Предмет: Математика, автор: ogog6625

ПОМОГИТЕ СРОЧНО ОЧЕНЬ!!!!!! решите производности.ПРОШУ

Приложения:

Ответы

Автор ответа: Miroslava227

Ответ:

$y' = - 4 \times 4 {x}^{3} + 5 + 1 = - 16 {x}^{3} + 6$

$y' = - 3 \sin(x) - 2x$

$y' = - 2 {x}^{ - 3} + ( {x}^{ - 3} ) '= \\ = - 2 {x}^{ - 3} - 3 {x}^{ - 4} = - \frac{2}{ {x}^{3} } - \frac{3}{ {x}^{4} }$

$y' = ( - {x}^{5} + 3x)'(1.5 {x}^{2} + 1) + (1.5 {x}^{2} + 1)'( - {x}^{5} + 3x) = \\ = ( - 5 {x}^{4} + 3)(1.5 {x}^{2} + 1) + 3x( - {x}^{5} + 3x) = \\ = - 7.5 {x}^{6} - 5 {x}^{4} + 4.5 {x}^{2} + 3 - 3 {x}^{6} + 9 {x}^{2} = \\ = - 10.5 {x}^{6} - 5 {x}^{4} + 13.5 {x}^{2} + 3$

$y' = \frac{1}{ \cos {}^{2} (4x - 5) } \times (4x - 5)' = \frac{4}{ \cos {}^{2} (4x - 5) } \\$

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

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

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

$y '= e {}^{2x - 4} \times (2x - 4) '= 2e {}^{2x - 4}$

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