Question

Помогите решить самостоятельную по теме "Производная сложной функции" как можно быстрее, пожалуйста)

Answer 1

Ответ:

$1)y' = 48 {x}^{7} - 11 {x}^{ - 2} + 34 {x}^{ - 3} + \frac{3}{4} {x}^{ - \frac{1}{4} } = \\ = 48 {x}^{7} - \frac{11}{ {x}^{2} } + \frac{34}{ {x}^{3} } + \frac{3}{x \sqrt[4]{x} }$

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

$3)y' = 15 \times ( - 3) {(x + 7)}^{ - 4} = - \frac{45}{ {(x + 7)}^{2} }$

$4)y' = \frac{1}{ {x}^{6} } \times 6 {x}^{5} \times { ln}^{6} (x) + 6 { ln}^{5} (x) \times \frac{1}{x} \times ln( {x}^{6} ) = \\ = \frac{6 { ln }^{6} (x)}{x} + \frac{6}{x} { ln}^{5} (x) \times ln( {x}^{6} )$

$5)y' = ln(33) \times {33}^{x} \times {x}^{33} + 33 {x}^{32} \times {33}^{x} = \\ = {33}^{x} \times {x}^{32} ( ln(33) \times x + 33)$

$6)y' = \frac{4 {e}^{4x} {(3x + 5)}^{ 3} - 3 {(3x + 5)}^{2} \times 3 \times {e}^{4x} }{ {(3x + 5)}^{6} } = \\ = \frac{ {(3x + 5)}^{2} {e}^{4x} (4(3x + 5) - 9) }{ {(3x + 5)}^{6} } = \\ = \frac{ {e}^{4x}(12x + 11) }{ {(3x + 5)}^{4} }$

$7)y '= \frac{8 { ln}^{7}(x - 12) \times \frac{1}{x - 12} \times tg( \frac{x}{2}) - \frac{1}{ { \cos}^{2}( \frac{x}{2}) } \times \frac{1}{2} \times { ln }^{8}(x -12) }{ {tg}^{2} ( \frac{x}{2}) } = \\ = \frac{ { ln}^{7}( x- 12) \times ( \frac{8tg( \frac{x}{2} )}{x - 12} - \frac{ ln(x - 12) }{2 { \cos }^{2} ( \frac{x}{2}) }) }{ {tg}^{2}( \frac{x}{2}) }$

$8)y '= \frac{1}{2} { (\frac{x - 16}{x + 16}) }^{ - \frac{1}{2} } \times \frac{x + 16 - (x - 16)}{ {(x + 16)}^{2} } = \\ = \frac{1}{2} \times \sqrt{ \frac{x + 16}{x - 16} } \times \frac{x + 16 - x + 16}{ {(x + 16)}^{2} } = \\ = \frac{1}{2} \times \frac{32}{ \sqrt{(x - 16) \times {(x + 16)}^{3} } } = \\ = \frac{16}{ \sqrt{(x - 16) {(x + 16)}^{3} } }$

$9)y= {(x + 5)}^{arccos(3x)}$

по формуле:

$y' = ( ln(y))' \times y$

$(ln(y)) ' = ( ln( {(x + 5)}^{arccos(3x)} )' = (arccos(3x) \times ln(x + 5)) '= \\ = - \frac{1}{ \sqrt{1 - 9 {x}^{2} } } \times 3 \times ln(x + 5) + \frac{arccos(3x)}{x + 5}$

$y '= {(x + 5)}^{arccos(3x)} \times ( - \frac{9 ln(x + 5) }{ \sqrt{1 - 9 {x}^{2} } } + \frac{arccos(3x)}{x + 5} )$