Найти производные dy/dx, пользуясь формулами дифференцирования.

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Ответ:

Саша73838

07.04.2021 22:09

$y = \frac{2x}{ \sqrt{ {x}^{3} - 5 {x}^{2} + 3} } \\$

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

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

$y' = 3 {( {3}^{ \cos(x) } + \sin {}^{2} (3x) )}^{2} \times \\ \times ( ln(3) \times {3}^{ \cos(x) } \times ( - \sin(x) ) + 2 \sin(3x) \cos(3x) \times 3) = \\ = 3 {( {3}^{ \cos(x) } + \sin {}^{2} (3x) )}^{2} \times( - ln(3) \times {3}^{ \cos(x) } \sin(x) + 3 \sin(6x))$

$y = arctg \frac{2x + 1}{2x - 1} \\$

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

$y = \sqrt{ ln( \frac{ {x}^{2} + 3 }{ {x}^{3} + 9x} ) } \\$

$y' = \frac{1}{2} {( ln( \frac{ {x}^{2} + 3 }{ {x}^{3} + 9x } ) )}^{ - \frac{1}{2} } \times ( ln( \frac{ {x}^{2} + 3}{ {x}^{3} + 9x} )) ' \times \frac{( {x}^{2} + 3)'( {x}^{3} + 9x) - ( {x}^{3} + 9x)'( {x}^{2} + 3) }{ {( {x}^{3} + 9x) }^{2} } = \\ = \frac{1}{2 \sqrt{ ln( \frac{ {x}^{2} + 3 }{ {x}^{3} + 9x } ) } } \times \frac{1}{ \frac{ {x}^{2} + 3}{ {x}^{3} + 9x} } \times \frac{2x( {x}^{3} + 9x) - (3 {x}^{2} + 9)( {x}^{2} + 3)}{ {( {x}^{3} + 9x)}^{2} } = \\ = \frac{1}{2 \sqrt{ ln( \frac{ {x}^{2} + 3}{ {x}^{3} + 9x } ) } } \times \frac{ {x}^{3} + 9x}{ {x}^{2} + 3} \times \frac{2 {x}^{4} + 18 {x}^{2} - 3 {x}^{4} - 9 {x}^{2} - 9 {x}^{2} - 27}{ {( {x}^{3} + 9x)}^{2} } = \\ = \frac{1}{2 \sqrt{ ln( \frac{ {x}^{2} + 3 }{ {x}^{3} + 9x } ) } } \times \frac{ (- {x}^{4} - 27) }{( {x}^{2} + 3)( {x}^{3} + 9x)}$

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

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

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