Answer
$\displaystyle \frac{dy}{dx}=-\frac{14}{(3x-1 )^{2}}$
Work Step by Step
$f(x)=2x+4,\displaystyle \ \ \quad g(x)=3x-1,\quad y=\frac{f(x)}{g(x)}$
$f^{\prime}(x)=2\qquad g^{\prime}(x)=3$
$\displaystyle \frac{dy}{dx}=\frac{d}{dx}[\frac{f(x)}{g(x)}]$= ... quotient rule ...
$= \displaystyle \frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{[g(x)]^{2}}$
$=\displaystyle \frac{2(3x-1) - (2x+4)(3)}{(3x-1 )^{2}}$
$=\displaystyle \frac{6x-2 -6x-12}{(3x-1 )^{2}}$
$= \displaystyle \frac{-14}{(3x-1 )^{2}}$
