Find the derivative of the following functions

Solution:

Let $f(x)=(a x+b)^{n}$. Accordingly, $f(x+h)=\{a(x+h)+b\}^{n}=(a x+a h+b)^{n}$

By first principle,

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f^{\prime}(x)}{h}$

$=\lim _{h \rightarrow 0} \frac{(a x+a h+b)^{n}-(a x+b)^{n}}{h}$

$=\lim _{h \rightarrow 0} \frac{(a x+b)^{n}\left(1+\frac{a h}{a x+b}\right)^{n}-(a x+b)^{n}}{h}$

$=(a x+b)^{n} \lim _{h \rightarrow 0} \frac{\left(1+\frac{a h}{a x+b}\right)^{n}-1}{h}$

$=(a x+b)^{n} \lim _{b \rightarrow 0} \frac{1}{n}\left[\left\{1+n\left(\frac{a h}{a x+b}\right)+\frac{n(n-1)}{\lfloor 2}\left(\frac{a h}{a x+b}\right)^{2}+\ldots\right\}-1\right]$

(Using binomial theorem)

$=(a x+b)^{n} \lim _{b \rightarrow 0} \frac{1}{h}\left[n\left(\frac{a h}{a x+b}\right)+\frac{n(n-1) a^{2} h^{2}}{\left\lfloor 2(a x+b)^{2}\right.}+\ldots(\right.$ Terms containing higher degrees of $\left.h)\right]$

$=(a x+b)^{n} \lim _{b \rightarrow 0}\left[\frac{n a}{(a x+b)}+\frac{n(n-1) a^{2} h}{\left\lfloor 2(a x+b)^{2}\right.}+\ldots\right]$

$=(a x+b)^{n}\left[\frac{n a}{(a x+b)}+0\right]$

$=n a \frac{(a x+b)^{a}}{(a x+b)}$

$=n a(a x+b)^{n-1}$