• Предмет: Алгебра
  • Автор: lipa45
  • Вопрос задан 2 года назад

Очень срочно плачу много​

Приложения:

Ответы

Ответ дал: Miroslava227
2

Ответ:

1.

y '= 8 \times 5 {x}^{4}  + 3 - 0 = 40 {x}^{4}  + 3

2.

y' =  \cos(x)  - 5 \sin(x)

3.

y' = 7 \times ( - 3) {x}^{ - 4}  +  \frac{6}{ \cos {}^{2} (x)  }  =  \\  =  -  \frac{21}{ {x}^{4} }  +  \frac{6}{ \cos {}^{2} (x) }

4.

y '= 12 {x}^{11}  + 24 {x}^{2}  - 4x +  \sin(x)

5.

y' =  - 14 \times ( - 4) {x}^{ - 5}  + 6 - 0   =  \frac{56}{ {x}^{5} }  + 6 \\

6.

y '= (2 {x}^{ \frac{1}{2} }  - 12 {x}^{ - 3} ) '= 2 \times  \frac{1}{2}  {x}^{ -  \frac{1}{2} }  - 12 \times ( - 3) {x}^{ - 4}  =  \\  =  \frac{1}{ \sqrt{x} }  +  \frac{36}{ {x}^{4} }

7.

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

8.

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

9.

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

10.

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

11.

y '=  \frac{(2 {x}^{4}  + 4) '\times 8x - (8x) '\times (2 {x}^{4} + 4) }{ {(8x)}^{2} }  =  \\  =  \frac{8 {x}^{3} \times 8x - 8(2 {x}^{4}  + 4) }{64 {x}^{2} }  =  \\  =  \frac{64 {x}^{4} - 16 {x}^{4} - 32  }{64 {x}^{2} }  =  \frac{48 {x}^{4} - 32 }{64{x}^{2} }  =  \\  =  \frac{3 {x}^{4} - 2 }{4{x}^{2} }

12.

y '= ( {x}^{7}  -6 )'( {x}^{3}  - 4x) + ( {x}^{3}  - 4x)'( {x}^{7}  - 6) =  \\  = 7 {x}^{6} ( {x}^{3}  - 4x) + (3 {x}^{2}  - 4)( {x}^{7}  - 6) =  \\  = 7 {x}^{9}  - 28 {x}^{7}  + 3 {x}^{9}  - 18 {x}^{2}  - 4 {x}^{7}  + 24 =  \\  = 10 {x}^{9}  - 32 {x}^{4}   - 18 {x}^{2}   + 24

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