Answer
$\displaystyle \frac{3x^{2}-2x-13}{(3x-1)^{2}}$
Work Step by Step
$\displaystyle \frac{d}{dx}[\frac{f(x)}{g(x)}]= \displaystyle \frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{[g(x)]^{2}}$
$f(x)=(x+3)(x+1)=x^{2}+4x+3$
$f^{\prime}(x)=2x+4=2(x+2)$
$g(x)=3x-1$
$g^{\prime}(x)=3$
$\displaystyle \frac{dy}{dx}=\frac{2(x+2)(3x-1)-(x+3)(x+1)(3)}{(3x+1)^{2}}$
$= \displaystyle \frac{2(3x^{2}+5x-2)-3(x^{2}+4x+3)}{(3x-1)^{2}}$
$=\displaystyle \frac{6x^{2}+10x-4-3x^{2}-12x-9}{(3x-1)^{2}}$
$= \displaystyle \frac{3x^{2}-2x-13}{(3x-1)^{2}}$
