Question

ДАЮ 15, НАЙТИ ПРОИЗВОДНУЮ

Answer 1

Ответ:

$b)y' = \frac{1}{ \sqrt{ \frac{8 - 3 {x}^{2} }{ {x}^{3} - 8x } } } \times \frac{1}{2} {( \frac{8 - 3 {x}^{2} }{ {x}^{3} - 8x}) }^{ - \frac{ 1}{2} } \times \frac{ - 6x( {x}^{3} - 8x) - (3 {x}^{2} - 8)(8 - 3 {x}^{2}) }{ {( {x}^{3} - 8x) }^{2} } = \sqrt{ \frac{ {x}^{3} - 8x}{8 - 3 {x}^{2} } } \times \frac{1}{2} \times \sqrt{ \frac{ {x}^{3} - 8x}{8 - 3 {x}^{2} } } \times \frac{ - 6 {x}^{4} + 48 {x}^{2} - 24 {x}^{2} + 9 {x}^{3} + 64 - 24 {x}^{2} }{ {( {x}^{3} - 8x) }^{2} } = \frac{1}{2} \times \frac{ {x}^{3} - 8x }{8 - 3 {x}^{2} } \times \frac{ - 6 {x}^{4} + 9 {x}^{3} + 64 }{ {( {x}^{3} - 8x) }^{2} } = \frac{9 {x}^{3} - 6 {x}^{4} + 64 }{(8 - 3 {x}^{2} )( {x}^{3} - 8x)}$

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