Question

differentiate by first principle k x^n

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Solution

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

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