Question

помогите найти вторую производную функции y(x):

Answer 1

Ответ:

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

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

$y't = - \sin(t)$

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

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

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

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