Answer
$ \displaystyle \frac{6x^{2}+24x-58}{(2x+4)^{2}}$
Work Step by Step
$\displaystyle \frac{d}{dx}[\frac{f(x)}{g(x)}]= \displaystyle \frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{[g(x)]^{2}}$
$f(x)=3x^{2}-9x+11,$
$f^{\prime}(x)=3(2x)-9=6x-9$
$g(x)=2x+4,$
$g^{\prime}(x)=2$
$\displaystyle \frac{dy}{dx}=\frac{(6x-9)(2x+4,)-(3x^{2}-9x+11)(2)}{(2x+4)^{2}}$
$=\displaystyle \frac{(12x^{2}+24x-18x-36)-(6x^{2}-18x+22)}{(2x+4)^{2}}$
$=\displaystyle \frac{6x^{2}+24x-58}{(2x+4)^{2}}$
