873. Найти производную функции.

Решение:

1. \( (\frac{x^3+1}{x^2+1})' = \frac{3x^2(x^2+1) - (x^3+1)2x}{(x^2+1)^2} = \frac{3x^4+3x^2 - 2x^4-2x}{(x^2+1)^2} = \frac{x^4+3x^2-2x}{(x^2+1)^2} \)
2. \( (\frac{x^2}{x^3+1})' = \frac{2x(x^3+1) - x^2(3x^2)}{(x^3+1)^2} = \frac{2x^4+2x - 3x^4}{(x^3+1)^2} = \frac{2x-x^4}{(x^3+1)^2} \)
3. \( (\frac{\sin x}{x+1})' = \frac{\cos x (x+1) - \sin x}{(x+1)^2} \)
4. \( (\frac{\ln x}{1-x})' = \frac{\frac{1}{x}(1-x) - \ln x (-1)}{(1-x)^2} = \frac{\frac{1}{x}-1 + \ln x}{(1-x)^2} = \frac{1-x+x\ln x}{x(1-x)^2} \)

Ответ: 1) \( \frac{x^4+3x^2-2x}{(x^2+1)^2} \); 2) \( \frac{2x-x^4}{(x^3+1)^2} \); 3) \( \frac{\cos x (x+1) - \sin x}{(x+1)^2} \); 4) \( \frac{1-x+x\ln x}{x(1-x)^2} \).