Question

СРОЧНО!!! знайти похідну: ​

Answer 1

Ответ:

а

$( \frac{8}{x} ) '= (8 {x}^{ - 1} ) '= - 8 {x}^{ - 2} = - \frac{8}{ {x}^{2} }$

б

$( {x}^{7} - tgx) '= 7 {x}^{6} - \frac{1}{ { \cos}^{2}x }$

с

$(4 \sin(x) + {x}^{ \frac{1}{2} } ) '= 4 \cos(x) + \frac{1}{2} {x}^{ - \frac{1}{2} } = \\ = 4 \cos(x) + \frac{1}{2 \sqrt{x} }$

д

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