Question

Даю 40 баллов.Найти производные первого порядка данных функций

Answer 1

Ответ:

г)

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

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

$x't = - 3 {e}^{ - t}$

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

д)

$y \sin(x) - \cos(x - y) = 0$

$y' \sin(x) + \cos(x) \times y + \sin(x - y) \times (x - y)' = 0 \\ y' \sin(x) + y\cos(x) + (1 - y') \sin(x - y) = 0 \\ y' \sin(x) + \sin(x - y) - y' \sin(x - y) = - y \cos(x) \\ y'( \sin(x) - \sin(x - y) ) = - y \cos(x) - \sin(x - y) \\ y' = - \frac{ y \cos(x) + \sin(x - y) }{ \sin(x) - \sin(x - y) }$