Решебник по алгебре 11 класс Никольский Параграф 4. Производная Задание 33

\[\boxed{\mathbf{33}\mathbf{.}}\]

\[\textbf{а)}\ y = \frac{1}{x}\]

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

\[= \frac{0 \cdot x - 1 \cdot 1}{x^{2}} = - \frac{1}{x^{2}}.\]

\[\textbf{б)}\ y = \frac{1}{x^{2}}\]

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

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

\[= \frac{0 \cdot x^{2} - 1 \cdot 2x}{x^{4}} = - \frac{2x}{x^{4}} = - \frac{2}{x^{3}}.\]

\[\textbf{в)}\ y = \frac{1}{x + 1}\]

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

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

\[= \frac{0 \cdot (x + 1) - 1 \cdot 1}{(x + 1)^{2}} =\]

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

\[\textbf{г)}\ y = \frac{x + 1}{x - 1}\]

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

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

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

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

\[\textbf{д)}\ y = \frac{x}{x^{2} + 1}\]

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

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

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

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

\[\textbf{е)}\ y = \frac{4 - x^{2}}{x}\]

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

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

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

\[= \frac{- 2x^{2} - 4 + x^{2}}{x^{2}} = - \frac{x^{2} + 4}{x^{2}}.\]

\[\textbf{ж)}\ y = \frac{x^{2} + 3x}{x + 1}\]

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

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

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

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

\[= \frac{x^{2} + 2x + 3}{(x + 1)^{2}}.\]

\[\textbf{з)}\ y = \frac{x^{2} + x - 7}{x^{2} + 1}\]

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

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

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

\[= \frac{2x^{3} + x^{2} + 2x + 1 - 2x^{3} - 2x^{2} + 14x}{\left( x^{2} + 1 \right)^{2}} =\]

\[= \frac{- x^{2} + 16x + 1}{\left( x^{2} + 1 \right)^{2}}.\]

\[\textbf{и)}\ y = \frac{- x^{2} + 7x - 8}{x^{2} - 7x + 8}\]

\[y^{'}(x) = \left( \frac{- x^{2} + 7x - 8}{x^{2} - 7x + 5} \right)^{'} =\]

\[= \frac{6x - 21}{\left( x^{2} - 7x + 5 \right)^{2}}.\]