Question

Помогите пожалуйста, очень надо!!

Answer 1

Ответ:

$1)(15 {x}^{ - \frac{1}{7} } + \frac{1}{8} \cos(x) + 3 {x}^{ - 2} )' = - \frac{15}{7} {x}^{ - \frac{8}{7} } - \frac{1}{8} \sin(x) - 6 {x}^{ - 3} = - \frac{15}{7x \sqrt[7]{x} } - \frac{1}{8} \sin(x) - \frac{6}{ {x}^{3} }$

$2)( 16{x}^{4} - 8x)'2 {x}^{3} + (2 {x}^{3})'(16 {x}^{4} - 8x) = \\ = (64 {x}^{3} - 8) \times 2 {x}^{3} + 6 {x}^{2} (16 {x}^{4} - 8x) = \\ = 128 {x}^{6} - 16 {x}^{3} + 96 {x}^{6} - 48 {x}^{3} = \\ = 224 {x}^{6} - 64 {x}^{3}$

$3) \frac{(5x + {x}^{10} )' \times 6 {x}^{7} - (6 {x}^{7})'(5 x{ + {x}^{10} }) }{36 {x}^{14} } = \\ \frac{(5 + 10 {x}^{9}) \times 6 {x}^{7} - 42 {x}^{6} (5x + {x}^{10} )}{36 {x}^{14} } = \\ \frac{ 6{x}^{6}(x(5 + 10 {x}^{9}) - 7(5x + {x}^{10} ) }{36 {x}^{14} } =\\ \frac{5x + 10 {x}^{10} - 35x - 7 {x}^{10} }{6 {x}^{8} } = \\ \frac{3 {x}^{10} - 30x }{6 {x}^{8} } = \\ \frac{3x( {x}^{9} - 10)}{6 {x}^{8} } = \frac{ {x}^{9} - 10}{2 {x}^{7} }$