Answer
$\frac{1}{\sqrt{x^2-1}}$
Work Step by Step
$\frac{d}{dx}\ln(x+\sqrt{x^2-1})$
$=\frac{1}{x+\sqrt{x^2-1}}*\frac{d}{dx}(x+\sqrt{x^2-1})$
$=\frac{1}{x+\sqrt{x^2-1}}*\frac{d}{dx}(x+(x^2-1)^\frac{1}{2})$
$=\frac{1}{x+\sqrt{x^2-1}}*(1+\frac{1}{2}(x^2-1)^{-\frac{1}{2}}\frac{d}{dx}(x^2-1))$
$=\frac{1}{x+\sqrt{x^2-1}}*(1+\frac{1}{2\sqrt{x^2-1}}*2x)$
$=\frac{1}{x+\sqrt{x^2-1}}*(1+\frac{x}{\sqrt{x^2-1}})$
$=\frac{1}{x+\sqrt{x^2-1}}*(\frac{\sqrt{x^2-1}}{\sqrt{x^2-1}}+\frac{x}{\sqrt{x^2-1}})$
$=\frac{1}{x+\sqrt{x^2-1}}*\frac{\sqrt{x^2-1}+x}{\sqrt{x^2-1}}$
$=\frac{1}{\sqrt{x^2-1}}$
