已知对于 $x > q$，有 $y = \frac{2x-1}{\sqrt{x+2}}$。

1. Determine the condition for the function to be defined

For $y = \frac{2x-1}{\sqrt{x+2}}$ to be defined and real, the expression inside the square root in the denominator must be strictly positive.
$x + 2 > 0$
$x > -2$

1. Apply the quotient rule to differentiate the function

Let $u = 2x - 1$ and $v = (x+2)^{\frac{1}{2}}$.
$\frac{du}{dx} = 2$
$\frac{dv}{dx} = \frac{1}{2}(x+2)^{-\frac{1}{2}}$

$\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$
$\frac{dy}{dx} = \frac{2(x+2)^{\frac{1}{2}} - (2x-1)\left[\frac{1}{2}(x+2)^{-\frac{1}{2}}\right]}{x+2}$

2. Simplify the derivative

Multiply the numerator and the denominator by $(x+2)^{\frac{1}{2}}$ to eliminate the negative exponent:
$\frac{dy}{dx} = \frac{2(x+2) - \frac{1}{2}(2x-1)}{(x+2)(x+2)^{\frac{1}{2}}}$
$\frac{dy}{dx} = \frac{2x + 4 - x + \frac{1}{2}}{(x+2)^{\frac{3}{2}}}$
$\frac{dy}{dx} = \frac{x + 4.5}{(x+2)^{\frac{3}{2}}}$
$\frac{dy}{dx} = \frac{2x+9}{2(x+2)^{\frac{3}{2}}}$

1. Recall the derivative from part (b) and analyse its sign for the given domain

From part (b), $\frac{dy}{dx} = \frac{2x+9}{2(x+2)^{\frac{3}{2}}}$.
For $x > -2$, $x + 2 > 0$, so the denominator $2(x+2)^{\frac{3}{2}} > 0$.
Also, $2x + 9 > 2(-2) + 9 = 5 > 0$, so the numerator is positive.

2. Conclude whether the function is increasing or decreasing

Since $\frac{dy}{dx} > 0$ for all $x > -2$, $y$ is an increasing function.