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I am looking for the proof of kth derivative $x^a$, where $a$ is rational, and $x>0$,(I do not need general form, because we haven't defined irrational powers yet) but all the proofs I have seen just considering integers $$\frac{d^{k}}{d x^{k}} x^{a}=a(a-1) \ldots(a-k+1) x^{a-k}$$

Tural Sukurov
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    It is less clear what $x^a$ means for arbitrary real numbers $x$ when $a$ is a rational number that is not an integer, than for integer exponents. In particular the cases $x\lt 0$ are not well-defined values, but when $x \gt 0$ the usual rules of calculus hold (and the values are well-defined if we are defining the principal (real) result). – hardmath Aug 31 '21 at 17:09
  • oh sorry I forgot x>0 – Tural Sukurov Aug 31 '21 at 17:10
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    $\frac{d}{dx}[x^a] = \frac{d}{dx}[e^{a\ln(x)}]=\frac{a}{x}e^{a\ln(x)}=\frac{a}{x}x^a=ax^{a-1}$ – JMoravitz Aug 31 '21 at 17:12
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    In that case you can prove the formula for repeated derivatives by induction on $k$ (once you prove it for the single differentiation rule). See here. – hardmath Aug 31 '21 at 17:13
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    This is overkill, but if you already know that this derivative holds for integers, you can deduce the general case with the implicit function theorem. – Alann Rosas Aug 31 '21 at 17:14

2 Answers2

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Let $m,n\in\Bbb Z\setminus \{0\}$ and $x, x+h>0$.

$$\frac{(x+h)^{m/n}-x^{m/n}}h=\frac{(x+h)^m-x^m}{h\sum_{j=0}^{n-1}(x+h)^{mj/n}x^{m(n-j-1)/n}}=\frac{\frac{(x+h)^m-x^m}h}{\sum_{j=0}^{n-1}(x+h)^{mj/n}x^{m(n-j-1)/n}}$$

Call $g(t)=t^m$. You have $$\lim_{h\to 0}\frac{(x+h)^{m/n}-x^{m/n}}h=\frac{g'(x)}{\sum_{j=0}^{n-1}x^{mj/n}x^{m(n-j-1)/n}}=\frac{mx^{m-1}}{nx^{m(n-1)/n}}=\frac mnx^{\frac mn-1}$$

Once you know that $(x^q)'=qx^{q-1}$ for $q\in\Bbb Q$ as well, it's easy.

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Consider the function $f(x) = x^{q}$, where $q$ is rational. Then one has that

\begin{align*} f(x) = x^{q} & \Rightarrow \ln f(x) = q\ln(x)\\\\ & \Rightarrow \frac{f'(x)}{f(x)} = \frac{q}{x}\\\\ & \Rightarrow f'(x) = \frac{qf(x)}{x}\\\\ & \Rightarrow f'(x) = qx^{q-1} \end{align*}

The same reasoning applies to real exponents.

Can you take it from here?

user0102
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