Differentiate the following from first principle: kx^n

Question

Differentiate the following from first principle:

$\text{k}\text{x}^\text{n}$

✓

Answer

We have,
$\text{f(x)}=\text{kx}^\text{n}$
$\because\text{f}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f(x+h)}-\text{f(x)}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{k}\text{(x+h)}^\text{n}-\text{kx}^\text{n}}{\text{h}}$
$=\text{k}\lim\limits_{\text{h}\rightarrow0}\frac{\bigg(\text{x}^\text{n}+\text{nx}^\text{n}-1\text{h}+\frac{\text{n(n-1)}}{2}\text{x}^\text{n-2}\text{h}^2+...\bigg)-\text{x}^\text{n}}{\text{h}}$ $\big[\because\text{(x+y)}^\text{n}=\text{x}+\text{nx}^\text{n-1}\text{y}...\big]$
$=\text{k}\lim\limits_{\text{h}\rightarrow0}\text{nx}^\text{n-1}+\frac{\text{n(n-1)}}{2!}\text{x}^{\text{n-2}}\text{h}+\frac{\text{n}(\text{n}-1){(\text{n}-2)}}{3!}\text{x}^\text{n-3}\text{h}^2...$
$=\text{k }\text{nx}^{\text{n}-1}+0+0...$
$=\text{k }\text{nx}^{\text{n}-1}$

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