Предмет: Математика, автор: Аноним

Знайдіть похідні наступних функцій:

Приложения:

Ответы

Автор ответа: Miroslava227
1

Ответ:

1.

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

y' = 3 {x}^{2}  + 2x + 2 + 0 - 2 {x}^{ - 2}     - 2 {x}^{ - 3}  =  \\  = 3 {x}^{2}  + 2x + 2 -  \frac{2}{ {x}^{2} }   -   \frac{2}{ {x}^{3} }

2.

y = ( {x}^{4}  -  \frac{1}{ {x}^{4} } )( {x}^{3}  +  \frac{1}{ {x}^{3} } ) =  \\  =  {x}^{7}  + x -  \frac{1}{x}  -  \frac{1}{ {x}^{7} } =  \\  =   {x}^{7}  + x -  {x}^{ - 1}  -  {x}^{ - 7}

y' = 7 {x}^{6}  + 1 +  {x}^{ - 2}  + 7 {x}^{ - 8}  =  \\  = 7 {x}^{6}  + 1 +  \frac{1}{ {x}^{2} }  +  \frac{7}{ {x}^{8} }

3.

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

4.

y =  {x}^{2}  \times  \sqrt{x}  =  {x}^{ \frac{5}{2} }  \\

y' =  \frac{5}{2}  {x}^{ \frac{3}{2} }  =  \frac{5}{2} x \sqrt{x}  \\

5.

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

6.

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

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