Question

How we will find derivative of $f\left(x\right)=(ax+b{)}^m(cx+d{)}^{rn}$

Answer 1

Let $f\left(x\right)=(ax+b{)}^m(cx+d{)}^{rn}$
By Leibnitz product rule,
$f^{\prime }\left(x\right)=(ax+b{)}^n\frac{d}{dx}(cx+d{)}^{\pi }(cx+d)m\frac{d}{dx}(ax+b{)}^m$ ...(I)
Now, let $F_1\left(fx\right)=(cx+d{)}^m$
$f_1\left(x\right)={lim}_{h\to 0}\frac{f_1(x+h)-f_1\left(x\right)}{h}$
$={lim}_{h\to 0}\frac{(cx+ch+d{)}^m-(cx+d{)}^m}{h}$
$=(cx+d{)}^m{lim}_{h\to 0}\frac{1}{h}[{(1+\frac{ch}{cx+d})}^m-1]$
$(cx+d^m){lim}_{h\to 0}\frac{1}{h}[(1+\frac{mch}{(cx+d)}+\frac{m(m-1)}{2}\frac{\left(c^2{h}^2\right)}{(cx+d{)}^2}+\dots )-1]$
$=(cx+d{)}^m{lim}_{h\to 0}[\frac{mch}{(cx+d)}+\frac{m(m-1)d^2{h}^2}{2(cx+d)}+\frac{m(m-1)c^{2}h}{2(cx+d{)}^2}+\dots ]$
$=(cx+d{)}^m[\frac{mc}{cx+d}+0]$
$=\frac{mc(cx+d{)}^m}{(cx+d)}$
$=mc(cx+d{)}^{m-1}$
$\frac{d}{dx}(cx+d{)}^m=mc(cx+d{)}^{m-1}$ ...(2)
Similarly, $\frac{d}{dx}(ax+b{)}^n=na(ax+b{)}^{n-1}$ ....(3)
Therefore, from (1), (2), and (3), we obtain
$f^{\prime }\left(x\right)=(ax+b{)}^n\left\{mc\right(cx+d{)}^{n-1}\}+(cx+d{)}^m\left\{na\right(ax+b{)}^{n-1}\}$
$=\left(a{x}_b{)}^{n-1}\right(cx+d{)}^{m-1}\left[mc\right(ax+b)+na(cx+d\left)\right]$