alternative proof of derivative of $x^{n}$

The typical derivative formula

\\frac{df}{dx}=\\lim_{h\\to 0}\\frac{f(x+h)-f(x)}{h}

combined with the binomial theorem yield an alternative way to prove that

\\frac{d}{dx}(x^{n})=nx^{n-1}

for any positive integer $n$ .

Proof.

$\\begin{array}[]{ll}\\displaystyle\\frac{d}{dx}(x^{n})&\\displaystyle=\\lim_{h\\to 0%}\\frac{(x+h)^{n}-x^{n}}{h}\\\\&\\\\&\\displaystyle=\\lim_{h\\to 0}\\frac{\\displaystyle\\left(\\sum_{j=0}^{n}{n\\choose j%}x^{j}h^{n-j}\\right)-x^{n}}{h}\\\\&\\\\&\\displaystyle=\\lim_{h\\to 0}\\frac{\\displaystyle x^{n}+nx^{n-1}h+h^{2}\\left(%\\sum_{j=0}^{n-2}{n\\choose j}x^{j}h^{n-2-j}\\right)-x^{n}}{h}\\\\&\\\\&\\displaystyle=\\lim_{h\\to 0}\\frac{\\displaystyle nx^{n-1}h+h^{2}\\sum_{j=0}^{n-2%}{n\\choose j}x^{j}h^{n-2-j}}{h}\\\\&\\\\&\\displaystyle=\\lim_{h\\to 0}\\left(nx^{n-1}+h\\sum_{j=0}^{n-2}{n\\choose j}x^{j}h%^{n-2-j}\\right)\\\\&\\\\&\\displaystyle=nx^{n-1}\\end{array}$

∎

alternative proof of derivative of xn

Proof.

alternative proof of derivative of $x^{n}$