Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 169

\[1)\ f(x) = \frac{2x - 1}{2x + 1};\]

\[f^{'}(x) = \frac{2(2x + 1) - 2(2x - 1)}{(2x + 1)^{2}};\]

\[f^{'}(1) = \frac{2(2 + 1) - 2(2 - 1)}{(2 + 1)^{2}} =\]

\[= \frac{2 \bullet 3 - 2 \bullet 1}{3^{2}} = \frac{6 - 2}{9} = \frac{4}{9}.\]

\[Ответ:\ \ \frac{4}{9}.\]

\[2)\ f(x) = \frac{x^{2} - 1}{x^{2} + 1};\]

\[f^{'}(x) = \frac{2x\left( x^{2} + 1 \right) - 2x\left( x^{2} - 1 \right)}{\left( x^{2} + 1 \right)^{2}};\]

\[f^{'}(1) = \frac{2(1 + 1) - 2(1 - 1)}{(1 + 1)^{2}} =\]

\[= \frac{2 \bullet 2 - 2 \bullet 0}{2^{2}\ } = \frac{4 - 0}{4} = 1.\]

\[Ответ:\ \ 1.\]

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

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

\[f^{'}(1) = \frac{2(5 - 4) + 4(2 - 3)}{(5 - 4)^{2}} =\]

\[= \frac{2 \bullet 1 + 4 \bullet ( - 1)}{1^{2}} = 2 - 4 = - 2.\]

\[Ответ:\ - 2.\]

\[4)\ f(x) = \frac{2x^{2}}{1 - 7x};\]

\[f^{'}(x) = \frac{2 \bullet 2x(1 - 7x) + 7 \bullet 2x^{2}}{(1 - 7x)^{2}};\]

\[f^{'}(1) = \frac{4(1 - 7) + 14 \bullet 1}{(1 - 7)^{2}} =\]

\[= \frac{4 \bullet ( - 6) + 14}{( - 6)^{2}} = \frac{14 - 24}{36} = - \frac{5}{18}.\]

\[Ответ:\ - \frac{5}{18}.\]