Question

Найдите производную второго порядка неявной функции y = ctg(x − y).

Answer 1

Ответ:

$y = ctg(x - y)$

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

$y' = \frac{y' - 1}{ { \sin }^{2} (x - y)} \\ y'' = - \frac{y''\times { \sin}^{2} (x - y) - 2 \sin(x - y) \cos(x - y) \times (1 - y') }{ { \sin}^{4} (x - y)} \\ y'' = \frac{y''}{ { \sin }^{2}(x - y) } - \frac{2(1 - y') \cos(x - y) }{ { \sin }^{3}(x - y) } \\ y''(1 - \frac{1}{ { \sin}^{2}(x - y) } ) = \frac{2(1 - y') \cos(x - y) }{ { \sin }^{3}(x - y) } \\ - y'' \times \frac{ { \cos }^{2} (x - y)}{ { \sin }^{2} (x - y)} = \frac{2(1 - y') \cos(x - y) }{ { \sin }^{3}(x - y) } \\ y'' = - \frac{ { \sin }^{2} (x - y)}{ { \cos }^{2}(x - y) } \times \frac{2(1 - y') \cos(x - y) }{ { \sin }^{3}(x - y) } \\ y''= - \frac{2(1 - y')}{ \cos(x - y) \sin(x - y) } \\ y''= \frac{2(y' - 1)}{ \sin(x - y) \cos(x - y) } \\ \\ y''= \frac{2( \frac{1}{ { \cos}^{2}(x - y) } - 1)}{ \sin(x - y) \cos(x - y) } \\ y'' = \frac{2{tg}^{2}(x - y)}{ \sin(x - y) \cos(x - y) } \\ y''= \frac{2\sin(x-y)}{ { \cos }^{3}(x - y) }$