## Thomas' Calculus 13th Edition

$\dfrac{1}{3}$
Use derivative formula; $f(x)= x^n \implies f'(x)= n x^{n-1}$ Consider $f(x) = 1 - \dfrac{1}{x}1-x^{-1}$ Now, $f'(x) = 0- (-1)x^{(-1-1)}=f'(x) = x^{-2}= \dfrac{1}{x^2}$ Then , we have $f'(\sqrt 3) = \dfrac{1}{(\sqrt 3)^{2}} = \dfrac{1}{3}$