Question

Здравствуйте решить задания под номерами 7.16, 1.16, 2.16

Answer 1

1.16

$y = \frac{8}{ {x}^{3} } + \frac{3}{x} - 4 \sqrt{ {x}^{3} } + 2 {x}^{7} = \\ = 8 {x}^{ - 3} + 3 {x}^{ - 1} - 4 {x}^{ \frac{3}{2} } + 2 {x}^{7}$

$y' = 8 \times ( - 3) {x}^{ - 4} - 3 {x}^{ - 2} - 4 \times \frac{3}{2} {x}^{ \frac{1}{2} } + 14 {x}^{6} = \\ = - \frac{24}{ {x}^{4} } - \frac{3}{ {x}^{2} } - 6 \sqrt{x} + 14 {x}^{6}$

2.16

$y = \sqrt[5]{ {(x - 2)}^{6} } - \frac{3}{7 {x}^{3} - {x}^{2} - 4} = \\ = {(x - 2)}^{ \frac{6}{5} } - 3 {(7 {x}^{3} - {x}^{2} - 4) }^{ - 1}$

$y' = \frac{6}{5} {(x - 2)}^{ \frac{1}{5} } \times (x - 2)' - 3 \times ( - 1) {(7 {x}^{3} - {x}^{2} - 4)}^{ - 2} \times (7 {x}^{3} - {x}^{2} - 4) '= \\ = \frac{6}{5} \sqrt[5]{x - 2} + \frac{3(21 {x}^{2} - 2x) }{ {(7 {x}^{3} - {x}^{2} - 4) }^{2} }$

7.16

$y = \frac{ {e}^{ - tg3x}}{4 {x}^{2} - 3x + 5} \\$

$y' = \frac{( {e}^{ - tg3x} )' \times (4 {x}^{2} - 3x + 5) - (4 {x}^{2} - 3x + 5)' {e}^{ - tg3x} }{ {(4 {x}^{2} - 3x + 5) }^{2} } = \\ = \frac{ {e}^{ - tg3x} \times ( - tg3x)' \times (3x)' \times (4 {x}^{2} - 3 x + 5) - (8x - 3) {e}^{ - tg3x} }{ {(4 {x}^{2} - 3x + 5)}^{2} } = \\ = \frac{ {e}^{ - tg3x}( - \frac{3(4 {x}^{2} - 3x + 5) }{ { \cos}^{2}(3x) } - (8x - 3))}{ {(4 {x}^{2} - 3x + 5) }^{2} } = \\ = - \frac{ {e}^{ - tg3x} }{ {(4 {x}^{2} - 3x + 5) }^{2} } \times ( \frac{12 {x}^{2} - 9x + 15 }{ { \cos }^{2} (3x)} + 8x - 3)$