Решебник по алгебре 11 класс Никольский Параграф 4. Производная Задание 19

\[\boxed{\mathbf{19}\mathbf{.}}\]

\[\textbf{а)}\ y = (x + 3)^{2} = x^{2} + 6x + 9;\ \]

\[\ x \in R\]

\[y^{'} = \left( x^{2} \right)^{'} + 6 \cdot x^{'} + 9^{'} = 2x +\]

\[+ 6 \cdot 1 + 0 = 2x + 6.\]

\[\textbf{б)}\ y = (x - 4)^{2} = x^{2} - 8x + 16;\ \ \ \]

\[x \in R\]

\[y^{'} = \left( x^{2} \right)^{'} - 8 \cdot x^{'} + 16^{'} = 2x -\]

\[- 8 \cdot 1 + 0 = 2x - 8.\]

\[\textbf{в)}\ y = (3x + 1)^{2} = 9x^{2} + 6x + 1;\ \ \]

\[x \in R\]

\[y^{'} = 9 \cdot \left( x^{2} \right)^{'} + 6 \cdot x^{'} + 1^{'} =\]

\[= 9 \cdot 2x + 6 \cdot 1 + 0 = 18x + 6.\]

\[\textbf{г)}\ y = (x + 1)^{3} = x^{3} + 3x^{2} +\]

\[+ 3x + 1;\ \ x \in R\]

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

\[+ 1^{'} = 3x^{2} + 3 \cdot 2x + 3 \cdot 1 +\]

\[+ 0 = 3x^{2} + 6x + 3.\]

\[\textbf{д)}\ y = (x - 2)^{3} = x^{3} - 6x^{2} +\]

\[+ 12x - 8;\ x \in R\]

\[y^{'} = \left( x^{3} \right)^{'} - 6 \cdot \left( x^{2} \right)^{'} + 12x^{'} -\]

\[- 8^{'} = 3x^{2} - 6 \cdot 2x +\]

\[+ 12 \cdot 1 - 0 =\]

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

\[\textbf{е)}\ y = (2x + 3)^{3} = 8x^{3} +\]

\[+ 36x^{2} + 54x + 27;\ \ x \in R\]

\[y^{'} = 8 \cdot \left( x^{3} \right)^{'} + 36 \cdot \left( x^{2} \right)^{'} +\]

\[+ 54x^{'} + 27^{'} =\]

\[= 8 \cdot 3x^{2} + 36 \cdot 2x + 54 \cdot 1 +\]

\[+ 0 = 24x^{2} + 72x + 54.\ \]