\[2)\ f^{'}(x) = x^{3} = \frac{1}{4} \bullet 4x^{3} + 0 =\]

\[= \frac{1}{4} \bullet \left( x^{4} \right)^{'} + (C)^{'} = \left( \frac{x^{4}}{4} + C \right)^{'};\]

\[Ответ:\ \ f(x) = \frac{x^{4}}{4} + C.\]

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

\[= - \frac{1}{2} \bullet \left( - 2x^{- 3} \right) + 0 =\]

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

\[= \left( - \frac{1}{2x^{2}} + C \right)^{'};\]

\[Ответ:\ \ f(x) = - \frac{1}{2x^{2}} + C.\]

\[4)\ f^{'}(x) = x^{- \frac{1}{2}} = 2 \bullet \frac{1}{2}x^{- \frac{1}{2}} + 0 =\]

\[= 2 \bullet \left( {x^{\frac{1}{2}}}^{'} \right) + (C)^{'} = \left( 2\sqrt{x} + C \right)^{'};\]

\[Ответ:\ \ f(x) = 2\sqrt{x} + C.\]