Предмет: Математика, автор: FreshMin7

Решить все примеры(там стоят штрихи, то бишь производные

Приложения:

Ответы

Автор ответа: Miroslava227

Ответ:

$(3 {x}^{6} - 14 {x}^{3} + 34)' = 18 {x}^{5} - 42 {x}^{2} \\$

$( - 13 {x}^{ - 4} - 4 {x}^{3} )' = 52 {x}^{ - 5} - 12 {x}^{2} = \\ = \frac{52}{ {x}^{5} } - 12 {x}^{2}$

$(4 {x}^{ \frac{3}{2} } - \frac{1}{3} {x}^{9} + 6)' = 4 \times \frac{3}{2} {x}^{ \frac{1}{2} } - \frac{1}{3} \times 9 {x}^{8} + 0 = \\ = 6 \sqrt{x} - 3 {x}^{8}$

$( \frac{3}{7} {x}^{21} - 2 {x}^{ \frac{3}{2} } + 1.3)' = \frac{3}{7} \times 21 {x}^{20} - 2 \times \frac{3}{2} {x}^{ \frac{1}{2} } + 0 = \\ = 9 {x}^{20} - 3 \sqrt{x}$

$( {x}^{ - 2} + 3 {x}^{4} )' = - 2 {x}^{ - 3} + 12 {x}^{3} = \\ = - \frac{2}{ {x}^{3} } + 12 {x}^{3}$

$( {x}^{ - 1} - 3 {x}^{4} )' = - {x}^{ - 2} - 12 {x}^{3} = - \frac{1}{ {x}^{2} } - 12 {x}^{3} \\$

$( - {x}^{ \frac{8}{7} } + 2 {x}^{5} + 14) '= - \frac{8}{7} {x}^{ \frac{1}{7} } + 10 {x}^{4} = \\ = - \frac{8}{7} \sqrt[7]{x} + 10 {x}^{4}$

$(ctg4x)' = - \frac{1}{ \sin {}^{2} (4x) } \times (4x) '= - \frac{4}{ \sin {}^{2} (4x) } \\$

$((6 + x) {}^{4} )' = 4 {(6 + x)}^{3}$

10.

$( {(5x - 8)}^{ \frac{1}{2} } )' = \frac{1}{2} {(5x - 8)}^{ - \frac{1}{2} } \times (5x - 8)' = \\ = \frac{5}{2 \sqrt{5x - 8} }$

11.

$(7 { \cos(3x)) }^{ - 2} )' = \\ = - 14 {( \cos(3x) )}^{ - 3} \times ( \cos(3x))' \times (3x) '= \\ = - \frac{14}{ \cos {}^{2} (3x) } \times ( - \sin(3x)) \times 3 = \frac{42 \sin(3x) }{ \cos {}^{2} (3x) }$

12.

$( {(3x + 4)}^{ \frac{2}{9} } )' = \frac{2}{9} {(3x + 4)}^{ - \frac{7}{9} } \times (3x + 4)' = \\ = \frac{2}{9 \sqrt[9]{ {(3x + 4)}^{7} } } \times 3 = \frac{2}{3 \sqrt[9]{ {(3x + 4)}^{7} } }$

13.

$( \cos {}^{5} (x)) ' = 5 \cos {}^{4} (x) \times ( - \sin(x)) = \\ = - 5 \sin(x) \cos {}^{4} (x)$

14.

$(4 + {x}^{7} )'(5 - 5x + {x}^{3} ) + (5 - 5x + {x}^{3} )'(4 + {x}^{7} ) = \\ = 7 {x}^{6} (5 - 5x + {x}^{3} ) + ( - 5 + 3{x}^{2} )(4 + {x}^{7} ) = \\ = 35 {x}^{6} - 35 {x}^{7} + 7 {x}^{9} - 20 - 5 {x}^{7} + 12 {x}^{2} + 3 {x}^{9} = \\ = 35 {x}^{6} - 40 {x}^{7} + 10 {x}^{9} + 12 {x}^{2} - 20$

15.

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

16.

$( \frac{3x + 1}{3 + 2x})' = \frac{(3x + 1)'(3 + 2x) - (2x + 3)'(3x + 1)}{ {(2x + 3)}^{2} } = \\ = \frac{3(2x + 3) - 2(3x + 1)}{ {(2x + 3)}^{2} } = \frac{6x + 9 - 6x - 2}{ {(2x + 3)}^{2} } = \\ = \frac{7}{ {(2x + 3)}^{2} }$

17.

$\frac{( \sqrt[3]{x} )'(4 + {x}^{5}) - ( {x}^{5} + 4)' \sqrt[3]{x} }{ {( {x}^{5} + 4)}^{2} } = \\ = \frac{ \frac{1}{3 \sqrt[3]{ {x}^{2} } }(4 + {x}^{5} ) - 5 {x}^{4} \sqrt[3]{x} }{ {( {x}^{5} + 4)}^{2} } = \\ = \frac{ \frac{4}{3 \sqrt[3]{ {x}^{2} } } + \frac{1}{3} {x}^{4} \sqrt[3]{x} - 5 {x}^{4} \sqrt[3]{x} }{ {( {x}^{5} + 4) }^{2} } = \\ = \frac{ \frac{4}{3 \sqrt[3]{ {x}^{2} } } - \frac{14}{3} {x}^{4} \sqrt[3]{x} }{ {( {x}^{5} + 4) }^{2} }$

18.

$( \frac{ {x}^{6} + 3}{ \sqrt[3]{x - 9} } )' = \frac{6 {x}^{5} \sqrt[ 3]{x - 9} - \frac{1}{3 \sqrt[3]{ {x - 2)}^{2} } }( {x}^{6} + 3)}{ \sqrt[3]{ {(x - 9)}^{2} } } \\$