Помогите найти производную пожалуйста!!! 35 БАЛЛОВ

Решите задачу:

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

$=\frac{\sqrt[3]{2}}{3\sqrt[3]{[(x-3)^2(x-8)]^2}}*(2(x-2)*[1]*(8-x)-(x-2)^2)=\\\\ =\frac{\sqrt[3]{2}}{3\sqrt[3]{[x-3)^4(x-8)^2}}*[2(x-2)(8-x)-(x-2)^2]=\\\\ =\frac{\sqrt[3]{2}*(x-2)}{3\sqrt[3]{[x-3)^4(x-8)^2}}*[2(8-x)-(x-2)]=\\\\ =\frac{\sqrt[3]{2}*(x-2)}{3\sqrt[3]{[x-3)^4(x-8)^2}}*[16-2x-x+2]=\\\\ =\frac{\sqrt[3]{2}*(x-2)*(18-3x)}{3\sqrt[3]{[x-3)^4(x-8)^2}}=\\\\ =\frac{\sqrt[3]{2}*(x-2)*(x-6)*(-3)}{3\sqrt[3]{[x-3)^4(x-8)^2}}=\\\\ =-\frac{\sqrt[3]{2}*(x-2)*(x-6)}{\sqrt[3]{[x-3)^4(x-8)^2}}\\\\$