Question

даю 80 балов, но очень срочно

Answer 1

Ответ:

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

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

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

$x't = \cos(t)$

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

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

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