Verified answer

1)

[tex]y = \frac{3x + 5}{x - 8} \\ y' = \frac{(3x + 5)'(x - 8) - (x - 8)'(3x + 5)}{(x - 8) {}^{2} } = \\ \frac{3(x - 8) - (3x + 5)}{(x - 8) {}^{2} } = \\ \frac{3x - 24 - 3x - 5}{(x - 8) {}^{2} } = - \frac{29}{(x - 8) {}^{2} } [/tex]

2)

[tex]y = \frac{2 {x}^{2} }{1 - 6x} \\ y' = \frac{(2 {x}^{2} )'(1 - 6x) - (1 - 6x) ' \times 2 {x}^{2} }{(1 - 6x) {}^{2} } = \\ \frac{4x(1 - 6x) - 2 {x}^{2} \times ( - 6) }{(1 - 6x) {}^{2} } = \\ \frac{4x - 24 {x}^{2} + 12 {x}^{2} }{(1 - 6x) {}^{2} } = \frac{4x - 12 {x}^{2} }{(1 - 6x) {}^{2} } [/tex]

3)

[tex]y = \frac{ \sin(x) }{x} \\ y' = \frac{( \sin(x) )' x - x'( \sin(x) )}{ {x}^{2} } = \\ \frac{x \cos(x) - \sin(x) }{x {}^{2} } [/tex]

4)

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