Let f (x) = (ax + b)n (cx + d)m ... (i) Differentiating (i) with respect to x, we get $\frac{d}{dx}$ (f (x)) = $[(ax+b)^n]' (cx+d)^m$ + $(ax+b)^n \cdot [(cx+d)^m]'$ = [$n(ax + b)^{n-1}$·(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] ∴ $\frac{d}{dx} [(ax+b)^n (cx+d)^m]$ = $(ax+b)^{n-1} (cx+d)^{m-1}$ [na(cx + d) + mc(ax + b)]