Single Variable Calculus, Chapter 3, 3.2, Section 3.2, Problem 24

Find the derivative of $\displaystyle f(x) = \frac{3 + x}{1 - 3x}$ using the definition and the domain of its derivative.

Using the definition of derivative


$
\begin{equation}
\begin{aligned}

\qquad f'(x) &= \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}
&&
\\
\\
\qquad f'(x) &= \lim_{h \to 0} \frac{\displaystyle \frac{3 + x + h}{1 - 3 (x + h)} - \left( \frac{3 + x}{1 - 3x} \right)}{h}
&& \text{Substitute $g(x + h)$ and $g(x)$}
\\
\\
\qquad f'(x) &= \lim_{h \to 0} \frac{\displaystyle \frac{3 + x + h}{1 - 3x - 3h} - \left( \frac{3 + x}{1 - 3x} \right)}{h}
&& \text{Get the LCD of the numerator}
\\
\\
\qquad f'(x) &= \lim_{h \to 0} \frac{(3 + x + h)(1 - 3x) - (1 - 3x - 3h)(3 + x)}{(h)(1 - 3x - 3h)(1 - 3x)}
&& \text{Expand the equation}
\\
\\
\qquad f'(x) &= \lim_{h \to 0} \frac{\cancel{3} + \cancel{x} h - \cancel{ 9x} - \cancel{3x^2} - \cancel{3xh} - \cancel{3} - \cancel{x} + \cancel{9x} + \cancel{3x^2} + 9h + \cancel{3xh}}{(h)(1 - 3x - 3h)(1 - 3x)}
&& \text{Combine like terms}
\\
\\
\qquad f'(x) &= \lim_{h \to 0} \frac{10 \cancel{h}}{\cancel{(h)} (1 - 3x - 3h)(1 - 3x)}
&& \text{Cancel out like terms}
\\
\\
\qquad f'(x) &= \lim_{h \to 0} \frac{10}{(1 - 3x - 3h)(1 - 3x)} = \frac{10}{(1 - 3x - 3(0))(1 - 3x)}
&& \text{Evaluate the limit}
\\
\\
\end{aligned}
\end{equation}
$


$\qquad \fbox{$f'(x) = \displaystyle \frac{10}{(1 - 3x)^2}$}$

Both $f(x)$ and $f'(x)$ are rational function that are continuous on $0 > 1 - 3x > 0$.

$\displaystyle \begin{array}{cc}
0 & > 1 -3x > 0 \\
x & > \frac{1}{3} \text{ and } x < \frac{1}{3}
\end{array} $

Therefore,

The domain of $f(x) = \displaystyle \frac{3 + x}{1 - 3x}$ is $\displaystyle \left(-\infty, \frac{1}{3} \right) \bigcup \left( \frac{1}{3}, \infty) \right)$

The domain of $f'(x) = \displaystyle \frac{10}{(1 - 3x)^2}$ is $\left(-\infty, \frac{1}{3} \right) \bigcup \left( \frac{1}{3}, \infty \right)$