Help clarifying the steps to find the derivative of $y=(3x+1)^3(2x+5)^{-4}$

Question

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?

Answer 1

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.

Answer 2

there is:

$$\left(\frac{u}{v}\right)' = \frac{u'v-uv'}{v^2}$$

your example can be rewritten like:

$$\frac{(3x+1)^3}{(2x+5)^4}$$

and therefore:

$$\frac{9(3x+1)^2\cdot (2x+5)^4-(3x+1)^3\cdot8(2x+5)^3}{(2x+5)^8}$$

Answer 3

You can chain the rules together. Each application of a rule gives you a simpler object to take the derivative of. I would start by using the product rule, giving $$\frac {dy}{dx}=(2x+5)^{-4}\frac d{dx}\left((3x+1)^3\right)+(3x+1)^{3}\frac d{dx}\left((2x+5)^{-4}\right)$$ Now you can use the chain rule on each term, getting $$\frac d{dx}\left((3x+1)^3\right)=3(3x+1)^2\frac d{dx}(3x+1)$$ and similarly for the other term. Probably you can do the last derivative on the right easily. Then just gather it all together. You could also use the quotient rule to start, but I find that a little more difficult.

Answer 4

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: $\frac{9(3x+ 1)^2(2x+ 4)}{(2x+ 5)^5}- \frac{8(3x+ 1)^3}{(2x+ 5)^5}= \frac{9(3x+ 1)^2(2x+ 5)- 8(3x+1)^3}{(2x+ 5)^5}$.