Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 815

\[\boxed{\mathbf{815}\mathbf{.}}\]

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

\[f^{'}(x) =\]

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

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

\[= \frac{2x^{3} + 2x - 2x^{3} + 2x}{\left( x^{2} + 1 \right)^{2}} =\]

\[= \frac{4x}{\left( x^{2} + 1 \right)^{2}}\]

\[f^{'}(1) = \frac{4}{\left( 1^{2} + 1 \right)^{2}} = \frac{4}{2^{2}} = \frac{4}{4} = 1.\]

\[Ответ:\ \ 1.\]

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

\[f^{'}(x) =\]

\[= \frac{\left( 2x^{2} \right)^{'} \bullet (1 - 7x) - 2x^{2} \bullet (1 - 7x)^{'}}{(1 - 7x)^{2}} =\]

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

\[= \frac{4x - 28x^{2} + 14x^{2}}{(1 - 7x)^{2}} =\]

\[= \frac{4x - 14x^{2}}{(1 - 7x)^{2}}\]

\[f^{'}(1) = \frac{4 - 14 \bullet 1^{2}}{(1 - 7)^{2}} = \frac{4 - 14}{( - 6)^{2}} =\]

\[= - \frac{10}{36} = - \frac{5}{18}.\]

\[Ответ:\ \ - \frac{5}{18}.\]