Question

Найти производные первого порядка для заданных функций.(С решением, пожалуйста)

Answer 1

Ответ:

а)

$y = {x}^{ \cos(x ) }$

по формуле:

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

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

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

б)

$y = arcsin( \frac{1}{ {x}^{2} } )$

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

в)

по формуле:

$y'x = \frac{y't}{x't}$

$y't = 1 - \frac{1}{1 - {t}^{2} } \times ( - 2t) = 1 + \frac{2t}{1 - {t}^{2} }$

$x't = \frac{2t( {t}^{2} - 1) - 2t \times {t}^{2} }{ {( {t}^{2} - 1)}^{2} } = \\ = \frac{2t( {t}^{2} - 1 - {t}^{2} ) }{ {( {t}^{2} - 1) }^{2} } = - \frac{2t}{ {( {t}^{2} - 1) }^{2} }$

$y'x = \frac{1 + \frac{2t}{1 - {t}^{2} } }{ - \frac{2t}{ {( {t}^{2} - 1) }^{2} } } = \frac{1 - {t}^{2} + 2t }{1 - {t}^{2} } \times ( - \frac{ {( {t}^{2} - 1)}^{2} }{2t} ) = \\ = \frac{ - ( {t}^{2} - 2t + 1) }{ - ( {t}^{2} - 1)} \times ( - \frac{ {( {t}^{2} - 1)}^{2} }{2t} ) = \\ = - \frac{ {(t - 1)}^{2} }{ {t}^{2} - 1} \times \frac{ {( {t}^{2} - 1) }^{2} }{2t} = - \frac{ {(t - 1)}^{2}( {t}^{2} - 1)}{2t} = \\ = - \frac{ {(t - 1)}^{3}(t + 1) }{2t}$