Given that $f(x)=-2\sqrt[\phantom{2}]{x3},$ find $f^{\text{'}}(x)$ by forming the difference quotient, $\frac{f(xh)-f(x)}{h},$ and taking the limit as $h0$

$f^{\text{'}}(x)$ does not exist.

$\frac{1}{\sqrt[\phantom{2}]{x3}}$

$-\frac{1}{\sqrt[\phantom{2}]{x3}}$

$\frac{2}{x3}$

$\frac{1}{2\sqrt[\phantom{2}]{x3}}$