Single Variable Calculus, Chapter 7, 7.4-1, Section 7.4-1, Problem 10

Determine the function $\displaystyle f(t) = \frac{1 + \ln t}{1 - \ln t}$

$\begin{equation} \begin{aligned} f'(t) &= \frac{d}{dt} \left( \frac{1 + \ln t}{1 - \ln t} \right)\\ \\ f'(t) &= \frac{(1- \ln t) \frac{d}{dt} (1 + \ln t) - (1 + \ln t) \frac{d}{dt} (1 - \ln t) }{(1 - \ln t)^2}\\ \\ f'(t) &= \frac{(1 - \ln t) \left( \frac{1}{t} \right) - (1+\ln t) \left( - \frac{1}{t} \right)}{(1-\ln t)^2}\\ \\ f'(t) &= \frac{\frac{1}{t} - \cancel{\frac{1}{t}\ln t} + \frac{1}{t} + \cancel{\frac{1}{t}\ln t} }{(1 - \ln t)^2}\\ \\ f'(t) &= \frac{\frac{2}{t}}{(1-\ln t)^2}\\ \\ f'(t) &= \frac{2}{t(1 - \ln t)^2} \end{aligned} \end{equation}$