Answer
$\frac{1}{3}$
Work Step by Step
$f(x)$ = $e^{x}+x$
$\frac{df}{dx}$ = $e^{x}+1$
$\frac{df^{-1}}{dx}$ = $\frac{1}{\frac{df}{dx}}$
$\frac{df^{-1}}{dx}$ = $\frac{1}{e^{x}+1}$
$(\frac{df^{-1}}{dx})_{x=f(\ln{2})}$ = $\frac{1}{e^{\ln2}+1}$ = $\frac{1}{3}$
