## Calculus: Early Transcendentals 8th Edition

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