Question

Нужна помощь. Пожалуйста. определить производные dy/dx формулами дифференцирования.

Answer 1

Ответ:

а

$y' = \frac{4 {x}^{3} \times \sqrt{ {x}^{5} - 8x + 2} - \frac{1}{2} {( {x}^{5} - 8x + 2) }^{ - \frac{1}{2} } \times (5 {x}^{4} - 8) \times ( {x}^{4} - 11) }{ {x}^{5} - 8x + 2 } = \\ = \frac{4 {x}^{3} \sqrt{ {x}^{5} - 8x + 2} - \frac{(5 {x}^{4} - 8)( {x}^{4} - 11) }{ 2\sqrt{ {x}^{5} - 8x + 2 } } }{ {x}^{5} - 8x + 2 } = \\ = \frac{4 {x}^{3} }{ \sqrt{ {x}^{5} - 8x + 2} } - \frac{(5 {x}^{4} - 8)( {x}^{4} - 11) }{2 \sqrt{ {( {x}^{5} - 8 + 2)}^{3} } }$

б

$y' = 4 {( {6}^{arctg2x} + arctg5x)}^{3} \times ( ln(6) \times {6}^{arctg2x} \times \frac{1}{1 + 4 {x}^{2} } \times 2 + \frac{5}{1 + 25 {x}^{2} } ) = \\ = 4 {( {6}^{arctg2x} + arctg5x) }^{3} \times ( \frac{2 ln(6) \times {6}^{arctg2x} }{4 {x}^{2} + 1 } + \frac{5}{1 + 25 {x}^{2} } )$

в

$y' = \frac{1}{tg( \frac{2}{ \sqrt{x} } )} \times \frac{1}{ { \cos }^{2} ( \frac{2}{ \sqrt{x} }) } \times 2 \times ( - \frac{1}{2} ) {x}^{ - \frac{3}{2} } = \\ = \frac{ \cos( \frac{2}{ \sqrt{x} } ) }{ \sin( \frac{2}{ \sqrt{x} } ) } \times \frac{1}{ { \cos }^{2}( \frac{ 2 }{ \sqrt{x} } )} - \frac{1}{x \sqrt{x} } = \\ = - \frac{1}{x \sqrt{x} \sin( \frac{2}{ \sqrt{x} } ) \cos( \frac{2}{ \sqrt{x} } ) } = - \frac{2}{2x \sqrt{x} \sin( \frac{2}{ \sqrt{x} } ) \cos( \frac{2}{ \sqrt{x} } ) } = \\ = - \frac{2}{x \sqrt{x} \sin( \frac{4}{ \sqrt{x} } ) }$

г

$y '= \frac{1}{ \sqrt[3]{ \frac{9 - 3 {x}^{4} }{ {x}^{3} + 13x } } } \times \frac{1}{3} {( \frac{9 - 3 {x}^{4} }{ {x}^{3} + 13x }) }^{ - \frac{2}{3} } \times \frac{( - 12 {x}^{3} )( {x}^{3} + 13x) - (3 {x}^{2} + 13)(9 - 3 {x}^{4} ) }{ {( {x}^{3} + 13x)}^{2} } = \\ = \sqrt[3]{ \frac{ {x}^ {3} + 13x }{9 - 3 {x}^{4} } } \times \frac{1}{3} \times \sqrt[3]{ {( \frac{ {x}^{3} + 13x }{9 - 3 {x}^{4} }) }^{2} } \times \frac{ - 12 {x}^{6} - 156 {x}^{4} - 27 {x}^{2} + 9 {x}^{6} - 117 + 39 {x}^{4} }{ {( {x}^{3} + 13x)}^{2} } = \\ = \frac{1}{3} \times \frac{ {x}^{3} + 13x }{9 - 3 {x}^{4} } \times \frac{ - 3{x}^{6} - 117 {x}^{4} - 27 {x}^{2} - 117 }{ {( {x}^{3} + 13x)}^{2} } = \\ = - \frac{3 {x}^{6} + 117 {x}^{4} + 27 {x}^{2} + 117 }{3( {x}^{3} + 13x)(9 - 3 {x}^{4} ) }$

д

$y' = {(x + 3 ln(x)) }^{ \frac{2}{x} }$

по формуле:

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

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

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