Question

Решите, пожалуйста, 35 баллов

Answer 1

Ответ:

а)

$y = \frac{x}{ {x}^{2} - 1}$

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

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

б)

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

$y't = 1 - \cos(t) \\ x't = - \sin( \frac{t}{2} ) \times \frac{1}{2}$

$y'x = \frac{1 - \ \cos(t) }{ - \frac{ \sin( \frac{t}{2} ) }{2} } = - \frac{2(1 - \cos(t)) }{ \sin( \frac{t}{2} ) }$

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

$y''x = \frac{ - 2 \sin(t) \sin( \frac{t}{2} ) + \cos( \frac{t}{2} )(1 - \cos(t)) }{ { \sin}^{2}( \frac{t}{2} )} \times ( - \frac{1}{ \frac{1}{2} \sin( \frac{t}{2} ) } ) = \\ = \frac{ 4 \sin(t) \sin( \frac{t}{2} ) - 2 \cos( \frac{t}{2} )(1 - \cos(t)) }{ { \sin }^{3}( \frac{t}{2} ) }$