#### Answer

$$1$$

#### Work Step by Step

$$\eqalign{
& f\left( x \right) = {x^{1/x}} \cr
& {\text{taking the natural logarithm of both sides of the equation}} \cr
& \ln f\left( x \right) = \ln {x^{1/x}} \cr
& {\text{using logarithmic properties}} \cr
& \ln f\left( x \right) = \frac{1}{x}\ln x \cr
& {\text{differentiate both sides}} \cr
& \frac{d}{{dx}}\left[ {\ln f\left( x \right)} \right] = \frac{d}{{dx}}\left[ {\frac{{\ln x}}{x}} \right] \cr
& {\text{use quotient rule}} \cr
& \frac{d}{{dx}}\left[ {\ln f\left( x \right)} \right] = \frac{{x\frac{d}{{dx}}\left[ {\ln x} \right] - \ln x\frac{d}{{dx}}\left[ x \right]}}{{{x^2}}} \cr
& {\text{solving derivatives}} \cr
& \frac{{f'\left( x \right)}}{{f\left( x \right)}} = \frac{{x\left( {1/x} \right) - \ln x\left( 1 \right)}}{{{x^2}}} \cr
& \frac{{f'\left( x \right)}}{{f\left( x \right)}} = \frac{{1 - \ln x}}{{{x^2}}} \cr
& f'\left( x \right) = f\left( x \right)\left( {\frac{{1 - \ln x}}{{{x^2}}}} \right) \cr
& {\text{where }}f\left( x \right) = {x^{1/x}} \cr
& f'\left( x \right) = {x^{1/x}}\left( {\frac{{1 - \ln x}}{{{x^2}}}} \right) \cr
& {\text{calculate }}f'\left( 1 \right) \cr
& f'\left( 1 \right) = {\left( 1 \right)^{1/1}}\left( {\frac{{1 - \ln 1}}{{{1^2}}}} \right) \cr
& {\text{Simplify}} \cr
& f'\left( 1 \right) = 1\left( {\frac{{1 - 0}}{1}} \right) \cr
& f'\left( 1 \right) = 1 \cr} $$