# 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 ) $(ax+b)^n$

let $f(x) = (ax+b)^n$. Accordingly, $f(x+h) = \{a(x+h)+b \}^n = (ax+ah+b)^n$
By first principle,
$f(x) = \lim\limits_{h \to 0} \large\frac{f(x+h)-f(x)}{h}$
$= \lim\limits_{ h \to 0} \large\frac{(ax+ah+b)^n-(ax+b)^n}{h}$
$= \lim\limits_{ h \to 0} \large\frac{(ax+b)^n \bigg( 1+ \Large\frac{ah}{ax+b} \bigg)^n-(ax+b)^n}{h}$
$= (ax+b)^n \lim\limits_{ h \to 0} \large\frac{\bigg( 1+ \Large\frac{ah}{ax+b} \bigg)^n-1}{h}$
$= (ax+b)^n \lim\limits_{h \to 0} \large\frac{1}{n}$$\bigg[ \bigg\{ 1+n \bigg(\large\frac{ah}{ax+b} \bigg)$$+\large\frac{n(n-1)}{2} $$\bigg( \large\frac{ah}{ax+b} \bigg)^2$$+.... \bigg \}-1 \bigg]$
$\qquad \qquad \qquad$( Using binomial theorem )
$= (ax+b)^n \lim\limits_{h \to 0} \large\frac{1}{h}$$\bigg[n \bigg( \large\frac{ah}{ax+b} \bigg)$$+ \large\frac{n(n-1)a^2h^2}{2 (ax+b)^2}$$+... ( Terms containing higher degrees of h) \bigg] = (ax+b)^n \lim\limits_{h \to 0} \bigg[ \large\frac{na}{ax+b}$$\large\frac{n(n-1)a^2h}{2(ax+b)^2}$$+... \bigg] = (ax+b)^n = \bigg[ \large\frac{na}{(ax+b)}$$ + 0 \bigg]$
$na\large\frac{(ax+b)^n}{(ax+b)}$
$= na(ax+b)^{n-1}$