Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 167

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

\[= \left( x^{2} - 6x + 9 \right)x^{3} =\]

\[= x^{5} - 6x^{4} + 9x^{3};\]

\[f^{'}(x) = 5x^{4} - 6 \bullet 4x^{3} + 9 \bullet 3x^{2} =\]

\[= 5x^{4} - 24x^{3} + 27x^{2}.\]

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

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

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

\[= 5x^{4} - 8x^{3} + 3x^{2} - 4x.\]

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

\[= x^{4} + 3x^{3};\]

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

\[= 4x^{3} + 9x^{2}.\]

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

\[= 3x^{3} - 12x^{2};\]

\[f^{'}(x) = 3 \bullet 3x^{2} - 12 \bullet 2x =\]

\[= 9x^{2} - 24x.\]