Answer
$f'(x) = -\frac{1}{x}$
Work Step by Step
$f(x) = \ln \frac{1}{x}$
$f(x) = \ln (x^{-1})$
Formula: $\ln x^{a} = a \ln x$
$f(x) = -\ln x$
Differentiate:
$f'(x) = \frac{d(-\ln x)}{dx}$
Formula: $\frac{d}{dx} (\ln x) = \frac{1}{x}$
$f'(x) = -\frac{1}{x}$
