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

Answered by Abhisek | 1 year ago