Question

помогите пожалуйста решить три примера даю 100 балов​

Answer 1

Ответ:

1.11

$y = 2 \sqrt{ {x}^{3} } - \frac{7}{x} + 3 {x}^{2} - \frac{2}{ {x}^{5} } = \\ = 2 {x}^{ \frac{3}{2} } - 7 {x}^{ - 1} + 3 {x}^{2} - 2 {x}^{ - 5}$

$y '= 2 \times \frac{3}{2} {x}^{ \frac{1}{2} } - 7 \times ( - 1) {x}^{ - 2} + 6x - 2 \times ( - 5) {x}^{ - 6} = \\ = 3 \sqrt{x} + \frac{7}{ {x}^{2} } + 6x + \frac{10}{ {x}^{6} }$

7.

$y = \sqrt[3]{3x + 2} ln(x) - \frac{x + 2}{ \cos(2x) }$

$y '= (( {3x + 2)}^{ \frac{1}{3} } ) ' ln(x) + ( ln(x)) ' \times \sqrt[3]{3x + 2} + \frac{(x + 2)' \times \cos(2x) - ( \cos(2x)) ' \times (x + 2) }{ { \cos }^{2} (2x)} = \\ = \frac{1}{3} {(3x + 2)}^{ - \frac{2}{3} } \times 3 \times ln(x) + \frac{1}{x} \times \sqrt[3]{3x + 2} + \frac{ \cos(2x) -( - 2 \sin(2x)) \times (x + 2) }{ \cos(2x) } = \\ = \frac{ ln(x) }{ \sqrt[3]{ {(3x + 2)}^{2} } } + \frac{ \sqrt[3]{3x + 2} }{x} + \frac{ \cos(2x) + (2x + 4) \sin(2x) }{ { \cos }^{2}(2x) }$

1.12

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

$y '= 7 \times ( - 3) {(x - 1)}^{ - 4} + \frac{1}{2} {(8 x- 3 + {x}^{2}) }^{ - \frac{1}{2} } \times (8 x- 3 + {x}^{2} )' = \\ = - \frac{21}{ {(x - 1)}^{4} } + \frac{8 - 2x}{2 \sqrt{8x - 3 + {x}^{2} } } = \\ = - \frac{21}{ {(x - 1)}^{4} } + \frac{4 - x}{ \sqrt{8x - 3 + {x}^{2} } }$