Алгебра: Найдите производную функции

Найдите производную функции

Ответ:

СнежаночкаФил

08.05.2021 19:20

$y' = ( {x}^{2} - 1)'( {x}^{5} + 2) + ( {x}^{5} + 2)'( {x}^{2} - 1) = \\ = 2x( {x}^{5} + 2) + 5 {x}^{4} ( {x}^{2} - 1) = \\ = 2 {x}^{6} + 4x + 5 {x}^{6} - 5 {x}^{4} = 7x {}^{6} - 5 {x}^{4} + 4x$

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

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

$y '= \frac{(2x + 3)'(3x - 2) - (3x - 2)'(2x + 3)}{ {(3x - 2)}^{2} } = \\ = \frac{2(3x - 2) - 3(2x + 3)}{ {(3x - 2)}^{2} } = \frac{6x - 4 - 6x - 9}{ {(3x - 2)}^{2} } = \\ = - \frac{13}{ {(3x - 2)}^{2} }$

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

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