How to prove that $\frac{\text{d}}{\text{d}x} x^n = nx^{n-1}$ without using the Binomial Theorem?

Question

How does one show that $\frac{\text{d}}{\text{d}x} x^n =nx^{n-1}$ without resorting to the Binomial Theorem?

Edited :

I'm interested in this approach as I've been tinkering with the proof for the Binomial Theorem using Taylor Series but then later realized that I've assumed $\frac{\text{d}}{\text{d}x} x^n = nx^{n-1}$ which I have proven earlier using the Binomial Theorem itself. So the proof ended up being circular in nature. I'm trying to circumvent that by proving it without the Binomial Theorem. Link to question : Binomial Theorem Proof from Taylor Series

Answer 1

$$ S=1+y+y^2+\cdots+y^{n-1} $$ is a geometric series; multiply $S$ by $y$ to get $$ Sy=y+y^2+\cdots+y^{n-1}+y^n $$ Subtract $Sy$ from $S$ to get $$ Sy-S=y^{n}-1 $$ Therefore $$ y^n-1=(y-1)(1+y+y^2+\cdots+y^{n-1}) $$ Put $y=\frac{x+h}{x}$ (provided $x\not=0$) to get $$ \frac{(x+h)^n}{x^n}-1=\left(\frac{x+h}{x}-1\right) \left( 1+\frac{x+h}{x}+\frac{(x+h)^2}{x^2}+\cdots+\frac{(x+h)^{n-1}}{x^{n-1}} \right)\tag{*} $$ Multiply both sides by $x^n$ to get $$ (x+h)^n-x^n=h(x^{n-1}+(x+h)x^{n-2}+(x+h)^2x^{n-3}+\cdots+(x+h)^{n-1}) $$

We wish to find the derivative of the function $f(x)=x^n$ from first principle. If $x=0$, then $$ f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{h^n}{h}=\lim_{h\to0}h^{n-1}=0 $$ If $x\not=0$, then $$ \begin{align} f'(x)&=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\ &=\lim_{h\to0}\frac{(x+h)^n-x^n}{h}\\ &=\lim_{h\to0}\frac{h(x^{n-1}+(x+h)x^{n-2}+(x+h)^2x^{n-3}+\cdots+(x+h)^{n-1})}{h}\\ &=\lim_{h\to0}(x^{n-1}+(x+h)x^{n-2}+(x+h)^2x^{n-3}+\cdots+(x+h)^{n-1})\\ &=x^{n-1}+xx^{n-2}+x^2x^{n-3}+\cdots+x^{n-1}\\ &=\underbrace{x^{n-1}+x^{n-1}+x^{n-1}+\cdots+x^{n-1}}_{n\text{ copies}}\\ &=nx^{n-1} \end{align} $$

Answer 2

Note that $$\dfrac{d(x^n)}{dx}_{(\text{at } x=a)}=\lim_{x\to a}\dfrac{x^n-a^n}{x-a} = \lim_{x\to a} \sum_{r=0}^{n-1}x^{n-1-r}a^r$$ Now set $x=a$ to find $$x^{n-1-r}a^r=a^{n-1}$$ for each $r$ such that $0\le r\le n-1$.

Answer 3

$$x^n = e^{n \log x}$$ Thus, $$\frac{dx^n}{dx} =e^{n\log x}\cdot \frac{n}{x}=nx^{n-1}$$

Answer 4

$(x^{n+1})'=(x\times x^n)' =x(x^n)'+(x^n) =xnx^{n-1}+x^n =nx^n+x^n=(n+1)x^n$.