Calculus and Its Applications, Chapter 1, 1.4, Section 1.4, Problem 36

Determine the $f'(x)$ of the function $\displaystyle f(x) = \frac{1}{1 - x}$

$
\begin{equation}
\begin{aligned}
\frac{f(x + h) - f(x)}{h} &= \frac{\left[ \frac{1}{1 - (x + h)} \right] - \left[ \frac{1}{1 - x} \right]}{h}\\
\\
&= \frac{\frac{1}{1 - x - h} - \frac{1}{1 - x}}{h}\\
\\
&= \frac{\frac{1 - x - (1 - x - h)}{(1 - x)(1 - x - h)}}{h}\\
\\
&= \frac{1 - x - 1 + x + h}{h(1 - x)(1 - x - h)}\\
\\
&= \frac{h}{h\left[ 1 - x - h - x + x^2 + xh \right]}\\
\\
&= \frac{1}{1 - 2x - h + x^2 + xh}
\end{aligned}
\end{equation}
$

Thus,

$
\begin{equation}
\begin{aligned}
f'(x) = \lim_{h \to 0} \frac{ f(x + h) - f(x) }{h} &= \frac{1}{ 1 - 2x - 0 + x^2 + x(0)}\\
\\
&= \frac{1}{1 - 2x + x^2}
\end{aligned}
\end{equation}
$