Answer
$\dfrac{3\sec^2(3x)(x+7)-4\tan(3x)}{(x+7)^5}$
Work Step by Step
$g'(x)=\dfrac{(\tan3x)'(x+7)^4-\tan3x((x+7)^4)'}{(x+7)^8}
$
Apply Chain Rule, we have $(\tan3x)'=(\sec^23x)(3x)'=3\sec^2(3x)$
and $[(x+7)^4]'=4(x+7)^3$
Thus, we get
$g'(x)=\dfrac{3\sec^2(3x)(x+7)^4-4\tan(3x)(x+7)^3}{(x+7)^8}=\dfrac{3\sec^2(3x)(x+7)-4\tan(3x)}{(x+7)^5}$
