Answer
$$y'=\frac{2}{t(1-\ln t)^2}$$
Work Step by Step
$$y=\frac{1+\ln t}{1-\ln t}$$
We have $$y'=\frac{(1+\ln t)'(1-\ln t)-(1+\ln t)(1-\ln t)'}{(1-\ln t)^2}$$
$$y'=\frac{\frac{1}{t}(1-\ln t)-(1+\ln t)\Big(-\frac{1}{t}\Big)}{(1-\ln t)^2}$$
$$y'=\frac{\frac{1}{t}(1-\ln t)+\frac{1}{t}(1+\ln t)}{(1-\ln t)^2}$$
$$y'=\frac{\frac{1}{t}(1-\ln t+1+\ln t)}{(1-\ln t)^2}$$
$$y'=\frac{\frac{1}{t}(2)}{(1-\ln t)^2}$$
$$y'=\frac{2}{t(1-\ln t)^2}$$
