Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 803

\[\boxed{\mathbf{803}\mathbf{.}}\]

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

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

\[= 3 \bullet 2x - 5 = 6x - 5\]

\[2)\ f(x) = 5x^{2} + 6x - 7\]

\[f^{'}(x) = 5 \bullet \left( x^{2} \right)^{'} + (6x - 7)^{'} =\]

\[= 5 \bullet 2x + 6 = 10x + 6\]

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

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

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

\[4)\ f(x) = x^{5} - 3x^{2}\]

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

\[= 5x^{4} - 3 \bullet 2x = 5x^{4} - 6x\]

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

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

\[6)\ f(x) = - 2x^{3} + 18x\]

\[f^{'}(x) = - 2 \bullet \left( x^{3} \right)^{'} + (18x)^{'} =\]

\[= - 2 \bullet 3x^{2} + 18 = 18 - 6x^{2}\]

\[7)\ f(x) = 2x^{3} - 3x^{2} + 6x + 1\]

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

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

\[= 2 \bullet 3x^{2} - 3 \bullet 2x + 6 =\]

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

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

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

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

\[= - 3 \bullet 3x^{2} + 2 \bullet 2x - 1 =\]

\[= - 9x^{2} + 4x - 1.\]