Question

АЛГЕБРА
Найти Производные второго порядка функции

Answer 1

Ответ:

81

$y' = 4 {( {x}^{3} - 2)}^{3} \times ( {x}^{3} - 2) '= \\ = 4 {( {x}^{3} - 2) }^{3} \times 3 {x}^{2} = 12 {x}^{2} {( {x}^{3} - 2)}^{3}$

82

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

83

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

84

$y '= ( {4}^{x} )'(x + 1) + (x + 1) '\times {4}^{x} = \\ = ln(4) \times {4}^{x} (x + 1) + {4}^{x} = {4}^{x} ( ln(4) (x + 1) + 1)$

85

$y = {x}^{x} \\ y' = ( ln(y)) ' \times y$

$( ln(y)) ' = ( ln( {x}^{x} ) ) '= (x ln(x)) '= \\ = ln(x) + x \times \frac{1}{x} = ln(x) + 1$

$y '= {x}^{x} ( ln(x) + 1)$

86

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

87

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

88

$y' = \frac{( {x}^{4} )'(2x - 1) - (2x - 1)' \times {x}^{4} }{ {(2x - 1)}^{2} } = \\ = \frac{4 {x}^{3}(2x - 1) - 2 {x}^{4} }{ {(2x - 1)}^{2} } = \frac{8 {x}^{4} - 4 {x}^{3} - 2 {x}^{4} }{ {(2x - 1)}^{2} } = \\ = \frac{6 {x}^{4} - 4 {x}^{3} }{ {(2x - 1)}^{2} }$

89

$y = {( ln(x)) }^{x}$

$( ln(y)) ' = ( ln( ln(x) {}^{x} ) )' = (x \times ln( ln(x) ) ) '= \\ = ln( ln(x) ) + x \times \frac{1}{ ln(x) } \times \frac{1}{x} = \\ = ln( ln(x) ) + \frac{1}{ ln(x) }$

$y' = {( ln(x)) }^{x} \times ( ln( ln(x) ) + \frac{1}{ ln(x) } )$

90

$y' = 3 \sin {}^{2} (x) \times ( \sin(x) )' = 3 \sin {}^{2} (x) \cos(x)$

91

$y '= 4 \times 3 \cos {}^{2} (x) \times ( - \sin(x) ) + 3 \sin(x) = \\ = 3 \sin(x) ( 1- 4 \cos {}^{2} (x) )$