Answer
$$y'=\frac{1}{x(1+\ln x)^2}$$
Work Step by Step
$$y=\frac{\ln x}{1+\ln x}$$
Recall the following Derivative Rules: $$\frac{d}{dt}(\ln t)=\frac{1}{t}$$ $$\frac{d}{dt}\Big(\frac{u}{v}\Big)=\frac{v\frac{du}{dt}-u\frac{dv}{dt}}{v^2}$$
Therefore, we have $$y'=\frac{(\ln x)'(1+\ln x)-(\ln x)(1+\ln x)'}{(1+\ln x)^2}=\frac{\frac{1}{x}(1+\ln x)-\ln x\times\frac{1}{x}}{(1+\ln x)^2}$$
$$y'=\frac{\frac{1}{x}+\frac{1}{x}\ln x-\frac{1}{x}\ln x}{(1+\ln x)^2}=\frac{\frac{1}{x}}{(1+\ln x)^2}$$
$$y'=\frac{1}{x(1+\ln x)^2}$$
