Question

помогите с самостоятельной срочно !!!!

Answer 1

Ответ:

а

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

б

$y '= \frac{(3x + 1)'(3 {x}^{2} - 1) - (3 {x}^{2} - 1)'(3x + 1) }{ {(3 {x}^{2} - 1)}^{2} } = \\ = \frac{3(3 {x}^{2} - 1) - 6x(3x + 1)}{ {(3 {x}^{2} - 1)}^{2} } = \frac{9 {x}^{2} - 3 - 18 {x}^{2} - 6x}{ {(3 {x}^{2} - 1) }^{2} } = \\ = \frac{ - 9 {x}^{2} - 6x - 3 }{ {(3 {x}^{2} - 1) }^{2} } = - \frac{9 {x}^{2} + 6x + 3 }{ {(3 {x}^{2} - 1) }^{2} }$

в

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