Question

Найти производную Y'x

Answer 1

Ответ:

$y'x = \frac{yt}{xt}$

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

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

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