Answer
$f'(x)=-nx^{-(n+1)}$
Work Step by Step
$f(x)=x^{-n}$ , where $n$ is a positive integer.
$f(x)=\frac{1}{x^n}$
Using the product rule to derivate:
$f'(x)=\frac{[0] * [x^n]-[1] * [n * x^{n-1}]}{[x^n]^2}$
$f'(x)=\frac{-n * x^{n-1}}{x^{2n}}$
$f'(x)=\frac{-n}{x^{2n} * x^{1-n}}$
$f'(x)=\frac{-n}{x^{2n+1-n}}$
$f'(x)=\frac{-n}{x^{n+1}}$
$f'(x)=-nx^{-(n+1)}$
