Question

differentiate by first principle k x^n

Answer 1

$$\begin{gathered} Letf\left(x\right)=k{x}^n \\ So, \\ \frac{d}{dx}f\left(x\right)=\underset{x\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ =\underset{x\to 0}{\lim }\frac{k{\left(x+h\right)}^n-k{x}^n}{h} \\ =\underset{x\to 0}{\lim }\frac{k{x}^n\left[{\left(1+{\displaystyle \frac{h}{x}}\right)}^n-1\right]}{h} \\ =\underset{x\to 0}{\lim }\frac{k{x}^n\left[\left(1+n\left({\displaystyle \frac{h}{x}}\right)+{\displaystyle \frac{n\left(n-1\right)}{2!}}{\left({\displaystyle \frac{h}{x}}\right)}^2+...\right).-1\right]}{h} \end{gathered}$$
$$\begin{gathered} =\underset{x\to 0}{\lim }\frac{k{x}^n\left[n\left({\displaystyle \frac{h}{x}}\right)+{\displaystyle \frac{n\left(n-1\right)}{2!}}{\left({\displaystyle \frac{h}{x}}\right)}^2+....\right]}{h} \\ =\underset{x\to 0}{\lim }\frac{hk{x}^n\left[n\left({\displaystyle \frac{1}{x}}\right)+{\displaystyle \frac{n\left(n-1\right)h}{2!}}{\left({\displaystyle \frac{1}{x}}\right)}^2+....\right]}{h} \end{gathered}$$
$=\underset{x\to 0}{\lim }k{x}^n\left[n\left({\displaystyle \frac{1}{x}}\right)+{\displaystyle \frac{n\left(n-1\right)h}{2!}}{\left({\displaystyle \frac{1}{x}}\right)}^2+....\text{higher power of h}\right]$
$=\underset{x\to 0}{\lim }k\left[n{x}^{n-1}+{\displaystyle \frac{n\left(n-1\right)h}{2!}}{\left({\displaystyle \frac{1}{x}}\right)}^2+....\text{higher power of h}\right]$
$=knxn-1Hence,\frac{d}{dx}\left(k{x}^n\right)=knxn-1$