Question

Find the derivative of $(ax+b{)}^n$ where $a,b$ are fixed non-zero constants and $n$ is an interger.

Answer 1

Method $1:$
Calculating derivative for $p^n$
Let $f\left(x\right)=p^n$ where $p$ be a function in $x$.
For $n=1$
$f^{\prime }\left(x\right)=p^{\prime }$
For $n=2$
$f^{\prime }\left(x\right)=(p^2{)}^{\prime }=(p\cdot p{)}^{\prime }=p^{\prime }p+p{p}^{\prime }=2p{p}^{\prime }$
For $n=3$
$f^{\prime }\left(x\right)=(p^3{)}^{\prime }=(p\cdot p^2{)}^{\prime }=p(p^2{)}^{\prime }+p^{\prime }{p}^2$
$\Rightarrow f^{\prime }\left(x\right)=p\left(2p{p}^{\prime }\right)+p^{\prime }{p}^2=2p^2{p}^{\prime }+p^2{p}^{\prime }$
$\Rightarrow f^{\prime }\left(x\right)=3p^2{p}^{\prime }$
Similarly, for $n=n$
$\Rightarrow f^{\prime }\left(x\right)=n{p}^{n-1}{p}^{\prime }$
Required derivative
Let $g\left(x\right)=(ax+b{)}^n$
Differentiating with respect to $x$
$\Rightarrow g^{\prime }\left(x\right)=\left(\right(ax+b{)}^n{)}^{\prime }$
From above formula, we get
$\Rightarrow g^{\prime }\left(x\right)=n(ax+b{)}^{n-1}(ax+b{)}^{\prime }$
$\Rightarrow g^{\prime }\left(x\right)=n(ax+b{)}^{n-1}(a)$
$\therefore g^{\prime }\left(x\right)=na(ax+b{)}^{n-1}$

Method $2:$
Let $g\left(x\right)=(ax+b{)}^n$
Using Chain Rule, we get
$\Rightarrow g^{\prime }\left(x\right)=n(ax+b{)}^{n-1}(ax+b{)}^{\prime }$
$\therefore g^{\prime }\left(x\right)=na(ax+b{)}^{n-1}$