Question

помогите пожалуйста очень срочно нужноо ​

Answer 1

Ответ:

а

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

б

$y' = \frac{(2x - 3 {x}^{2})'(3x - 4) - (3x - 4)'(2x - 3 {x}^{2}) }{ {(3x - 4)}^{2} } = \\ = \frac{(2 - 6x)(3x - 4) - 3(2x - 3 {x}^{2}) }{ {(3x - 4)}^{2} } = \\ = \frac{6x - 8 - 18 {x}^{2} + 24x - 6x + 9 {x}^{2} }{ {(3x - 4)}^{2} } = \\ = \frac{ - 9 {x}^{2} + 24x - 8 }{ {(3x - 4)}^{2} } = - \frac{9 {x}^{2} - 24x + 8}{ {(3x - 4)}^{2} }$

в

$y' = 6 {(3 {x}^{5} - x)}^{5} \times (3 {x}^{5} - x) '= \\ = 6(3 {x}^{5} - x) {}^{5} (15 {x}^{4} - 1)$