Finding the derivative of an integral $$g(x) = \int_{2x}^{6x} \frac{u+2}{u-4}du $$
For finding the $ g'(x)$, would I require to find first the derivative of $\frac{u+2}{u-4}$ 
then Replace the $u$ with 6x and 2x and add them ?
(the 2x would have to flip so the whole term is negative)
If the previous statement is true would the final showdown be the following:
$$ \frac{6}{(2x-4)^2} - \frac{6}{(6x-4)^2}$$  

Let $f(u)=\frac{u+2}{u-4}$, and let $F(u)$ be the antiderivative of $f(u)$. Then
$$
g'(x)=\frac{d}{dx}\int_{2x}^{6x}f(u)du=\frac{d}{dx}\left(F(u)\bigg\vert_{2x}^{6x}\right)=\frac{d}{dx}[F(6x)-F(2x)]=6F'(6x)-2F'(2x)
$$
But $F'(u)=f(u)$. So the above evaluates to
$$
6f(6x)-2f(2x)=6\frac{6x+2}{6x-4}-2\frac{2x+2}{2x-4}=\cdots.
$$
In general,
$$
\frac{d}{dx}\int_{a(x)}^{b(x)}f(u)du=f(b(x))\cdot b'(x)-f(a(x))\cdot a'(x).
$$
