Calculus and Its Applications, Chapter 1, 1.3, Section 1.3, Problem 52

Determine the simplified difference quotient of the function $\displaystyle f(x) = \frac{1}{1 - x}$
For $\displaystyle f(x) = \frac{1}{1 - x}$

$
\begin{equation}
\begin{aligned}
f(x + h) &= \frac{1}{1- (x + h)}\\
\\
&= \frac{1}{1 - x - h}
\end{aligned}
\end{equation}
$

Then,

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


Thus,

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