Question

Помогите! Нужно решить номер 3,если не сложно ,то с подробным описанием

Answer 1

Ответ:

а

$y = 3 {x}^{4} + 5 {x}^{ \frac{1}{2} } - 2 {x}^{ - 1} - 4 {x}^{ - 2} + 9$

$y' = 3 \times 4 {x}^{3} + 5 \times \frac{1}{2} {x}^{ - \frac{1}{2} } - 4 \times ( - 2) {x}^{ - 3} + 0 = \\ = 12 {x}^{3} + \frac{5}{2 \sqrt{x} } + \frac{8}{ {x}^{3} }$

б

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

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

в

$y = {( ln(x) + tgx + {x}^{ - 1} )}^{3}$

$y' = 3 {( ln(x) + tgx + {x}^{ - 1}) }^{2} \times ( ln(x) + tgx + {x}^{ - 1} )' = \\ = 3( ln(x) + tgx + \frac{1}{x} ) ^{2} \times ( \frac{1}{ x} + \frac{1}{ { \cos}^{2} x} - \frac{1}{ {x}^{2} } )$

г

$y = arcctg \sqrt{1 - {x}^{2} }$

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

д

$y = \sin( {2}^{x} + 4) - {(tgx - {x}^{3} )}^{2}$

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