(N/A) Let $f(x) = (ax + b)^n (cx + d)^m$.
Using the product rule for differentiation,$\frac{d}{dx}[u(x)v(x)] = u(x)v'(x) + v(x)u'(x)$.
Let $u(x) = (ax + b)^n$ and $v(x) = (cx + d)^m$.
Using the chain rule,$u'(x) = n(ax + b)^{n-1} \cdot \frac{d}{dx}(ax + b) = n(ax + b)^{n-1} \cdot a = na(ax + b)^{n-1}$.
Similarly,$v'(x) = m(cx + d)^{m-1} \cdot \frac{d}{dx}(cx + d) = m(cx + d)^{m-1} \cdot c = mc(cx + d)^{m-1}$.
Now,substitute these into the product rule formula:
$f'(x) = (ax + b)^n \cdot [mc(cx + d)^{m-1}] + (cx + d)^m \cdot [na(ax + b)^{n-1}]$.
Factor out the common terms $(ax + b)^{n-1}$ and $(cx + d)^{m-1}$:
$f'(x) = (ax + b)^{n-1} (cx + d)^{m-1} [mc(ax + b) + na(cx + d)]$.
Thus,the derivative is $(ax + b)^{n-1} (cx + d)^{m-1} [mc(ax + b) + na(cx + d)]$.