Answer
$$f'(x)=\frac{-1}{x^2}$$
$$f''(x)=\frac{3}{x^3}$$
$$f'''(x)=\frac{-12}{x^4}$$
Work Step by Step
The original function $f(x)=\frac{1}x$ can be rewritten as $f(x)=x^{-1}$ due to the definition of the derivatives, $f'(x)=-x^{-2}$ the definition of the derivatives can be applied again: $f''(x)=3x^{-3}$ and for the third derivative, it is applied one last time: $f'''(x)=-12x^{-4}$
