$y={x}^{-n}$ where $n$ is any positive integer. Earlier we had found the derivative $\frac{dy}{dx}$ using the definition of the derivative and the fact that the power rule is true for positive integer exponents. Here we will show the same using implicit differentiation and the fact that the power rule is true for positive integer exponents.

$y={x}^{-n}=\frac{1}{{x}^{n}}$

or, ${x}^{n}y=1$

Differentiating both sides with respect to $x$ we get

$(n{x}^{n-1)}(y)+({x}^{n})\frac{dy}{dx}=0$

or, $n{x}^{n-1}{x}^{-n}+{x}^{n}\frac{dy}{dx}=0$

$n{x}^{-1}+{x}^{n}\frac{dy}{dx}=0$

or, ${x}^{n}\frac{dy}{dx}=-n{x}^{-1}$

or, $\frac{dy}{dx}=-n{x}^{-1}{x}^{-n}=-n{x}^{-n-1}$