• Предмет: Алгебра
  • Автор: vladliss90
  • Вопрос задан 3 года назад

Найдите производные dy/dx заданных функций.

Приложения:

Ответы

Ответ дал: Miroslava227
0

Объяснение:

1

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

2

y '= 4 {( {5}^{arctg3x}  + ctg \frac{x}{2} )}^{3}  \times ( ln(5)  \times  {5}^{arctg3x}  \times  \frac{1}{1 + 9 {x}^{2} }  -  \frac{1}{2 \sin {}^{2} ( \frac{x}{2} ) } ) \\

3

y' =  \frac{1}{2 \sqrt{ {a}^{2}   +  {x}^{2} } }  \times 2x  +   \frac{a}{ \sqrt{1  - \frac{ {x}^{2} }{ {a}^{2} } } }  \times  \frac{1}{a}  =  \\  =  \frac{x}{ \sqrt{ {a}^{2}  +  {x}^{2} } }   +   \frac{a}{ \sqrt{a {}^{2} -  {x}^{2}  }}

4

y' =  \frac{1}{ \sqrt[3]{ \frac{x - 5}{ {x}^{2}  + 4 } } }  \times  \frac{1}{3}  {( \frac{x - 5}{ {x}^{2}  + 4} )}^{ -  \frac{2}{3} }  \times  \frac{(x - 5)'( {x}^{2}  + 4) - ( {x}^{2} + 4)'(x - 5) }{ {( {x}^{2} + 4) }^{2} }  =  \\  =  \sqrt[3]{ \frac{ {x}^{2} + 4 }{x - 5} }  \times  \frac{1}{3}  \times  \sqrt[3]{ {( \frac{ {x}^{2} + 4 }{x - 5} )}^{2} }  \times  \frac{ {x}^{2} + 4 - 2x(x - 5) }{ {( {x}^{2} + 4) }^{2} }  =  \\  =  \frac{1}{3}  \times  \frac{ {x}^{2}  + 4}{x - 5}  \times  \frac{ {x}^{2} + 4 - 2 {x}^{2} + 10x  }{ {( {x}^{2} + 4) }^{2} }  =  \\  =  \frac{ -  {x}^{2}  + 10x + 4}{3(x - 5)( {x}^{2} + 4) }

5

y =  {(x +  {e}^{3x}) }^{ \frac{x + 2}{2} }  \\

y' = ( ln(y)) ' \times y

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

6

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

7

y'_x =  \frac{y'_t}{x'_t}  \\

y'_t = 2t + 1 \\ x'_t = 3 {t}^{2}  \\  \\ y'_x =  \frac{2t + 1}{ 3t {}^{2} }

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