Очень нужна определить производные , пользуясь формулами дифференцирования.

Ответ:

SashaKitten0606

08.02.2021 19:09

а)

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

б)

$y' = 5 {( {2}^{arcsin2x} + arccosx)}^{4} \times ( ln(2) \times {2}^{arcsin2x} \times \frac{1}{ \sqrt{1 - 4 {x}^{2} } } \times 2 - \frac{1}{ \sqrt{1 - {x}^{2} } } ) \\$

в)

$y' = \frac{1}{arctg \sqrt{x - 1} } \times \frac{1}{1 + x - 1} \times \frac{1}{2 \sqrt{x - 1} } = \\ = \frac{1}{ 2x\sqrt{x - 1} \times arctg \sqrt{x - 1} }$

г)

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

д)

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

формула:

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

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