Question

найти производные функции

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

Answer 1

1.
$y'=5(3x^4- \frac{5}{ \sqrt[4]{x} }+2 )^{5-1}*(3*4x^{4-1}-5*(- \frac{1}{4} )x^{- \frac{1}{4}- \frac{4}{4}}+0)= \\ \\ =5(3x^4- \frac{5}{ \sqrt[4]{x} }+2 )^4*(12x^3+ \frac{5}{4}x^{- \frac{5}{4} } )= \\ \\ =5(3x^4- \frac{5}{ \sqrt[4]{x} }+2 )^4*(12x^3+ \frac{5}{4x \sqrt[4]{x} } )= \\ \\ =(60x^3+ \frac{25}{4x \sqrt[4]{x} } )(3x^4- \frac{5}{ \sqrt[4]{x} }+2 )^4$

2.
$y'= \frac{1}{ \sqrt[5]{( \frac{1-5x}{1+5x} )^3} }*( \sqrt[5]{( \frac{1-5x}{1+5x} )^3} )'*( \frac{1-5x}{1+5x} )'= \\ \\ = \frac{1}{ \sqrt[5]{( \frac{1-5x}{1+5x} )^3} }* \frac{3}{5}( \frac{1-5x}{1+5x} )^{ \frac{3}{5}- \frac{5}{5} } * \frac{(1-5x)'(1+5x)-(1+5x)'(1-5x)}{(1+5x)^2} = \\ \\ = \frac{1}{ \sqrt[5]{( \frac{1-5x}{1+5x} })^3 }* \frac{3}{5}( \frac{1-5x}{1+5x} )^{- \frac{2}{5} }* \frac{-5(1+5x)-5(1-5x)}{(1+5x)^2}=$
$= \frac{1}{( \frac{1-5x}{1+5x} )^{ \frac{3}{5} }}* \frac{3}{5( \frac{1-5x}{1+5x} )^{ \frac{2}{5} }}* \frac{-5-25x-5+25x}{(1+5x)^2}= \\ \\ = \frac{3}{5( \frac{1-5x}{1+5x} )^{ \frac{3}{5}+ \frac{2}{5} }}* \frac{-10}{(1+5x)^2}= \\ \\ =- \frac{6}{( \frac{1-5x}{1+5x} )*(1+5x)^2}=- \frac{6}{(1-5x)(1+5x)}= \\ \\ =- \frac{6}{1-25x^2}= \frac{6}{25x^2-1}$