Help clarifying the steps to find the derivative of $y=(3x+1)^3(2x+5)^{-4}$ For the problem $y=(3x+1)^3(2x+5)^{-4}$ do I use the chain, quotient and product rules? If so how do I know what parts to break up and where the rules apply?   For instance would I consider $f(x)$ to be $(3x+1)^3$ and $g(x)$ to be $(2x+5)^{-4}$? If so do I then need to take the product rule of $f(x)$ and the quotient of $g(x)$ and then combine them? Or do I just use one rule?

Hint: You are free to apply either the  quotient rule or the product rule, both of them are convenient to use.

Product rule: Set
  $$f(x)=(3x+1)^3\qquad\text{and}\qquad
g(x)=(2x+5)^{-4}
$$ and consider
  \begin{align*}
\left(f(x)\cdot g(x)\right)^\prime=\left((3x+1)^3\cdot(2x+5)^{-4}\right)^\prime
\end{align*}
Quotient rule: Set 
  \begin{align*}
f(x)=(3x+1)^3\qquad\text{and}\qquad g(x)=(2x+5)^4
\end{align*} and consider
  \begin{align*}
\left(\frac{f(x)}{g(x)}\right)^\prime=\left(\frac{(3x+1)^3}{(2x+5)^4}\right)^\prime
\end{align*}
In  both cases we have  to apply the chain rule.

Recall:
\begin{align*}
\color{blue}{\left(f(x)(g(x))^{-1}\right)^\prime}
&=f^\prime(x)\left(g(x)\right)^{-1}+f(x)(-1)(g(x))^{-2} g^\prime(x)\\
&=\frac{f^\prime(x)}{g(x)}-\frac{f(x)g^\prime(x)}{g(x)^2}\\
&=\frac{f^\prime(x)g(x)-f(x)g^\prime(x)}{g^2(x)}\\
&\color{blue}{=\left(\frac{f(x)}{g(x)}\right)^\prime}
\end{align*}
Tip:  Try both ways and decide which one you prefer.
