Question

Помогите найти производную
$\sqrt[3]{2 {(x - 2)}^{2}(8 - x) - 1 }$

Answer 1

$[\sqrt[3]{2(x-2)^2(8-x)-1}]'=[(2(x-2)^2(8-x)-1})^{\frac{1}{3}}]'=\\\\ =\frac{1}{3}*(2(x-2)^2(8-x)-1})^{\frac{1}{3}-1}*[2(x-2)^2(8-x)-1]'=\\\\ =\frac{1}{3}*(2(x-2)^2(8-x)-1})^{-\frac{2}{3}}*([2(x-2)^2(8-x)]'-[1]')=\\\\ =\frac{1}{3}*(2(x-2)^2(8-x)-1})^{-\frac{2}{3}}*(2*[(x-2)^2(8-x)]'-0)=\\\\ =\frac{2}{3}*(2(x-2)^2(8-x)-1})^{-\frac{2}{3}}*[(x-2)^2(8-x)]'=\\\\ =\frac{2}{3}*\frac{1}{(2(x-2)^2(8-x)-1)^{\frac{2}{3}}}*[((x-2)^2)'*(8-x)+(x-2)^2*(8-x)']=$

$=\frac{2}{3\sqrt[3]{[2(x-2)^2(8-x)-1]^2}}* [2(x-2)*1*(8-x)+(x-2)^2*(-1)]=\\\\ =\frac{2}{3\sqrt[3]{[2(x-2)^2(8-x)-1]^2}}* [2(x-2)(8-x)-(x-2)^2]=\\\\ =\frac{2(x-2)}{3\sqrt[3]{[2(x-2)^2(8-x)-1]^2}}* [2(8-x)-(x-2)]=\\\\ =\frac{2(x-2)}{3\sqrt[3]{[2(x-2)^2(8-x)-1]^2}}* [16-2x-x+2]=\\\\ =\frac{2(x-2)(18-3x)}{3\sqrt[3]{[2(x-2)^2(8-x)-1]^2}}=\\\\ =\frac{2(x-2)(x-6)*(-3)}{3\sqrt[3]{[2(x-2)^2(8-x)-1]^2}}=\\\\ =-\frac{2(x-2)(x-6)}{\sqrt[3]{[2(x-2)^2(8-x)-1]^2}}.$