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