Question

Помогите пожалуйста решить Номера 3 ,7,10,12 , 18 Даю 80 баллов

Answer 1

Ответ:

3

$(0.8 {x}^{15} + 12 \sqrt[3]{ {x}^{5} } ) '= \frac{4}{5} \times 15 {x}^{14} + 12 \times \frac{5}{3} {x}^{ \frac{2}{3} } = \\ = 12 {x}^{14} + 20 \sqrt[3]{ {x}^{2} }$

7

$(2 {x}^{6} + x \sqrt[3]{ x } )' = (2 {x}^{6} + {x}^{ \frac{4}{3} } ) '= \\ = 2 \times 6 {x}^{5} + \frac{4}{3} {x}^{ \frac{1}{3} } = \\ = 12 {x}^{5} + \frac{4}{3} \sqrt[3]{x}$

10

$( \sqrt{13 {x}^{8} + 3} ) '= ( {(13 {x}^{8} + 3)}^{ \frac{1}{2 } } ) '= \\ = \frac{1}{2} {(13 {x}^{8} + 3) }^{ - \frac{1}{2} } \times (13 {x}^{8} + 3) = \\ = \frac{104 {x}^{7} }{2 \sqrt{13 {x}^{8} + 3} } = \frac{52 {x}^{7} }{ \sqrt{13 {x}^{8} + 3} }$

12

$( \sqrt[9]{ {(3x + 4)}^{2} } ) '= ( {(3x + 4)}^{ \frac{2}{9} } ) '= \\ = \frac{2}{9} {(3x + 4)}^{ - \frac{7}{9} } \times (3x + 4) '= \\ = \frac{2 \times 3}{9 \sqrt[9]{ {(3x + 4)}^{7} } } = \frac{2}{3 \sqrt[9]{ {(3x + 4)}^{7} } }$

18

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