Question

Помогите пожалуйста с математикой! Найти вторую производную функции y(x):

Answer 1

Ответ:

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

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

$y't = - {t}^{ - 2} = - \frac{1}{ {t}^{2} }$

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

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

$(y'x)'t = \frac{2( {t}^{2} + 1) \times 2t \times 2 {t}^{3} - 6 {t}^{2} {( {t}^{2} + 1)}^{2} }{4 {t}^{6} } = \\ = \frac{( {t}^{2} + 1) \times 2 {t}^{2}(4 {t}^{2} - 3( {t}^{2} + 1))}{4 {t}^{6} } = \\ = \frac{( {t}^{2} + 1)(4 {t}^{2} - 3 {t}^{2} - 3) }{2 {t}^{4} } = \\ = \frac{ ({t}^{2} + 1) ( {t}^{2} - 3) }{2 {t}^{4} }$

$y''xx = \frac{( {t}^{2} + 1)( {t}^{2} - 3)}{2 {t}^{4} } \times ( - \frac{ {( {t}^{2} + 1)}^{2} }{2t} ) = \\ = - \frac{ {( {t}^{2} + 1)}^{3} ( {t}^{2} - 3)}{4 {t}^{5} }$