Differentiate each of the following from first principle: `k x^n`

Explanation

`frac{d}{dx}(f(x))=lim_(h->0)frac{f(x+h)-f(x)}{h}`

`=lim_(h->0)frac{k(x+h)^n-kx^n}{h}`

`=lim_(h->0)frac{k[(x+h)^n-x^n]}{(x+h)-x}`

As we know that `lim_(x->a)frac{x^m-a^m}{x-a}=ma^(m-1)`

Hence, `f'(x)=kna^(n-1)`