How do you differentiate y=sqrtx+1+1/sqrtx?

$y = \sqrt{x} + 1 + \frac{1}{\sqrt{x}}$
$y = {x}^{\frac{1}{2}} + 1 + {x}^{- \frac{1}{2}}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\frac{1}{2}\right) {x}^{- \frac{1}{2}} - \left(\frac{1}{2}\right) {x}^{- \frac{3}{2}}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \left[\frac{1}{2 \sqrt{x}}\right] - \left[\frac{1}{2 {\left(\sqrt{x}\right)}^{3}}\right]$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\frac{1}{2 \sqrt{x}}\right) \left(1 - \frac{1}{x}\right)$