Let f(x) = (ax + b)n (cx + d)m By Leibnitz product rule,
\( f'(x)=(ax+b)^n\dfrac{d}{dx}(cx+b)^m+(cx+d)^m\dfrac{d}{dx}(ax+b)^n\)........(1)
Now let
\( f_1(x)=(cx+d)^m\)
\( f_1(x+h)=(cx+ch+d)^m\)
\( f'(x)=\lim\limits_{h \to 0}\dfrac{f(x+h)-f(x)}{h} \)
\(\lim\limits_{h \to 0}\dfrac{(cx+ch+d)^m-(cx+d)^m}{h} \)
\( (cx+d)^m \lim\limits_{h \to 0}\dfrac{1}{h}[(1+\dfrac{ch}{cx+d})^m-1]\)
\( (cx+d)^m [\dfrac{mc}{(cx+d)}+0]\)
\( \dfrac{mc(cx+d)^m}{(cx+d)}\)
\( mc(cx+d)^{m-1}\)
\(\dfrac{d}{dx} mc(cx+d)^m= mc(cx+d)^{m-1}\) ............(2)
\( \dfrac{d}{dx}(ax+b)^n=na(Ax+b)^{n-1}\)....................(3)
Therefore, from (1), (2), and (3), we obtain