Question

Qual é o valor desta derivada aqui ?$\sqrt[3]{(3x^{2}+6x-2) ^{2} }$

Answer 1

$\\ Y = \sqrt[3]{(3x^2+6x-2)^2} = (3x^2+6x-2)^ \frac{2}{3} \\ \\ Y' = \frac{2}{3} * (3x^2+6x-2)^ \frac{2}{3} ^-^1(3x^2+6x-2)' \\ \\ Y' = \frac{2}{3} (3x^2+6x-2)^ \frac{2}{3} ^-^ \frac{3}{3} (2*3x^2^-^1+6*1-0) \\ \\ Y' = \frac{2}{3} (3x^2+6x-2)^-^ \frac{1}{3}(6x+6) \\ \\ Y' = \frac{2}{3}* \frac{1}{(3x^2+6x-2)^ \frac{1}{3}} *(6x+6) \\ \\ Y' = \frac{2(6x+6)}{3(3x^2+6x-2)^\frac{1}{3}} \\ \\ Y' = \frac{2(6)(x+1)}{3 \sqrt[3]{(3x^2+6x-2)^1} }$

$\\ Y' = \frac{4(x+1)}{\sqrt[3]{3x^2+6x-2} }$