Question

помогите взять производные, на кону стоит мое отчисление :(

Answer 1

Ответ:

1.

$1)y' = 3 {tg}^{2} (2x) \times \frac{2}{ { \cos }^{2}(2x) } \times { \cos}^{2} (2x) + 2 \cos(2x) \times ( - \sin(2x) ) \times 2 \times {tg}^{3} (2x) = \\ = {tg}^{2} (2x) \times (6 - 4 { \sin }^{2}(2x) )$

$2)y' = \frac{1}{ \frac{ {x}^{3} - 9}{ {x}^{3} - 1 } } \times \frac{3 {x}^{2} ( {x}^{3} - 1) - 3 {x}^{2} ( {x}^{3} - 9)}{ {( {x}^{3} - 1)}^{2} } = \\ = \frac{ {x}^{3} - 1}{ {x}^{3} - 9} \times \frac{3 {x}^{2}( {x}^{3} - 1 - {x}^{3} + 9)}{ {( {x}^{3} - 1)}^{2} } = \\ = \frac{3 {x}^{2} \times 8}{( {x}^{3} - 9)( {x}^{3} - 1) } = \frac{24 {x}^{2} }{( {x}^{3} - 1)( {x}^{3} - 9) }$

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

2.

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

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