Question

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

Answer 1

Ответ:

а)

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

последним действием почленно разделила

в)

$y' = 2 ln(x + \sqrt{ {x}^{2} + 4 } ) \times \frac{1}{x + \sqrt{ {x}^{2} + 4 } } \times (1 + \frac{1}{2 \sqrt{ {x}^{2} + 4} } \times 2x) = \\ = \frac{2 ln(x + \sqrt{ {x}^{2} + 4} ) }{x + \sqrt{ {x}^{2} + 4} } \times (1 + \frac{x}{ \sqrt{ {x}^{2} + 4} } )$

д)

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