Решебник по алгебре 11 класс Никольский Параграф 4. Производная Задание 67

\[\boxed{\mathbf{67}\mathbf{.}}\]

\[\textbf{а)}\ y = (x - 2)^{20};\ \ x_{1} = 1;\ \]

\[\ x_{2} = 3\]

\[y^{'}(x) = 20 \cdot (x - 2)^{19};\]

\[y^{'}(1) = 20 \cdot (1 - 2)^{19} =\]

\[= 20 \cdot ( - 1) = - 20.\]

\[y^{'}(3) = 20 \cdot (3 - 2)^{19} =\]

\[= 20 \cdot 1 = 20.\]

\[\textbf{б)}\ y = (x + 5)^{21};\ \ x_{1} = - 6;\ \ \]

\[x_{2} = - 4\]

\[y^{'}(x) = 21 \cdot (x + 5)^{20};\]

\[y^{'}( - 6) = 21 \cdot ( - 6 + 5)^{20} =\]

\[= 21 \cdot 1 = 21;\]

\[6^{'}( - 4) = 21 \cdot ( - 4 + 5)^{20} =\]

\[= 21 \cdot 1 = 21.\]

\[\textbf{в)}\ y = (2x - 11)^{100};\ \ x_{1} = 5;\]

\[\text{\ \ }x_{2} = 6\]

\[y^{'}(x) = 2 \cdot 100 \cdot (2x - 11)^{99} =\]

\[= 200 \cdot (2x - 11)^{99};\]

\[y^{'}(5) = 200 \cdot (2 \cdot 5 - 11)^{99} =\]

\[= 200 \cdot ( - 1) = - 200;\]

\[y^{'}(6) = 200 \cdot (2 \cdot 6 - 11)^{99} =\]

\[= 200 \cdot 1 = 200.\]

\[\textbf{г)}\ y = (2x - 3)^{1001};\ \ x_{1} = 1;\ \ \]

\[x_{2} = 2\]

\[y^{'}(x) = 2 \cdot 1001 \cdot (2x - 3)^{1000} =\]

\[= 2002 \cdot (2x - 3)^{1000};\]

\[y^{'}(1) = 2002 \cdot (2 \cdot 1 - 3)^{1000} =\]

\[= 2002 \cdot 1 = - 2002;\]

\[y^{'}(2) = 2002 \cdot (2 \cdot 2 - 3)^{1000} =\]

\[= 2002 \cdot 1 = 2002.\]