Предмет: Алгебра, автор: lipa45

Очень срочно плачу много

Приложения:

Ответы

Автор ответа: Miroslava227

Ответ:

$y '= 8 \times 5 {x}^{4} + 3 - 0 = 40 {x}^{4} + 3$

$y' = \cos(x) - 5 \sin(x)$

$y' = 7 \times ( - 3) {x}^{ - 4} + \frac{6}{ \cos {}^{2} (x) } = \\ = - \frac{21}{ {x}^{4} } + \frac{6}{ \cos {}^{2} (x) }$

$y '= 12 {x}^{11} + 24 {x}^{2} - 4x + \sin(x)$

$y' = - 14 \times ( - 4) {x}^{ - 5} + 6 - 0 = \frac{56}{ {x}^{5} } + 6 \\$

$y '= (2 {x}^{ \frac{1}{2} } - 12 {x}^{ - 3} ) '= 2 \times \frac{1}{2} {x}^{ - \frac{1}{2} } - 12 \times ( - 3) {x}^{ - 4} = \\ = \frac{1}{ \sqrt{x} } + \frac{36}{ {x}^{4} }$

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

$y' = ( {x}^{ \frac{5}{3} } + {x}^{ - \frac{2}{4} } )' = \frac{5}{3} {x}^{ \frac{2}{3} } - \frac{2}{4} {x}^{ \frac{6}{4} } = \\ = \frac{5}{3} \sqrt[3]{ {x}^{2} } - \frac{1}{2} x \sqrt{x}$

$y' = ( {x}^{ \frac{4}{5} } + {x}^{ - 1} + 3x)' = \frac{4}{5} {x}^{ - \frac{1}{5} } - {x}^{ - 2} + 3 = \\ = \frac{4}{5 \sqrt[5]{x} } - \frac{1}{ {x}^{2} } + 3$

10.

$y' = (3 {( - 5x + 3)}^{ - 1} ) '= \\ = - 3 {( - 5x + 3)}^{ - 2} \times ( - 5x + 3) '= \\ = \frac{ - 3}{ {(3 - 5x)}^{2} } \times ( - 5) = \frac{15}{ {(3 - 5x)}^{2} }$

11.

$y '= \frac{(2 {x}^{4} + 4) '\times 8x - (8x) '\times (2 {x}^{4} + 4) }{ {(8x)}^{2} } = \\ = \frac{8 {x}^{3} \times 8x - 8(2 {x}^{4} + 4) }{64 {x}^{2} } = \\ = \frac{64 {x}^{4} - 16 {x}^{4} - 32 }{64 {x}^{2} } = \frac{48 {x}^{4} - 32 }{64{x}^{2} } = \\ = \frac{3 {x}^{4} - 2 }{4{x}^{2} }$

12.

$y '= ( {x}^{7} -6 )'( {x}^{3} - 4x) + ( {x}^{3} - 4x)'( {x}^{7} - 6) = \\ = 7 {x}^{6} ( {x}^{3} - 4x) + (3 {x}^{2} - 4)( {x}^{7} - 6) = \\ = 7 {x}^{9} - 28 {x}^{7} + 3 {x}^{9} - 18 {x}^{2} - 4 {x}^{7} + 24 = \\ = 10 {x}^{9} - 32 {x}^{4} - 18 {x}^{2} + 24$