已知函数定义为 $g(x) = \frac{x^2 - 4}{x^2 + 6}$，$x > 0$。

1. Differentiate $g(x)$ using the quotient rule

$g'(x) = \frac{(x^2 + 6)\frac{d}{dx}(x^2 - 4) - (x^2 - 4)\frac{d}{dx}(x^2 + 6)}{(x^2 + 6)^2}$
$g'(x) = \frac{(x^2 + 6)(2x) - (x^2 - 4)(2x)}{(x^2 + 6)^2}$

2. Simplify the derivative

$g'(x) = \frac{2x^3 + 12x - (2x^3 - 8x)}{(x^2 + 6)^2}$
$g'(x) = \frac{20x}{(x^2 + 6)^2}$

3. Analyse the sign of the derivative for the given domain

For $x > 0$, the numerator $20x > 0$.
The denominator $(x^2 + 6)^2 > 0$ for all real $x$.
Therefore, $g'(x) = \frac{20x}{(x^2 + 6)^2} > 0$ for all $x > 0$.

4. Conclude the nature of the function

Since $g'(x) > 0$ for $x > 0$, $g(x)$ is an increasing function.

1. Find the derivative of $g(x)$ using the quotient rule (from part i)

$g(x) = \frac{x^2 - 4}{x^2 + 6}$
$g'(x) = \frac{(x^2 + 6)(2x) - (x^2 - 4)(2x)}{(x^2 + 6)^2}$
$g'(x) = \frac{2x^3 + 12x - 2x^3 + 8x}{(x^2 + 6)^2} = \frac{20x}{(x^2 + 6)^2}$

2. Evaluate $g'(x)$ at $x = 2$

When $x = 2$,
$\frac{dy}{dx} = g'(2) = \frac{20(2)}{(2^2 + 6)^2} = \frac{40}{100} = 0.4$

3. Use the chain rule to find the rate of increase of the $x$-coordinate

$\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}$
$0.04 = 0.4 \times \frac{dx}{dt}$
$\frac{dx}{dt} = \frac{0.04}{0.4} = 0.1$