Let f(x) = (ax+b)(cx+d)2 ... (i)
Differentiating (i) with respect to x, we get

d

dx

(f(x)) = (ax + b) [(cx+d)2]′ + (cx+d)2 (ax + b)′ = (ax + b) 2c (cx + d) + (cx+d)2 a
= 2c(ax + b) (cx + d) + a(cx+d)2