Question

Помогите пожалуйста, дам 35 баллов
(x^3 ln x)'=
(x^2/x^3+1)'=
(8^cos x)'=

Answer 1

Ответ:

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

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

$( {8}^{ \cos(x) } )' = ln(8) \times {8}^{ \cos(x) } \times (\cos(x) ) '= \\ = ln(8) \times {8}^{ \cos(x) } \times ( - \sin(x))$