\(\lim\limits_{x\rightarrow0}\dfrac{f\left(x\right)-f\left(0\right)}{x-0}=\lim\limits_{x\rightarrow0}\dfrac{\sqrt[3]{4x^2+8}-\sqrt[]{8x^2+4}}{x^2}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt[3]{4x^2+8}-2+2-\sqrt[]{8x^2+4}}{x^2}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{4x^2}{\sqrt[3]{\left(4x^2+8\right)^2}+2\sqrt[3]{4x^2+8}+4}-\dfrac{8x^2}{2+\sqrt[]{8x^2+4}}}{x^2}\)
\(=\lim\limits_{x\rightarrow0}\left(\dfrac{4}{\sqrt[3]{\left(4x^2+8\right)^2}+2\sqrt[3]{4x^2+8}+4}-\dfrac{8}{2+\sqrt[]{8x^2+4}}\right)\)
\(=\dfrac{4}{\sqrt[3]{8^2}+2\sqrt[3]{8}+4}-\dfrac{8}{2+2}=-\dfrac{5}{3}\)
Vậy \(f'\left(0\right)=-\dfrac{5}{3}\)
