Предмет: Алгебра, автор: maksimkadada

Помогите пожалуйста!!!!! Найти производную функции :

Приложения:

Ответы

Автор ответа: Miroslava227
2

Ответ:

36.5

1

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

2

y '= (5 {(3x - 2)}^{ - 1} ) '=  - 5 {(3x - 2)}^{ - 2}   \times (3x - 2)' =  \\  =  -  \frac{5}{ {(3x - 2)}^{2} }  \times 3 =  -  \frac{15}{ {(3x - 2)}^{2} }

3

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

4

y '=  \frac{3 {x}^{2}  \times  \cos(x)  - ( -  \sin(x)) \times  {x}^{3}  }{ \cos {}^{2} (x) }  =  \\  =  \frac{3 {x}^{2}  \cos(x)  +  {x}^{3} \sin(x)  }{ \cos {}^{2} (x) }

5

y'  = \frac{(3 -  {x}^{2})'(4 + 2x) - (4 + 2x)'(3 -  {x}^{2})  }{ {(4 + 2x)}^{2} }  =  \\  =  \frac{ - 2x(4 + 2 {x}^{} ) - 2(3 -  {x}^{2}) }{ {(2x + 4)}^{2} }  =  \\  =  \frac{ - 8x - 4 {x}^{2} - 6 + 2 {x}^{2}  }{ {(2x + 4)}^{2} }  =  \frac{ - 2 {x}^{2}  - 8x - 6}{ {(2x + 4)}^{2} }  =  \\  =  -  \frac{2( {x}^{2}  + 4x + 3)}{4 {(x + 2)}^{2} }  =  -  \frac{ {x}^{2} + 4x + 3 }{2 {(x + 2)}^{2} }

6

y' =  \frac{(2x - 5)(x - 7) - 1 \times ( {x}^{2} - 5x) }{ {(x - 7)}^{2} }  =  \\  =  \frac{2 {x}^{2} - 14x - 5x + 35 -  {x}^{2}  + 5x }{ {(x - 7)}^{2} }  =  \\  =  \frac{ {x}^{2}  - 14x + 35}{ {(x - 7)}^{2} }

36.6

1

y' =  \frac{3(x - 8) - 1 \times (3x + 5)}{(x - 8) {}^{2} }  =  \\  =  \frac{3x - 24 - 3x - 5}{ {(x - 8)}^{2} }  =  -  \frac{29}{ {(x - 8)}^{2} }

2

y '= (7 {(10x - 3)}^{ - 1} )' =  - 7 {(10x - 3)}^{ - 2}  \times (10x - 3)' =  \\  =  -  \frac{70}{ {(10x - 3)}^{2} }

3

y '=  \frac{4x(1 - 6x) - ( - 6) \times 2 {x}^{2} }{ {(1 - 6x)}^{2} }  =  \\  =  \frac{4x - 24 {x}^{2}  + 12 {x}^{2} }{ {(1 - 6x)}^{2} }  =  -  \frac{12 {x}^{2} - 4x }{ {(1 - 6x)}^{2} }

4

y '=  \frac{ \cos(x) \times x -  \sin(x)  }{ {x}^{2} }  \\

5

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

6

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

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