Answer
Does not exist.
Work Step by Step
$g(x)=(x+3)^{\frac{1}{3}}$
$g'(c)=\lim\limits_{x \to c}\frac{f(x)-f(c)}{x-c}$
$g'(-3)=\lim\limits_{x \to -3}\frac{(x+3)^{\frac{1}{3}}-((-3)+3)^{\frac{1}{3}}}{x-(-3)}$
$g'(-3)=\lim\limits_{x \to -3}\frac{(x+3)^{\frac{1}{3}}}{x+3}$
$g'(-3)=\lim\limits_{x \to -3}\frac{1}{(\sqrt[3]{x+3})^2}$
$g'(-3)=\frac{1}{(\sqrt[3]{(-3)+3})^2}$
$g'(-3)=\frac{1}{0}$
Does not exist.
