Answer
$\frac{dy}{dx} = -\frac{1}{x^2} $
Work Step by Step
$y = \frac{1}{x} $
$\frac{dy}{dx} = \lim\limits_{ Δx\to 0} \frac{f(x+Δx) - f(x)}{Δx} $
$\frac{dy}{dx} = \lim\limits_{ Δx\to 0} \frac{\frac{1}{x+Δx}-\frac{1}{x}}{Δx} $
$\frac{dy}{dx} = \lim\limits_{ Δx\to 0} \frac{1}{Δx}[\frac{1}{x+Δx} - \frac{1}{x}] $
$\frac{dy}{dx} = \lim\limits_{ Δx\to 0} \frac{1}{∆x} [\frac{x-(x+∆x)}{x(x+∆x)}] $
$\frac{dy}{dx} = \lim\limits_{ Δx\to 0} \frac{1}{∆x} [\frac{-∆x}{x(x+∆x)}] $
$\frac{dy}{dx} = \lim\limits_{ Δx\to 0} [\frac{-1}{x(x+∆x)}] $
$\frac{dy}{dx} = \frac{-1}{x(x+0)} $
$\frac{dy}{dx} = \frac{-1}{x^2} $
$\frac{dy}{dx} = -\frac{1}{x^2} $