Calculus and Its Applications, Chapter 1, 1.8, Section 1.8, Problem 66

Find the first through the fourth derivatives of the function $\displaystyle f(x) = \frac{x + 3}{x - 2}$. Be sure to
simplify at each stage before continuing.

By applying Quotient Rule,

$
\begin{equation}
\begin{aligned}
f'(x) &= \frac{(x - 2) \cdot \frac{d}{dx} (x + 3) - (x + 3) \cdot \frac{d}{dx} (x - 2) }{(x - 2)^2}\\
\\
&= \frac{(x - 2)(1) - (x + 3)(1)}{(x - 2)^2}\\
\\
&= \frac{x -2 - x - 3}{(x - 2)^2}\\
\\
&= \frac{-5}{(x - 2)^2}
\end{aligned}
\end{equation}
$


We have $f'(x) = -5(x -2 )^{-2}$, so by applying Chain Rule,

$
\begin{equation}
\begin{aligned}
f''(x) = -5(-2)(x - 2)^{-2 -1} \cdot \frac{d}{dx} ( x- 2) &= 10 ( x - 2)^{-3} (1)\\
\\
&= 10 (x - 2)^{-3}
\end{aligned}
\end{equation}
$


Again,

$
\begin{equation}
\begin{aligned}
f'''(x) = 10 (-3)(x - 2)^{- 3- 1} \cdot \frac{d}{dx} (x - 2) &= -30 (x - 2)^{-4}(1)\\
\\
&= -30 (x - 2)^{-4}
\end{aligned}
\end{equation}
$


Then,

$
\begin{equation}
\begin{aligned}
f^{(4)}x = -30 (-4)(x - 2)^{-4-1} \cdot \frac{d}{dx} (x - 2) &= 120 (x -2)^{-5} (1) \\
\\
&= 120(x - 2)^{-5} \text{ or } \frac{120}{(x - 2)^5}
\end{aligned}
\end{equation}
$