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?

Instead of the "quotient rule" you could write the function as $y= (3x+ 1)^3(2x+ 5)^{-4}$ and use the product rule: $y'= [(3x+ 1)^3]'(2x+ 5)^{-4}+ (3x+ 1)^3[(2x+ 5)^{-3}]'$.  Now, using the "chain rule", $[(3x+ 1)^3]'= 3(3x+ 1)^2(3)= 9(3x+ 1)^2$ and $[(2x+ 5)^{-4}]'= (-4)(2x+ 5)^{-5}(2)= -8(2x+ 5)^{-5}$ so that $y'= 9(3x+ 1)^2(2x+ 5)^{-4}- 8(3x+ 1)^3(2x+ 5)^{-5}$.  We can then write that in terms of fractions again: $\frac{9(3x+ 1)^2}{(2x+ 5)^4}-\frac{8(3x+ 1)^3}{(2x+ 5)^5}$.  We need to multiply both numerator and denominator of the first fraction by 2x+ 5 to get denominator $(2x+ 5)^5$ for both: $\frac{9(3x+ 1)^2(2x+ 4)}{(2x+ 5)^5}- \frac{8(3x+ 1)^3}{(2x+ 5)^5}= \frac{9(3x+ 1)^2(2x+ 5)- 8(3x+1)^3}{(2x+ 5)^5}$. 
