Решение

1. $$y = \frac{3x + 5}{x - 8}$$
   $$y' = \frac{(3x + 5)'(x - 8) - (3x + 5)(x - 8)'}{(x - 8)^2} = \frac{3(x - 8) - (3x + 5) \cdot 1}{(x - 8)^2} = \frac{3x - 24 - 3x - 5}{(x - 8)^2} = \frac{-29}{(x - 8)^2}$$
2. $$y = \frac{7}{10x - 3}$$
   $$y' = \frac{(7)'(10x - 3) - 7(10x - 3)'}{(10x - 3)^2} = \frac{0 \cdot (10x - 3) - 7 \cdot 10}{(10x - 3)^2} = \frac{-70}{(10x - 3)^2}$$
3. $$y = \frac{2x^2}{1 - 6x}$$
   $$y' = \frac{(2x^2)'(1 - 6x) - 2x^2(1 - 6x)'}{(1 - 6x)^2} = \frac{4x(1 - 6x) - 2x^2 \cdot (-6)}{(1 - 6x)^2} = \frac{4x - 24x^2 + 12x^2}{(1 - 6x)^2} = \frac{4x - 12x^2}{(1 - 6x)^2} = \frac{4x(1 - 3x)}{(1 - 6x)^2}$$
4. $$y = \frac{\sin x}{x}$$
   $$y' = \frac{(\sin x)' \cdot x - \sin x \cdot (x)'}{x^2} = \frac{\cos x \cdot x - \sin x \cdot 1}{x^2} = \frac{x \cos x - \sin x}{x^2}$$
5. $$y = \frac{x^2 - 1}{x^2 + 1}$$
   $$y' = \frac{(x^2 - 1)'(x^2 + 1) - (x^2 - 1)(x^2 + 1)'}{(x^2 + 1)^2} = \frac{2x(x^2 + 1) - (x^2 - 1) \cdot 2x}{(x^2 + 1)^2} = \frac{2x^3 + 2x - 2x^3 + 2x}{(x^2 + 1)^2} = \frac{4x}{(x^2 + 1)^2}$$
6. $$y = \frac{x^2 + 6x}{x + 2}$$
   $$y' = \frac{(x^2 + 6x)'(x + 2) - (x^2 + 6x)(x + 2)'}{(x + 2)^2} = \frac{(2x + 6)(x + 2) - (x^2 + 6x) \cdot 1}{(x + 2)^2} = \frac{2x^2 + 4x + 6x + 12 - x^2 - 6x}{(x + 2)^2} = \frac{x^2 + 4x + 12}{(x + 2)^2}$$