Answer
$\displaystyle \frac{1}{\sqrt{x}(\sqrt{x}+1)^{2}}$
Work Step by Step
$\displaystyle \frac{d}{dx}[\frac{f(x)}{g(x)}]= \displaystyle \frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{[g(x)]^{2}}$
$f(x)=\sqrt{x}-1=x^{1/2}-1,$
$f^{\prime}(x)=\displaystyle \frac{1}{2}x^{-1/2}=\frac{1}{2\sqrt{x}}$
$g(x)=\sqrt{x}+1=x^{1/2}+1,$
$g^{\prime}(x)=\displaystyle \frac{1}{2}x^{-1/2}=\frac{1}{2\sqrt{x}}$
$\displaystyle \frac{dy}{dx}=\frac{(\frac{1}{2\sqrt{x}})(\sqrt{x}+1)-(\sqrt{x}-1)(\frac{1}{2\sqrt{x}})}{(\sqrt{x}-1)^{2}}$
$=\displaystyle \frac{1}{2\sqrt{x}}\cdot\frac{\sqrt{x}+1-(\sqrt{x}-1)}{(\sqrt{x}+1)^{2}}$
$=\displaystyle \frac{2}{2\sqrt{x}(\sqrt{x}+1)^{2}}$
$=\displaystyle \frac{1}{\sqrt{x}(\sqrt{x}+1)^{2}}$
