Question

Помогите решить пожалуйста Срочно

Answer 1

Ответ:

a

$y' = \frac{3}{4} \times 4 {x}^{3} + 4 \times 2x - 5 = \\ = 3 {x}^{3} + 8x - 5$

$y'' = 3 \times 3 {x}^{2} + 8 = 9 {x}^{2} + 8$

б

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

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

$y ''= - 6x \times ( - 2) {( {x}^{2} + 1) }^{ - 3} \times ( {x}^{2} + 1) '= \\ = \frac{12x}{ {( {x}^{2} + 1)}^{3} } \times 2x = \frac{24 {x}^{2} }{ {( {x}^{2} + 1) }^{3} }$

в

$y' = \frac{1}{ \cos {}^{2} (2x) } \times (2x)' = \frac{2}{ \cos {}^{2} (2x) }$

$y ''= 2 \times ( - 2) {( \cos(2x)) }^{ - 3} \times ( \cos(2x))' \times (2x)' = \\ = - \frac{4}{ \cos {}^{3} (2x) } \times ( - \sin(2x)) \times 2 = \\ = \frac{8 \sin(2x) }{ \cos {}^{3} (2x) }$