Предмет: Алгебра, автор: isqqqzhin

ДАМ МАКСИМАЛЬНЫЕ БАЛЛЫ. Алгебра

Приложения:

Ответы

Автор ответа: Miroslava227
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)


isqqqzhin: Огромное спасибо !
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