Question

Срочно!!! даю 35 балов!!!! 2 вложения
Знайти похідну функцій:

Answer 1

7

$y '= ( {x}^{ \frac{5}{6} } ) '= \frac{5}{6} {x}^{ - \frac{1}{6} } = \frac{5}{6 \sqrt[6]{x} }$

8

$y' = - 7\cos(x)$

9

$y' = ( {x}^{ \frac{4}{3} } + 6 {x}^{ - 6} ) '= \frac{4}{3} {x}^{ \frac{1}{3} } - 36 {x}^{ - 7} = \frac{4}{3} \sqrt[3]{x} - \frac{36}{ {x}^{7} }$

10

$y' = ( {x}^{2} + 4) '\cos(x) + (\cos(x) )' \times ( {x}^{2} + 4) = \\ = 2x \cos(x) - \sin(x) \times ( {x}^{2} + 4)$

11

$y' = (4 {x}^{ - 5} - 7 {x}^{ - 4} )' = - 20 {x}^{ - 6} + 28 {x}^{ - 5} = \\ = - \frac{20}{ {x}^{6} } + \frac{28}{ {x}^{5} }$

13

$y' = \frac{( {x}^{2} + 3)'( {x}^{2} - 2x) - ( {x}^{2} - 2x)'( {x}^{2} + 3) }{ {(x {}^{2} - 2x)}^{2} } = \\ = \frac{2x( {x}^{2} - 2x) - (2x - 2)( {x}^{2} + 3) }{ {( {x}^{2} - 2x) }^{2} } = \\ = \frac{2 {x}^{3} - 4 {x}^{2} - 2 {x}^{3} - 6x + 2 {x}^{2} + 6}{ {( {x}^{2} - 2x)}^{2} } = \\ = \frac{ - 2 {x}^{2} - 6x + 6 }{ {( {x}^{2} - 2x)}^{2} }$