Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 791

\[\boxed{\mathbf{791}\mathbf{.}}\]

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

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

\[= 8(4x - 3) = 32x - 24.\]

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

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

\[= - 15(5x + 2)^{- 4} = - \frac{15}{(5x + 2)^{4}}.\]

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

\[f^{'}(x) = - 6 \bullet ( - 2) \bullet (1 - 2x)^{- 6 - 1} =\]

\[= 12(1 - 2x)^{- 7} = \frac{12}{(1 - 2x)^{7}}.\]

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

\[f^{'}(x) = 4 \bullet ( - 5) \bullet (2 - 5x)^{4 - 1} =\]

\[= - 20(2 - 5x)^{3}.\]

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

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

\[= 6 \bullet (2x)^{2} = 6 \bullet 4x^{2} = 24x^{2}.\]

\[6)\ f(x) = ( - 5x)^{4}\]

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

\[= - 20 \bullet ( - 5x)^{3} =\]

\[= - 20 \bullet - 125x^{3} = 2500x^{3}.\]