Question

Найдите производные 1 порядка, прошу

Answer 1

Ответ:

1.

а)

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

б)

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

в)

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

г)

$y = {x}^{ \sqrt{x} }$

по формуле:

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

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

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

д)

$x \sin(y) - y \cos(x) = 0$

$\sin(y) + x \times \cos(y) \times y' - y' \cos(x) + y \sin(x) = 0 \\ y'(x \cos(y) - \cos(x)) = - \sin(y) - y \sin(x) \\ y '= - \frac{ \sin(y) + y \sin(x) }{x \cos(y) - \cos(x) } \\ y' = \frac{ \sin(y) + y \sin(x) }{ \cos(x) - x \cos(y) }$

2.

а)

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

б)

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

в)

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

г)

$y = {x}^{5x}$

$( ln(y)) '= ( ln( {x}^{5x} ) )' = (5x \times ln(x)) ' = \\ = 5 ln(x) + 5x \times \frac{1}{x} = \\ = 5 ln(x) + 5$

$y' = {x}^{5x} \times 5( ln(x) + 1)$

д)

$( {e}^{x} - 1)( {e}^{2y} - 1) = 1$

${e}^{x} ( {e}^{2y} - 1) + 2 {e}^{2y} \times y'( {e}^{x} - 1) = 0 \\ 2 {e}^{2y} y'( {e}^{x} - 1) = - {e}^{x} ( {e}^{2y} - 1) \\ y' = - \frac{ {e}^{x} ( {e}^{2y} - 1)}{2 {e}^{2y} ( {e}^{x} - 1) }$