Question

Оооооооочень прошу пожалуйста помогите найти производную!!)))

Answer 1

1)

$y = ln \sqrt[4]{ \frac{ {x}^{4} + 1}{2 {x}^{4} + 3 } }$

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

2)

$y = ln(3 - x)$

$y''' = ((y')')'$

$y' = ( ln(3 - x) )' = (3 - x)' \frac{1}{3 - x} = - \frac{1}{3 - x} = \frac{1}{x - 3}$

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

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

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