Find the derivative of the following functions

Solution:

Let $f(x)=(a x+b)^{n}(c x+d)^{m}$

By Leibnitz product rule,

$f^{\prime}(x)=(a x+b)^{\prime} \frac{d}{d x}(c x+d)^{m}+(c x+d)^{m} \frac{d}{d x}(a x+b)^{n}$ $\ldots(1)$

Now, let $f_{1}(x)=(c x+d)^{m}$

$f_{1}(x+h)=(c x+c h+d)^{m}$

$f_{1}^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f_{1}(x+h)-f_{1}(x)}{h}$

$=\lim _{h \rightarrow 0} \frac{(c x+c h+d)^{m}-(c x+d)^{m}}{h}$

$=(c x+d)^{m} \lim _{h \rightarrow 0} \frac{1}{h}\left[\left(1+\frac{c h}{c x+d}\right)^{m}-1\right]$

$=(c x+d)^{m} \lim _{h \rightarrow 0} \frac{1}{h}\left[\left(1+\frac{m c h}{(c x+d)}+\frac{m(m-1)}{2} \frac{\left(c^{2} h^{2}\right)}{(c x+d)^{2}}+\ldots\right)-1\right]$

$=(c x+d)^{m} \lim _{h \rightarrow 0}\left[\frac{m c}{(c x+d)}+\frac{m(m-1) c^{2} h}{2(c x+d)^{2}}+\ldots\right]$

$=(c x+d)^{m}\left[\frac{m c}{c x+d}+0\right]$

$=m c(c x+d)^{m-1}$

Therefore, from (1), (2), and (3), we obtain

$f^{\prime}(x)=(a x+b)^{n}\left\{m c(c x+d)^{m-1}\right\}+(c x+d)^{m}\left\{n a(a x+b)^{n-1}\right\}$

$=(a x+b)^{n-1}(c x+d)^{m-1}[m c(a x+b)+n a(c x+d)]$