Single Variable Calculus, Chapter 3, Review Exercises, Section Review Exercises, Problem 44

Determine $f^{(n)} (x)$ if $\displaystyle f(x) = \frac{1}{(2-x)}$
Rewrite $f(x)$ to $f(x) = (2-x)^{-1}$, then solve for the 1st derivative.
We have,



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

Solving for the 2nd derivative




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


Solving for the 3rd derivative



$
\begin{equation}
\begin{aligned}
f'''(x) &= 2 \frac{d}{dx} (2-x)^{-3}\\
\\
f'''(x) &= (2)(-3)(2-x)^{-4}\frac{d}{dx}(2-x)\\
\\
f'''(x) &= -6(2-x)^{-4}(-1)\\
\\
f'''(x) &= 6(2-x)^{-4}
\end{aligned}
\end{equation}
$


By solving the first, second and third derivative of the function. We get the pattern,
$f^{(n)} = n!(2-x)^{(-1-n)}$