• Предмет: Алгебра
  • Автор: ilyakiselev97
  • Вопрос задан 1 год назад

Необходимо решить карточку за сегодняшний день, желательно быстрее!)

Приложения:

Ответы

Ответ дал: Miroslava227
1

Ответ:

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

2)y' =  -  \frac{15}{3}  {x}^{14}  =  - 5 {x}^{14}

3)y' =  - 6 {x}^{2}  + 12 \times  \frac{1}{2}  {x}^{ -  \frac{1}{2} }  =  - 6 {x}^{2}  +  \frac{6}{ \sqrt{x} }

4)y' = 7 \times ( -  \frac{1}{4} ) {x}^{ -  \frac{5}{4} }   + 3 {x}^{ - 2}  =  -  \frac{7}{4 \sqrt[4]{ {x}^{5} } }  +  \frac{3}{ {x}^{2} }

5)y' = 1 \times  {x}^{3}  + 3 {x}^{2} (x - 6) =  {x}^{3}  + 3 {x}^{3}  - 18 {x}^{2}  = 4 {x}^{3}  - 18 {x}^{2}

6)y' = ( {x}^{ \frac{5}{2}  }  + 1)' =  \frac{5}{2}  {x}^{ \frac{3}{2} }  = 2.5x \sqrt{x}

7)y' =  \frac{1}{2}  {(6x + 1)}^{ -  \frac{1}{2} } \times 6 \times  ( {x}^{4}  - 5) + 4 {x}^{3}  \sqrt{6x + 1}  =  \frac{3({x}^{4} - 5) }{ \sqrt{6x + 1} }  + 4 {x}^{3}  \sqrt{6x + 1}

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

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

10)y' =  \frac{3 {x}^{2}(2x - 3) - 2 {x}^{3}  }{ {(2x - 3)}^{2} }  =  \frac{6 {x}^{3} - 9 {x}^{2}   - 2 {x}^{3} }{ {(2x - 3)}^{2} }  =  \frac{4 {x}^{3} - 9 {x}^{2}  }{ {(2x - 3)}^{2} }

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

12)y' =  \frac{2 {x}^{5} (3x - 2) - 3( \frac{ {x}^{6} }{3}  + 2)}{ {(3x - 2)}^{2} }  =  \frac{6 {x}^{6}  - 4 {x}^{5} -  {x}^{6}  - 6 }{ {(3x - 2)}^{2} }  =  \frac{5 {x}^{6}  - 4 {x}^{5}  - 6}{ {(3x - 2)}^{2} }

13)y' =  \frac{15 {x}^{2} {(x - 4) - 2(x - 4) }^{2}5 {x}^{3}   }{ {(x - 4)}^{4} }  =  \frac{(x - 4)(15 {x}^{2} - 10 {x}^{4} - 40 {x}^{3} )  }{ {(x - 4)}^{4} }  =  \frac{15 {x}^{2} - 10 {x}^{4} - 40 {x}^{3}   }{ {(x - 4)}^{3} }

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

15)y' = ((3x - 6 -  {x}^{2}  + 2x)( {x}^{2}  + 2x + 3x + 6))' = (( -  {x}^{2}  + 5x - 6)( {x}^{2}  + 5x + 6))' = ( - 2x + 5)( {x}^{2}  + 5x + 6) + (2x + 5)( -  {x}^{2}  + 5x - 6) =  - 2 {x}^{3}  - 10 {x}^{2}  - 12x + 5 {x}^{2}  + 25x + 30 - 2 {x}^{3}  + 10 {x}^{2}  - 12x - 5 {x}^{2}  + 25x - 30 =  - 4 {x}^{3}  + 10 {x}^{2}  + 26x

16)f'(x) =  \frac{2}{ \sqrt{x} }  -  \frac{1}{10 {x}^{2} }

f'( \frac{1}{9}) =   \frac{2}{ \sqrt{ \frac{1}{9} } }  -  \frac{1}{10 \times  \frac{1}{81} }  = 2 \times 3 -  \frac{81}{10}  = 6 -  \frac{81}{10}  =  \frac{60 - 81}{10}  =  -  \frac{21}{10}  =  - 2.1

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