Question

ДАМ МАКСИМАЛЬНЫЕ БАЛЛЫ. Алгебра

Answer 1

Ответ:

1

$y = 4 {x}^{9} - 3 \sqrt[3]{ {x}^{5} } + \frac{2}{ {x}^{4} } - \frac{5}{ \sqrt[4]{x} } = \\ = 4 {x}^{9} - 3 {x}^{ \frac{5}{3} } + 2 {x}^{ - 4} - 5 {x}^{ - \frac{1}{4} }$

$y' = 4 \times 9 {x}^{8} - 3 \times \frac{5}{3} {x}^{ \frac{2}{3} } + 2 \times ( - 4) {x}^{ - 5} - 5 \times ( - \frac{1}{4} ) {x}^{ - \frac{5}{4} } = \\ = 36 {x}^{8} - 5 \sqrt[3]{ {x}^{2} } - \frac{8}{ {x}^{5} } + \frac{5}{4x \sqrt[4]{x} }$

2

$y' = ( \sqrt{x} + x)' \times 3 {x}^{4} + (3 {x}^{4} )' \times ( \sqrt{x} + x) = \\ = ( \frac{1}{2} {x}^{ - \frac{1}{2} } + 1) \times 3 {x}^{4} + 12 {x}^{3} ( \sqrt{x} + x) = \\ = ( \frac{1}{2 \sqrt{x} } + x) \times 3 {x}^{4} + 12 {x}^{3} \sqrt{x} + 12 {x}^{4} = \\ = \frac{3}{2} {x}^{3} \sqrt{x} + 3 {x}^{5} + 12 {x}^{3} \sqrt{x} + 12 {x}^{4} = \\ = 3 {x}^{5} + 12 {x}^{4} + 13.5 {x}^{3} \sqrt{x}$

3

$y' = ( \frac{7 {x}^{9} + {x}^{4} - 3 {x}^{8} + 1}{ {x}^{2} } ) '= \\ = ( \frac{7 {x}^{9} }{ {x}^{2} } + \frac{ {x}^{4} }{ {x}^{2} } - \frac{3 {x}^{8} }{ {x}^{2} } + \frac{1}{ {x}^{2} } ) '= \\ = (7 {x}^{7} + {x}^{2} - 3 {x}^{6} + {x}^{ - 2} ) '= \\ = 49 {x}^{6} + 2x - 18 {x}^{5} - \frac{2}{ {x}^{3} }$

4

$y' = \frac{( {x}^{4} - 3)'( {x}^{4} + 5) - ( {x}^{4} + 5)'( {x}^{4} - 3) }{ {( {x}^{4} + 5)}^{2} } = \\ = \frac{4 {x}^{3} ( {x}^{4} + 5) - 4 {x}^{3} ( {x}^{4} - 3)}{ {( {x}^{4} + 5) }^{2} } = \\ = \frac{4 {x}^{3}( {x}^{4} + 5 - {x}^{4} + 3) }{ {( {x}^{4} + 5)}^{2} } = \frac{4 {x}^{3} \times 8}{ {( {x}^{4} + 5)}^{2} } = \\ = \frac{32 {x}^{3} }{ {( {x}^{4} + 5)}^{2} }$

6

$y = \sqrt{3x - 7} = {(3x - 7)}^{ \frac{1}{2} }$

$y' = \frac{1}{2} {(3x - 7)}^{ - \frac{1}{2} } \times (3x - 7)' = \\ = \frac{1}{2 \sqrt{3x - 7} } \times 3 = \frac{3}{2 \sqrt{3x - 7} }$

7

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

$y' = \frac{1}{2} {( \frac{ {x}^{2} - 4}{x + 3} )}^{ - \frac{1}{2} } \times \frac{( {x}^{2} - 4)'(x + 3) - (x + 3)'( {x}^{2} - 4) }{ {(x + 3)}^{2} } = \\ = \frac{1}{2} \times \sqrt{ \frac{x + 3}{ {x}^{2} - 4} } \times \frac{2x(x + 3) -( {x}^{2} - 4)}{ {(x + 3)}^{2} } = \\ = \frac{1}{2} \times \sqrt{ \frac{x + 3}{ {x}^{2} - 4 } } \times \frac{2 {x}^{2} + 6x - {x}^{2} + 4}{ {(x + 3)}^{2} } = \\ = \frac{ {x}^{2} + 6x + 4}{2 \sqrt{( {x}^{2} - 4) {(x + 3)}^{3} } }$

8

$y' = 3 {( \frac{ {x}^{2} + 1 }{ {x}^{2} - 1} )}^{2} \times \frac{( {x}^{2} + 1)' ({x}^{2} - 1) - ( {x}^{2} - 1)'( {x}^{2} + 1)}{ {( {x}^{2} - 1) }^{2} } = \\ = 3 \times \frac{ {( {x}^{2} + 1 )}^{2} }{ {( {x}^{2} - 1)}^{2} } \times \frac{2x( {x}^{2} - 1) - 2x( {x}^{2} + 1) }{ {( {x}^{2} - 1) }^{2} } = \\ = \frac{3 {( {x}^{2} + 1)}^{2} \times 2x( {x}^{2} - 1 - {x}^{2} - 1) }{ {( {x}^{2} - 1) }^{4} } = \\ = \frac{ - 12x {( {x}^{2} + 1) }^{2} }{ {( {x}^{2} - 1)}^{4} }$

9

$y' = \cos(7x) \times (7x)' - ( - \sin(19x)) \times (19x)' = \\ = 7 \cos(7x) + 19\sin(19x)$