# Find the derivative of the following functions ( it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers ) $(ax+b)^n(cx+d)^m$

Let $f(x) = (ax+b)^n(cx+d)^m$
By Leibnitx product rule,
$f'(x) =(ax+b)^n \large\frac{d}{dx}$$(cx+d)^m +(cx+d)^m \large\frac{d}{dx}$$(ax+b)^n$---------(1)
Now, let $f_1(x) = (cx+d)^m$
$f_1(x+h)=(cx+ch+d)^m$
$= f'_1(x) = \lim\limits_{h \to 0} \large\frac{f_1(x+h)-f_1(x)}{h}$
$= \lim\limits_{h \to 0} \large\frac{(cx+ch+d)^m-(cx+d)^m}{h}$
$= (cx+d)^m \lim\limits_{h \to 0} \large\frac{1}{h}$$\bigg[ \bigg(1+\Large\frac{ch}{cx+d}$$ \bigg)^m -1\bigg]$
$= (cx+d)^m \lim\limits_{h \to 0} \large\frac{1}{h}$$\bigg[ \bigg( 1+ \Large\frac{mch}{cx+d}$$+ \Large\frac{m(m-1)}{2}$$\Large\frac{c^2h^2}{(cx+d)^2}$$+.... \bigg)-1 \bigg]$
$= (cx+d)^m \lim\limits_{h \to 0} \large\frac{1}{h}$$\bigg[\Large\frac{mch}{cx+d}$$+\large\frac{m(m-1)c^2h^2}{2(cx+d)^2}$.....(Terms containing higher degrees of $h) \bigg]$
$= (cx+d)^m \lim\limits_{h \to 0} \bigg[\Large\frac{mc}{cx+d} $$\Large\frac{m(m-1)c^2h}{2(cx+d)^2}$$+.. \bigg]$
$= (cx+d)^m \bigg[ \large\frac{mc}{cx+d} $$+0\bigg] = \large\frac{mc(cx+d)^m}{cx+d} = mc(cx+d)^{m-1} \large\frac{d}{dx}$$(cx+d)^m = mc(cx+d)^{m-1}$--------------(2)
Similarly, $\large\frac{d}{dx}$$(ax+b)^n=na(ax+b^{n-1}$------(3)
Therefore, from (1), (2) and (3) , we obtain,
$f'(x) =(ax+b)^n \{ mc (cx+d)^{m-1} \}+ (cx+d)^m \{ na(ax+b)^{n-1} \}$
$= (ax+b)^{n-1} (cx+d)^{m-1} [mc(ax+b)+na(cx+d)]$
answered Apr 14, 2014