Алгебра: Найдите производные dy/dx заданных функций.

Объяснение:1234567...

Найдите производные dy/dx заданных функций.

Ответ:

Mymir06

23.04.2021 10:26

Объяснение:

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

$y '= 4 {( {5}^{arctg3x} + ctg \frac{x}{2} )}^{3} \times ( ln(5) \times {5}^{arctg3x} \times \frac{1}{1 + 9 {x}^{2} } - \frac{1}{2 \sin {}^{2} ( \frac{x}{2} ) } ) \\$

$y' = \frac{1}{2 \sqrt{ {a}^{2} + {x}^{2} } } \times 2x + \frac{a}{ \sqrt{1 - \frac{ {x}^{2} }{ {a}^{2} } } } \times \frac{1}{a} = \\ = \frac{x}{ \sqrt{ {a}^{2} + {x}^{2} } } + \frac{a}{ \sqrt{a {}^{2} - {x}^{2} }}$

$y' = \frac{1}{ \sqrt[3]{ \frac{x - 5}{ {x}^{2} + 4 } } } \times \frac{1}{3} {( \frac{x - 5}{ {x}^{2} + 4} )}^{ - \frac{2}{3} } \times \frac{(x - 5)'( {x}^{2} + 4) - ( {x}^{2} + 4)'(x - 5) }{ {( {x}^{2} + 4) }^{2} } = \\ = \sqrt[3]{ \frac{ {x}^{2} + 4 }{x - 5} } \times \frac{1}{3} \times \sqrt[3]{ {( \frac{ {x}^{2} + 4 }{x - 5} )}^{2} } \times \frac{ {x}^{2} + 4 - 2x(x - 5) }{ {( {x}^{2} + 4) }^{2} } = \\ = \frac{1}{3} \times \frac{ {x}^{2} + 4}{x - 5} \times \frac{ {x}^{2} + 4 - 2 {x}^{2} + 10x }{ {( {x}^{2} + 4) }^{2} } = \\ = \frac{ - {x}^{2} + 10x + 4}{3(x - 5)( {x}^{2} + 4) }$

$y = {(x + {e}^{3x}) }^{ \frac{x + 2}{2} } \\$

$y' = ( ln(y)) ' \times y$

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

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