Математика 12 дней назад vladzorkov17

Найдите производные \frac{dx}{dx} заданных функций

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Ответ
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Miroslava227

Ответ:

1

y '=  \frac{(5x + 4)' \times ( {x}^{2}  + 3x + 1) - ( {x}^{2}  + 3x + 1)'(5x + 4)}{ {( {x}^{2} + 3x + 1) }^{2} }  =  \\  =  \frac{5( {x}^{2} + 3x + 1) - (2x + 3)(5x + 4) } { {( {x}^{2}  + 3x +1 )}^{2} }  =  \\  =  \frac{5 {x}^{2} + 15x + 5 - 10 {x}^{2}  - 8x - 15x - 12 }{ {( {x}^{2}  + 3x +1 )}^{2} }  =  \\  =  \frac{ - 5 {x}^{2}  - 8x - 7}{ {( {x}^{2} + 3x + 1) }^{2} }  =  -  \frac{5 {x}^{2} + 8x + 7 }{ {( {x}^{2}  + 3x + 1)}^{2} }

2

y' =  4{( {8}^{ctg {}^{2} x} -  \frac{3}{ \sin(x) } ) }^{3}  \times ( {8}^{ {ctg}^{2} x}  - (3 {( \sin(x)) }^{ - 1} ) '=  \\  = 4 {( {8}^{ {ctg}^{2}x }  -  \frac{3}{ \sin(x) }) }^{3}  \times ( ln(8)  \times  {8}^{ {ctg}^{2}x }  \times 2ctgx \times ( -  \frac{1}{ \sin {}^{2} (x) } ) + 3 {( \sin(x) )}^{ - 2}  \times  \cos(x))   =  \\  = 4 {( {8}^{ {ctg}^{2} x}  -  \frac{3}{ \sin(x) } )}^{3}  (  -  \frac{ ln(8) \times  {8}^{  {ctg}^{2} x } ctgx }{ \sin {}^{2} (x) }  +  \frac{3 \cos(x) }{ \sin {}^{2} (x) } )

3

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

4

y =  ln( \sqrt[7]{ \frac{ {x}^{3}   +  2}{ {x}^{3}  - 2} } )  \\

y '=  \frac{1}{ \sqrt[7]{ \frac{ {x}^{3}  + 2}{ {x}^{3} - 2 } } }  \times  \frac{1}{7}  {( \frac{ {x}^{3} + 2 }{ {x}^{3} - 2 } )}^{ -  \frac{6}{7} }  \times  \frac{( {x}^{3} + 2)'( {x}^{3} - 2) - ( {x}^{3}    - 2)'( {x}^{3} + 2) }{ {( {x}^{3}  - 2)}^{2} }  =  \\  =  \sqrt[7]{ \frac{ {x}^{3}  - 2}{ {x}^{3}  + 2} }  \times  \frac{1}{7}  \times  \sqrt[7]{ {( \frac{ {x}^{3}  - 2}{ {x}^{3}  + 2} )}^{6} }  \times  \frac{3 {x}^{2}( {x}^{3}  - 2) - 3 {x}^{2} ( {x}^{3}   + 2)}{ {( {x}^{3}  - 2)}^{2} }  =  \\  =  \frac{1}{7}  \times  \frac{ {x}^{3} - 2 }{ {x}^{3}  + 2}  \times  \frac{3 {x}^{2}( {x}^{3}  - 2 -  {x}^{3}  - 2) }{ {( {x}^{3}  - 2)}^{2} }  =  \\  =  \frac{3 {x}^{2}  \times ( - 4)}{7( {x}^{3} - 2)( {x}^{3}   + 2)}  =  -  \frac{12 {x}^{2} }{ {x}^{6} - 4 }

5

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

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

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

6.

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

7.

y'_x =  \frac{yt}{xt}  \\  \\ y'_t = 2 \cos(2t)  - 4 \sin(2t)  \\  \\ x'_t =  \frac{1}{ \cos {}^{2} (t) }  \\  \\ y'_x = (2 \cos(2t)  - 4 \sin(2t) )  \times  \cos {}^{2} (t)  =  \\  = (2 \cos {}^{2} (t)   - 2\sin {}^{2} (t)  - 8 \sin(t)  \cos(t) ) \times  \cos {}^{2} (t)  =  \\  = 2 \cos {}^{4} (t)  - 2 \sin {}^{2} (t)  \cos {}^{2} (t)  - 8 \sin(t)  \cos {}^{3} (t)