Calculus and Its Applications, Chapter 1, 1.8, Section 1.8, Problem 60

Find $y'''$ of $\displaystyle y = \frac{3x - 1}{2x + 3}$
By applying Quotient Rule,

$
\begin{equation}
\begin{aligned}
y' &= \frac{(2x + 3) \cdot \frac{d}{dx} (3x - 1) - (3x - 1) \cdot \frac{d}{dx} (2x + 3) }{(2x + 3)^2}\\
\\
y' &= \frac{(2x + 3)(3) - (3x - 1)(2)}{(2x + 3)^2} = \frac{6x + 9 -6x + 2}{(2x + 3)^2} = \frac{11}{(2x + 3)^2}
\end{aligned}
\end{equation}
$


We have $y' = 11(2x + 3)^{-2}$, so

$
\begin{equation}
\begin{aligned}
y'' &= \frac{d}{dx} \left[ 11(2x + 3)^{-2} \right] \\
\\
&= 11(-2)(2x +3)^{-2-1} \cdot \frac{d}{dx} (2x +3) \\
\\
&= -22 (2x + 3)^{-3} (2) \\
\\
&= -44 (2x + 3)^{-3}
\end{aligned}
\end{equation}
$


Again, by applying Chain Rule,

$
\begin{equation}
\begin{aligned}
y''' &= \frac{d}{dx} \left[ -44 ( 2x + 3 )^{-3} \right]\\
\\
&= -44(-3)(2x + 3)^{-3-1} \cdot \frac{d}{dx} (2x + 3)\\
\\
&= 132 (2x + 3)^{-4} (2) = 264(2x + 3)^{-4} \text{ or } \frac{264}{(2x + 3)^4}
\end{aligned}
\end{equation}
$