Упр.873 ГДЗ Алимов 10-11 класс (Алгебра)

1. $$\frac{x^3+1}{x^2+1}$$
2. $$\frac{x^2}{x^3+1}$$
3. $$\frac{\sin x}{x+1}$$
4. $$\frac{\ln x}{1-x}$$

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

Найдём производную по формуле производной частного:

$$f'(x)=\frac{(x^3+1)'(x^2+1)-(x^3+1)(x^2+1)’}{(x^2+1)^2}$$

$$f'(x)=\frac{3x^2(x^2+1)-(x^3+1)\cdot 2x}{(x^2+1)^2}$$

$$f'(x)=\frac{3x^4+3x^2-2x^4-2x}{(x^2+1)^2} =\frac{x^4+3x^2-2x}{(x^2+1)^2}$$

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

$$f'(x)=\frac{(x^2)'(x^3+1)-x^2(x^3+1)’}{(x^3+1)^2}$$

$$f'(x)=\frac{2x(x^3+1)-x^2\cdot 3x^2}{(x^3+1)^2}$$

$$f'(x)=\frac{2x^4+2x-3x^4}{(x^3+1)^2} =\frac{2x-x^4}{(x^3+1)^2}$$

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

$$f'(x)=\frac{(\sin x)'(x+1)-\sin x\cdot (x+1)’}{(x+1)^2}$$

$$f'(x)=\frac{\cos x\,(x+1)-\sin x}{(x+1)^2}$$

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

$$f'(x)=\frac{(\ln x)'(1-x)-\ln x\cdot (1-x)’}{(1-x)^2}$$

$$f'(x)=\frac{\frac{1}{x}(1-x)-\ln x\cdot(-1)}{(1-x)^2} =\frac{\frac{1}{x}(1-x)+\ln x}{(1-x)^2}$$

$$f'(x)=\frac{1-x+x\ln x}{x(1-x)^2}$$

Ответ:

$$1)\ f'(x)=\frac{x^4+3x^2-2x}{(x^2+1)^2},\quad 2)\ f'(x)=\frac{2x-x^4}{(x^3+1)^2},\quad 3)\ f'(x)=\frac{\cos x\,(x+1)-\sin x}{(x+1)^2},\quad 4)\ f'(x)=\frac{1-x+x\ln x}{x(1-x)^2}$$