Question

помогите, пожалуйста

Answer 1

Ответ:

а

$y' = 6 {x}^{2} - \frac{1}{2} \times 2x + 0 = 6 {x}^{2} - x$

б

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

в

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

г

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