Answer
$$g'(x)=\frac{3\sec^23x(x+7)-4\tan3x}{(x+7)^5}$$
Work Step by Step
$$g(x)=\frac{\tan 3x}{(x+7)^4}$$
The derivative of $g(x)$ is $$g'(x)=\frac{(\tan3x)'(x+7)^4-\tan3x\Big((x+7)^4\Big)'}{(x+7)^8}$$
According to the Chain Rule, we have $(\tan3x)'=\sec^23x(3x)'=3\sec^23x$
and $\Big((x+7)^4\Big)'=4(x+7)^3(x+7)'=4(x+7)^3$
Therefore, $$g'(x)=\frac{3\sec^23x(x+7)^4-4\tan3x(x+7)^3}{(x+7)^8}$$
$$g'(x)=\frac{(x+7)^3\Big(3\sec^23x(x+7)-4\tan3x\Big)}{(x+7)^8}$$
$$g'(x)=\frac{3\sec^23x(x+7)-4\tan3x}{(x+7)^5}$$
