Question

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

Answer 1

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.

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