Answer
$\displaystyle \frac{x}{x^{2}-4}$
Work Step by Step
By theorem 5.3/2: $\displaystyle \frac{d}{dx}[\ln u]=\frac{1}{u}\cdot\frac{du}{dx}=\frac{u^{\prime}}{u},\ \ \ (u>0)$
Here, $u(x)=\sqrt{x^{2}-4}= (x^{2}-4)^{1/2}$
$\displaystyle \frac{du}{dx}$=...chain rule...=$\displaystyle \frac{1}{2}(x^{2}-4)^{-1/2}\cdot 2x=\frac{x}{\sqrt{x^{2}-4}}$
So,
$\displaystyle \frac{d}{dx}[\ln(\sqrt{x^{2}-4})]=\frac{1}{\sqrt{x^{2}-4}}\cdot\frac{x}{\sqrt{x^{2}-4}}=\frac{x}{x^{2}-4}$
