Chain rule with a quotient raised to a power

Write $f(x) = u^5$ with $u = \dfrac{2x+1}{x-3}$. By the chain rule, $f'(x) = 5u^4 \cdot u' = 5\left(\dfrac{2x+1}{x-3}\right)^4 \cdot u'$, so we need $u'$ next.

Using the quotient rule on $u = \dfrac{2x+1}{x-3}$: $u' = \dfrac{2(x-3) - (2x+1)(1)}{(x-3)^2} = \dfrac{2x-6-2x-1}{(x-3)^2} = \dfrac{-7}{(x-3)^2}$. Substituting back gives $f'(x) = 5\left(\dfrac{2x+1}{x-3}\right)^4 \cdot \dfrac{-7}{(x-3)^2} = \dfrac{-35(2x+1)^4}{(x-3)^6}$.

The outer function was u raised to the fifth power, since f was written as u to the fifth with u equal to the quotient. The chain rule required differentiating this power first, giving five u to the fourth times u prime, before substituting the quotient rule result for u prime. The answer is the fifth power, or simply five.