Question

Помогите! Тема: производные​

Answer 1

Ответ:

4 задание

1

$y '= \frac{(x + 4)'(x - 3) - (x - 3)'(x + 4)}{ {(x - 3)}^{2} } = \\ = \frac{1 \times (x - 3) - 1 \times (x + 4)}{ {(x - 3)}^{2} } = \\ = \frac{x - 3 - x - 4}{ {(x - 3)}^{2} } = - \frac{7}{ {(x - 3)}^{2} }$

2

$y' = \frac{(4x - 7)'(3x + 9) - (3x + 9)'(4x -7 )}{ {(3x + 9)}^{2} } = \\ = \frac{4(3x + 9) - 3(4 x - 7)}{ {(3x + 9)}^{2} } = \\ = \frac{12x + 36 - 12x + 21}{ {(3x + 9)}^{2} } = \frac{57}{ {(3x + 9)}^{2} }$

3

$y '= \frac{( {(x - 4)}^{2})'(3x + 1) - (3x + 1) '{(x - 4)}^{2} }{ {(3 x + 1)}^{2} } = \\ = \frac{2(x - 4)(3x + 1) - 3 {(x - 4)}^{2} }{ {(3x + 1)}^{2} } = \\ = \frac{(x - 4)(2(3x + 1) - 3(x - 4))}{ {(3x + 1)}^{2} } = \\ = \frac{(x - 4)(6x + 2 - 3x + 12)}{ {(3x + 1)}^{2} } = \\ = \frac{(x - 4)(3x + 14)}{ {(3x + 1)}^{2} }$

4

$y '= \frac{( {x}^{3} + 4 {x}^{2})'(3x - 1) - (3x - 1)'( {x}^{3} + 4 {x}^{2} ) }{ {(3x - 1)}^{2} } = \\ = \frac{(3 {x}^{2} + 8x)(3x - 1) - 3( {x}^{3} + 4 {x}^{2}) }{ {(3x - 1)}^{2} } = \\ = \frac{9 {x}^{3} - 3 {x}^{2} + 24 {x}^{2} - 8x - 3 {x}^{3} - 12 {x}^{2} }{ {(3x - 1)}^{2} } = \\ = \frac{6 {x}^{3} + 9 {x}^{2} - 8x }{ {(3x - 1)}^{2} }$

5

$y' = \frac{(8x - 2)(2x - 1) - 2(4 {x}^{2} - 2x - 4)}{ {(2x - 1)}^{2} } = \\ = \frac{16 {x}^{2} - 8x - 4x + 2 - 8 {x}^{2} + 4x + 8 }{ {(2x - 1)}^{2} } = \\ = \frac{8 {x}^{2} - 8x + 10 }{ {(2x - 1)}^{2} }$

6

$y' = ( \frac{4x + 1}{ {x}^{2} + 3x } ) '= \\ = \frac{4( {x}^{2} + 3x) - (2x + 3)(4x + 1)}{ {( {x}^{2} + 3x) }^{2} } = \\ = \frac{4 {x}^{2} + 12x - 8 {x}^{2} - 2x - 12x - 3 }{ {x}^{2} {(x + 3)}^{2} } = \\ = \frac{ - 4 {x}^{2} - 2x - 3 }{ {x}^{2} {(x + 3)}^{2} } = - \frac{4 {x}^{2} + 2x + 3 }{ {x}^{2} {(x + 3)}^{2} }$

Задание 5

1

$y' = 7 \times 7 {(4 {x}^{2} - 4x + 4)}^{6} \times (4 {x}^{2} - 4x + 4)' = \\ = 49 {(4 {x}^{2} - x + 4) }^{6} \times (8x - 4)$

2

$y '= (5 {(1 + 2 {x}^{3} - {x}^{2}) }^{ \frac{1}{2} } ) '= \\ = 5 \times \frac{1}{2} {(1 + 2 {x}^{3} - {x}^{2} )}^{ - \frac{1}{2} } \times (1 + 2 {x}^{3} - {x}^{2} ) '= \\ = \frac{5(6 {x}^{2} - 2x) }{2 \sqrt{1 + 2 {x}^{3} - {x}^{2} } } = \frac{5(3 {x}^{2} - x)}{ \sqrt{1 + 2 {x}^{3} - {x}^{2} } }$

3

$y '= ( \sqrt[3]{6 - 10x - 3x + 5 {x}^{2} } ) '= ( {(5 {x}^{2} - 13x + 6)}^{ \frac{1}{3} } ) '= \\ = \frac{1}{3} {(5 {x}^{2} - 13x + 6) }^{ - \frac{2}{3} } \times (5 {x}^{2} - 13x + 6)' = \\ = \frac{10x - 13}{3 \sqrt[3]{ {(5 {x}^{2} - 13x + 6)}^{2} } }$

4

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

5

$y' = ( \sqrt[3]{ \frac{7}{2 {x}^{2} + 1} } ) '= ( \frac{ \sqrt[3]{7} }{ \sqrt[3]{2 {x}^{2} + 1} } )' = \sqrt[3]{7} ( {(2 {x}^{2} + 1) }^{ - \frac{1}{3} } )' = \\ = \sqrt[3]{7} \times ( - \frac{1}{3} ) ({2 {x}^{2} + 1)}^{ - \frac{4}{3} } \times (2 {x}^{2} + 1)' = \\ = - \frac{ \sqrt[3]{7} }{ 3 } \times \frac{1}{ \sqrt[3]{ {(2 {x}^{2} + 1)}^{4} } } \times 4x = - \frac{4x}{3} \times \sqrt[ 3]{ \frac{7}{ {(2 {x}^{2} + 1)}^{4} } }$