Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 873

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

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

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

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

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

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

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

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

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

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

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

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

\[3)\ f(x) = \frac{\sin x}{x + 1}\]

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

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

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

\[4)\ f(x) = \frac{\ln x}{1 - x}\]

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

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

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

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

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