Question

Найти производные $\frac{dy}{dx}$ заданных функций.

Answer 1

Ответ:

б)

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

г)

По формуле:

$y' = ( ln(y)) ' \times y$

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

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

е)

По формуле:

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

$y't = \frac{ {e}^{t} (t - 1) - {e}^{t} }{ {(t - 1)}^{2} } = \frac{ {e}^{t} (t - 2)}{ {(t - 1)}^{2} }$

$x't = \frac{ {e}^{t} (t + 1) - {e}^{t} }{ {(t + 1)}^{2} } = \frac{ {e}^{t} (t + 1 - 1)}{ {(t + 1)}^{2} } = \\ = \frac{ {e}^{t} t}{ {(t + 1)}^{2} }$

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