Question

derivada da raiz cúbica de (3x²+6-2)²

Answer 1

$f(x)~=~\sqrt[3]{(3x^2+6x-2)^2}\\\\\\Reescrevendo~o~radical~como~um~expoente~fracionario:\\\\\\f(x)~=~(3x^2+6x-2)^{\frac{2}{3}}\\\\\\\frac{df(x)}{dx}~=~\left((3x^2+6x-2)^{\frac{2}{3}}\right)'\\\\\\Aplicando~a~regra~da~cadeia\\\\Seja:~u(x)~=~3x^2+6x-2\\\\\\\frac{df(x)}{dx}~=~\frac{df(u)}{du}.\frac{u(x)}{dx}\\\\\\\frac{df(x)}{dx}~=~\left(\left(u(x)\right)^{\frac{2}{3}}\right)'.\,(3x^2+6x-2)'\\\\\\\frac{df(x)}{dx}~=~\frac{2}{3}.u^{\frac{2}{3}-1}.\,(3~.~2x^1+6)$

$\frac{df(x)}{dx}~=~\frac{2}{3}.u^{-\frac{1}{3}}.\,(6x+6)\\\\\\\frac{df(x)}{dx}~=~\frac{2}{3}.(3x^2+6x-2)^{-\frac{1}{3}}.\,(6x+6)\\\\\\\frac{df(x)}{dx}~=~\frac{2}{3.(3x^2+6x-2)^{\frac{1}{3}}}.\,(6x+6)\\\\\\\frac{df(x)}{dx}~=~\frac{2.(6x+6)}{3.\sqrt[3]{(3x^2+6x-2)}}\\\\\\\frac{df(x)}{dx}~=~\frac{2.(2x+2)}{\sqrt[3]{(3x^2+6x-2)}}\\\\\\\boxed{\frac{df(x)}{dx}~=~\frac{4x+4}{\sqrt[3]{(3x^2+6x-2)}}}$