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Prove that $\frac d {dx} x^n=nx^{n-1}$ for all $n \in \mathbb R$.

I saw some proof of $\frac d {dx} x^n=nx^{n-1}$ using binomial theorem, which is only available for $n \in\mathbb N$. Do anyone have the proof of $\frac d {dx} x^n=nx^{n-1}$ for all real $n$? Thank you.

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Note that, if $n$ is viewed as a real number, you should compensate by viewing $x$ as an element of $[0,\infty).$ – goblin Sep 9 '13 at 8:02
up vote 4 down vote accepted

Writing $x^r=e^{r\ln x}:$

$$\begin{align*}(x^r)'&=\lim_{h\to 0}\frac{e^{r\ln (x+h)}-e^{r\ln x}}{h}\\&=\lim_{h\to 0}\left(\frac{e^{r\ln (x+h)}-e^{r\ln x}}{r\ln(x+h)-r\ln x}\right)\left(\frac{r\ln (x+h)-r\ln x}{h}\right)\end{align*}$$

$(\ln $ is continuous and injective)

Let $r\ln x =w,\;\;r\ln (x+h)=w+i:$

$$\begin{align*}\cdots &=\left(\lim_{i\to 0}\frac{e^{w+i}-e^w}{i}\right)\cdot \lim_{h\to 0}r\ln\left(1+\frac{h}{x}\right)^{\frac{1}{h}}\quad\qquad\quad\\&=\left(\lim_{i\to 0}\frac{e^{w+i}-e^w}{i}\right)\left(\lim_{h\to 0}\frac{r}{x}\ln\left(1+\frac{h}{x}\right)^{\frac{x}{h}}\right)\\&=\displaystyle\lim_{i\to 0}e^w\left(\frac{e^{i}-1}{i}\right)\cdot \frac{r}{x}\end{align*}$$

Let $i=\ln \left(1+\frac{1}{a}\right):\qquad (i\to0 \Rightarrow a\to \infty)$

$$\begin{align*}\cdots &=\lim_{a\to \infty}e^w\left(\frac{1}{a \ln \left(1+\frac{1}{a}\right)}\right)\cdot \frac{r}{x}\quad\qquad\qquad\qquad\qquad\\&=\lim_{a\to \infty}e^w\left(\frac{1}{\ln \left(1+\frac{1}{a}\right)^{a}}\right)\cdot \frac{r}{x}\\&=\dfrac{re^w}{x}\\&=rx^{r-1}\end{align*}$$

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From the definition of derivatives, we have

$$ f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h} $$ Let's assume here that $f(x) = x^n$. Then,

  • Case 1, $n \in \mathbb{N}$

$$ \begin{align} f'(x) &= \lim_{h \to 0} \frac{(x + h)^n - x^n}{h} \\ &= \lim_{h \to 0} \frac{x^n + nhx^{n-1} + \frac{n(n-1)}{2!}h^2x^{n-2} + \cdots - x^n}{h}\\ &= \lim_{h \to 0} \text{ } h \text{ } \frac{nx^{n-1} + \frac{n(n-1)}{2!}hx^{n-2} + \cdots \text{ higher powers in h}}{h} \\ &= nx^{n-1} + \lim_{h \to 0} \text{ } h \cdot \left( \frac{n(n-1)}{2!}x^{n-2} + \cdots \text{multiples of h}\right)\\ &= nx^n + 0\\ &= nx^n \end{align} $$

  • Case 2, $r \in \mathbb{R}$: A similar expansion of binomials exists for real powers, as given by Issac Newton.

$$ \begin{align} f'(x) &= \lim_{h \to 0} \frac{(x + h)^r - x^r}{h} \\ &= \lim_{h \to 0} \frac{x^r + hrx^{r-1} + \frac{r(r-1)}{2!}h^2x^{r-2} + \cdots - x^r}{h}\\ &= \lim_{h \to 0} \text{ } h \text{ } \frac{rx^{r-1} + \frac{r(r-1)}{2!}hx^{r-2} + \cdots \text{ higher powers in h}}{h} \\ &= rx^{r-1} + \lim_{h \to 0} \text{ } h \cdot \left( \frac{r(r-1)}{2!}x^{r-2} + \cdots \text{multiples of h}\right)\\ &= rx^r + 0\\ &= rx^r \end{align} $$

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Binomial expansion for real powers requires the derivative of $x^r$ doesn't it? – Ishan Banerjee Mar 2 '13 at 11:55
A few $-1$'s missing in those exponents there! – L. F. Mar 4 '13 at 14:05

Assuming you know the derivatives of $e^x$ and $\log(x)$, $$\dfrac{dx^n}{dx}=\dfrac{de^{n\log(x)}}{dx}=n\dfrac{x^n}{x}=nx^{n-1}$$ Here's a proof for $n\in Q$. Let n=$\frac pq$ and $y=x^{\frac pq}$.


With implicit differentiation we get the result.
For irrational n, $x^n=\lim_{r\to n}x^r \quad r \in Q$

$\dfrac{dx^n}{dx}=\dfrac{d}{dx}\lim_{r\to n}x^r$

Not sure if we can interchange the limit and the differentiation, though.

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Can it be proved just by the first principle? – ᴊ ᴀ s ᴏ ɴ Mar 2 '13 at 11:24
@jasoncube Can you define $x^n$ for $n\in \mathbb R$ without $\exp$ and $\ln$? – Hagen von Eitzen Mar 2 '13 at 11:49
Well, you can with a limiting argument. – Ishan Banerjee Mar 2 '13 at 11:54
@jasoncube I believe evaluating derivative directly from this definition is really tedious and tough. That is, $x^{\alpha}=\inf \{x^q|\alpha<q, q\in\mathbb{Q}\}$ where $x>1$. – Jj- Mar 2 '13 at 12:05

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