Question

Помогите пожалуйста!
Нужно с решением!

Answer 1

Ответ:

а)

$y = 5 {x}^{2} - \frac{2}{ \sqrt{x} } + \sin( \frac{\pi}{4} ) = \\ = 5 {x}^{2} - 2 {x}^{ - \frac{1}{2} } + \sin( \frac{\pi}{4} )$

$y' = 10x - 2 \times ( - \frac{1}{2} ) {x}^{ - \frac{3}{2} } + 0 = \\ = 10x + \frac{1}{x \sqrt{x} }$

б)

$y = \frac{ \cos(x) }{1 + \sin(x) }$

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

в)

$y = {2}^{x} \cos(x)$

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

г)

$y = ln( \sqrt{ {x}^{3} + 4} )$

$y' = \frac{1}{ \sqrt{ {x}^{3} + 4 } } \times ( \sqrt{ {x}^{3} + 4 } )' \times ( {x}^{3} + 4)' = \\ = \frac{1}{ \sqrt{ {x}^{3} + 4} } \times \frac{1}{2 \sqrt{ {x}^{3} + 4} } \times 3 {x}^{2} = \\ = \frac{3 {x}^{2} }{2( {x}^{3} + 4) }$