Let f (x) = (ax + b)n (cx + d)m ... (i)
Differentiating (i) with respect to x, we get

d

dx

(f (x)) = [(ax+b)n]′(cx+d)m + (ax+b)n.[(cx+d)m]′
= [n(ax+b)n–1·(a·1 + 0)]·(cx+d)m + (ax+b)n·[m(cx+d)m–1·(c·1 + 0)]
= [n(ax+b)n–1·a] [cx+d]m + [ax+b]n [m(cx+d)m–1 c]
∴

d

dx

[(ax+b)n(cx+d)m]
= (ax+b)n−1(cx+d)m−1 [na(cx + d) + mc(ax + b)]