Answer
\[2\]
Work Step by Step
It ia given that :- \[f(x)=\ln (x+\ln x)\;\;\;...(1)\]
Differentiating (1) with respect to $x$ using chain rule
\[f'(x)=\frac{1}{x+\ln x}\cdot (x+\ln x)'\]
\[\Rightarrow f'(x)=\frac{1}{x+\ln x}\cdot \left(1+\frac{1}{x}\right)\]
\[\Rightarrow f'(x)=\frac{1}{x+\ln x}\cdot \left(\frac{x+1}{x}\right)\]
\[\Rightarrow f'(x)=\frac{x+1}{x(x+\ln x)}\]
\[\Rightarrow f'(1)=\frac{1+1}{1(1+\ln 1)}\]
\[\Rightarrow f'(1)=\frac{2}{1}\]
Hence $f'(1)=2$.
