ГДЗ по алгебре 11 класс Никольский Параграф 4. Производная Задание 34

\[\boxed{\mathbf{34}\mathbf{.}}\]

\[\textbf{а)}\ f(x) = \frac{5}{x^{2} + 1};\ \ x_{0} = 0;\]

\[f^{'}(x) = \left( \frac{5}{x^{2} + 1} \right)^{'} =\]

\[= \frac{5^{'} \cdot \left( x^{2} + 1 \right) - 5 \cdot \left( x^{2} + 1 \right)^{'}}{\left( x^{2} + 1 \right)^{2}} =\]

\[= \frac{0 - 5 \cdot 2x}{\left( x^{2} + 1 \right)^{2}} = - \frac{10x}{\left( x^{2} + 1 \right)^{2}};\]

\[f^{'}(0) = - \frac{10 \cdot 0}{(0 + 1)^{2}} = 0.\]

\[\textbf{б)}\ f(x) = \frac{- 2x}{x^{2} + 2};\ \ \ x_{0} = 1;\]

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

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

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

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

\[f^{'}(1) = \frac{2 \cdot 1 - 4}{(1 + 2)^{2}} = - \frac{2}{9}.\]

\[\textbf{в)}\ f(x) = \frac{x^{2} - 1}{x^{2} + 3};\ \ \ x_{0} = - 1;\]

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

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

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

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

\[= \frac{8x}{\left( x^{2} + 3 \right)^{2}};\]

\[f( - 1) = \frac{8 \cdot ( - 1)}{(1 + 3)^{2}} = - \frac{8}{16} = - \frac{1}{2}.\]

\[\textbf{г)}\ f(x) = \frac{x^{2} - 4}{x^{2} + 4};\ \ \ x_{0} = - 2;\ \]

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

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

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

\[= \frac{2x\left( x^{2} + 4 - x^{2} + 4 \right)}{\left( x^{2} + 4 \right)^{2}} = \frac{16x}{\left( x^{2} + 4 \right)^{2}};\]

\[f^{'}( - 2) = \frac{16 \cdot ( - 2)}{(4 + 4)^{2}} =\]

\[= - \frac{32}{64} = - \frac{1}{2}.\]