(i) To find \(f'(x)\), use the quotient rule: if \(f(x) = \frac{u}{v}\), then \(f'(x) = \frac{u'v - uv'}{v^2}\).

Here, \(u = 5\) and \(v = 1 - 3x\).

\(u' = 0\) and \(v' = -3\).

So, \(f'(x) = \frac{0 imes (1 - 3x) - 5 imes (-3)}{(1 - 3x)^2} = \frac{15}{(1 - 3x)^2}\).

Since the derivative is negative, \(f'(x) = \frac{-15}{(1 - 3x)^2}\).

(ii) Since \(15 > 0\) and \((1 - 3x)^2 > 0\) for \(x \geq 1\), \(f'(x) > 0\).

Therefore, \(f\) is an increasing function.