Find the derivative of the function $(ax+b)(cx+d)^{2}$ with respect to $x$,where $a, b, c, d$ are fixed non-zero constants.

Let $f(x) = (ax+b)(cx+d)^{2}$.
Using the product rule $\frac{d}{dx}[u(x)v(x)] = u(x)v'(x) + v(x)u'(x)$:
$f'(x) = (ax+b) \frac{d}{dx}(cx+d)^{2} + (cx+d)^{2} \frac{d}{dx}(ax+b)$
Applying the chain rule to $\frac{d}{dx}(cx+d)^{2} = 2(cx+d) \cdot c = 2c(cx+d)$:
$f'(x) = (ax+b)[2c(cx+d)] + (cx+d)^{2}(a)$
Factoring out $(cx+d)$:
$f'(x) = (cx+d)[2c(ax+b) + a(cx+d)]$
$f'(x) = (cx+d)[2acx + 2bc + acx + ad]$
$f'(x) = (cx+d)(3acx + 2bc + ad)$