2. Найдите производные следующих функций: y = x+5; x-1 y = 3x-7; 2x+9 y = (x-3)2. 2x+1 y = x3+3x2, 3x-1 y = 3x² - 2x - 4 2x - 1

Для нахождения производных функций будем использовать правило дифференцирования частного:

$$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$$

1. $$y = \frac{x+5}{x-1}$$

   $$u = x+5, u' = 1$$

   $$v = x-1, v' = 1$$

   $$y' = \frac{1(x-1) - (x+5)1}{(x-1)^2} = \frac{x-1 - x - 5}{(x-1)^2} = \frac{-6}{(x-1)^2}$$

2. $$y = \frac{3x-7}{2x+9}$$

   $$u = 3x-7, u' = 3$$

   $$v = 2x+9, v' = 2$$

   $$y' = \frac{3(2x+9) - (3x-7)2}{(2x+9)^2} = \frac{6x+27 - 6x + 14}{(2x+9)^2} = \frac{41}{(2x+9)^2}$$

3. $$y = \frac{(x-3)^2}{2x+1}$$

   $$u = (x-3)^2, u' = 2(x-3)$$

   $$v = 2x+1, v' = 2$$

   $$y' = \frac{2(x-3)(2x+1) - (x-3)^2 \cdot 2}{(2x+1)^2} = \frac{2(x-3)(2x+1 - (x-3))}{(2x+1)^2} = \frac{2(x-3)(2x+1 - x + 3)}{(2x+1)^2} = \frac{2(x-3)(x+4)}{(2x+1)^2} = \frac{2(x^2 + 4x - 3x - 12)}{(2x+1)^2} = \frac{2(x^2 + x - 12)}{(2x+1)^2} = \frac{2x^2 + 2x - 24}{(2x+1)^2}$$

4. $$y = \frac{x^3+3x^2}{3x-1}$$

   $$u = x^3+3x^2, u' = 3x^2 + 6x$$

   $$v = 3x-1, v' = 3$$

   $$y' = \frac{(3x^2+6x)(3x-1) - (x^3+3x^2)3}{(3x-1)^2} = \frac{9x^3 - 3x^2 + 18x^2 - 6x - 3x^3 - 9x^2}{(3x-1)^2} = \frac{6x^3 + 6x^2 - 6x}{(3x-1)^2} = \frac{6x(x^2 + x - 1)}{(3x-1)^2}$$

5. $$y = \frac{3x^2 - 2x - 4}{2x-1}$$

   $$u = 3x^2 - 2x - 4, u' = 6x - 2$$

   $$v = 2x-1, v' = 2$$

   $$y' = \frac{(6x-2)(2x-1) - (3x^2-2x-4)2}{(2x-1)^2} = \frac{12x^2 - 6x - 4x + 2 - 6x^2 + 4x + 8}{(2x-1)^2} = \frac{6x^2 - 6x + 10}{(2x-1)^2}$$

Ответ:

1. $$\frac{-6}{(x-1)^2}$$
2. $$\frac{41}{(2x+9)^2}$$
3. $$\frac{2x^2 + 2x - 24}{(2x+1)^2}$$
4. $$\frac{6x(x^2 + x - 1)}{(3x-1)^2}$$
5. $$\frac{6x^2 - 6x + 10}{(2x-1)^2}$$