Answer
$$y'=\frac{1}{2\sqrt{x^2-x}}$$
Work Step by Step
$y'=\frac{d}{dx}\cosh^{-1}{\sqrt{x}}$
Using the chain rule:
$y'=\frac{d\cosh^{-1}{\sqrt{x}}}{d\sqrt{x}}
\times \frac{d\sqrt{x}}{dx}$
$=\frac{1}{\sqrt{(\sqrt x)^2-1}}
\times \frac{1}{2\sqrt x}$
$=\frac{1}{2\sqrt x\sqrt{x-1}}$
$=\frac{1}{2\sqrt{x^2-x}}$
