Question

Помогите пожалуйста найти производную!!! 25 БАЛЛОВ
$sqrt[3]{2 {(x - 2)}^{2}(8 - x) } - 1$

Answer 1

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

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