ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 241

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

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

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

\[= \frac{3x^{4} + 6x^{2} - 2x^{4} - 2x}{\left( x^{2} + 2 \right)^{2}} =\]

\[= \frac{x^{4} + 6x^{2} - 2x}{\left( x^{2} + 2 \right)^{2}} =\]

\[= \frac{x\left( x^{3} + 6x - 2 \right)}{\left( x^{2} + 2 \right)^{2}}.\]

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

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

\[= \frac{2x^{4} + 2x - 3x^{4}}{\left( x^{3} + 1 \right)^{2}} =\]

\[= \frac{2x - x^{4}}{\left( x^{3} + 1 \right)^{2}} = \frac{x\left( 2 - x^{3} \right)}{\left( x^{3} + 1 \right)^{2}}.\]

\[3)\ y = \frac{\sin x}{x + 1};\]

\[y^{'}(x) = \frac{\cos x \bullet (x + 1) - \sin x}{(x + 1)^{2}}.\]

\[4)\ y = \frac{\ln x}{1 - x};\]

\[y^{'}(x) = \frac{\frac{1}{x} \bullet (1 - x) - \ln x \bullet ( - 1)}{(1 - x)^{2}} =\]

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