## University Calculus: Early Transcendentals (3rd Edition)

$r' = \frac{1}{\sqrt x}(1-\frac{1}{x})$
$r = 2(\frac{1}{\sqrt x} +\sqrt x)$ ($\theta$ is replaced with x for convinience.) $r = 2(x^{-\frac{1}{2}}+ x^{\frac{1}{2}})$ $r' = 2(-\frac{1}{2}x^{-\frac{3}{2}} + \frac{1}{2}x^{-\frac{1}{2}})$ $r' = -x^{-\frac{3}{2}} + x^{-\frac{1}{2}}$ $r' = x^{-\frac{1}{2}}(1-x^{-1})$ $r' = \frac{1}{\sqrt x}(1-\frac{1}{x})$