Question

Помогите решить пожалуйста

Answer 1

Ответ:

a

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

б

$y' = \frac{(1 - \cos(3x))' (1 + \cos(3x)) - (1 + \cos(3x)) '(1 - \cos(3x) }{ {(1 + \cos(3)) }^{2} } = \\ = \frac{3 \sin(3x) (1 + \cos(3x)) + 3 \sin(3x) (1 - \cos(3x)) }{ {(1 + \cos(3x)) }^{2} } = \\ = \frac{ 3\sin(3x)( 1 + \cos(3x) + 1 - \cos(3x)) }{ {(1 + \cos(3x)) }^{2} } = \\ = \frac{3 \sin(3x) \times 2 }{ {(1 + \cos( {}^{} 3x)) }^{2} } = \frac{6 \sin(3x) }{ {(1 + \cos(3x)) }^{2} }$

в

$y'_x = \frac{y'_t}{x'_t}$

$y'_t = 2t + 5$

$x'_t = 4 {t}^{3} + 2$

$y'_x = \frac{2t + 5}{4t {}^{3} + 2}$