Single Variable Calculus, Chapter 7, 7.4-1, Section 7.4-1, Problem 32

Differentiate $\displaystyle f(x) = \frac{1}{1 + \ln x}$ and find the domain of $f$
The denominator of the given function should be greater than zero. So,

$
\begin{equation}
\begin{aligned}
1 + \ln x & > 0 \\
\\
\ln x & > - 1\\
\\
e^{\ln x} & > e^{-1}\\
\\
x &> e^{-1}\\
\\
x &> \frac{1}{e}
\end{aligned}
\end{equation}
$

Therefore, the domain is $\displaystyle \left[ 0, \frac{1}{e} \right)\bigcup \left( \frac{1}{e}, \infty \right)$
Solving for $f'$

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