Алгебра: найти производную функции

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найти производную функции

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Ответ:

228alex777

05.05.2021 10:08

y' = - 5

$y' = - \frac{1}{5} \\$

y '= 0

$y '= 10 {x}^{9}$

$y' = ( {x}^{ - 4} ) '= - 4 {x}^{ - 5} = - \frac{4}{ {x}^{5} } \\$

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

$y = (x \sqrt{x} ) = ( {x}^{ \frac{3}{2} } ) = \frac{3}{2} {x}^{ \frac{1}{2} } = 1.5 \sqrt{x} \\$

$y' = 6 {x}^{2} - 5 \times ( - 1) {x}^{ - 2} + 8 \times \frac{1}{2} {x}^{ - \frac{1}{2} } = \\ = 6 {x}^{2} + \frac{5}{ {x}^{2} } + \frac{4}{ \sqrt{x} }$

$y' = 0.5 \cos(x) - \frac{3}{ \cos {}^{2} (x) } - 8 \sin(x) + \frac{1}{3 \sin {}^{2} (x) } \\$

$y' = ((x + 1) \sqrt{x} ) '= (x \sqrt{x} + \sqrt{x} ) '= \\ = ( {x}^{ \frac{3}{2} } + {x}^{ \frac{1}{2} } ) '= \frac{3}{2} {x}^{ \frac{1}{2} } + \frac{1}{2} {x}^{ - \frac{1}{2} } = 1.5 \sqrt{x} + \frac{1}{2 \sqrt{x} }$

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

$y' = \frac{( {x}^{2} - 5x)'(x - 7) - (x - 7)'( {x}^{2} - 5x)}{ {(x - 7)}^{2} } = \\ = \frac{(2x - 5)(x - 7) - ( {x}^{2} - 5x) }{ {(x - 7)}^{2} } = \\ = \frac{2 {x}^{2} - 14x - 5x + 35 - {x}^{2} + 5x}{ {(x - 7)}^{2} } = \\ = \frac{ {x}^{2} - 14x + 35}{ {(x - 7)}^{2} }$

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

$y '= 3 {(3x - 5)}^{2} \times (3x - 5)' = \\ = 9 {(3x - 5)}^{2}$

$y '= \cos( \frac{x}{5} ) \times ( \frac{x}{5} )' = \frac{1}{5} \cos( \frac{x}{5} ) \\$

$y '= ( {(6x + 1)}^{ \frac{1}{4} } )' = \frac{1}{4} {(6x + 1)}^{ - \frac{3}{4} } \times (6x + 1) '= \\ = \frac{1}{4 \sqrt[4]{ {(6x + 1)}^{3} } } \times 6 = \frac{3}{2 \sqrt[4]{ {6x + 1)}^{3} } }$

$y '= 2 \cos(x) \times ( \cos(x)) ' = \\ = 2 \cos(x) \times ( - \sin(x)) = - \sin(2x)$

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