Regularity of the function $|x|^ax$

Obviously, the only possible problem is in $x=0$. Using the definition of derivative: $$ f'(0) = \lim_{x\to 0}\frac{f(x)-f(0)}{x-0} = \lim_{x\to 0}\frac{|x|^ax}x = \lim_{x\to 0}|x|^a = \cases{ 0&$a>0$\cr 1&$a=0$\cr \infty&$a<0$} $$ And for $x\ne 0$ $$f'(x)=\cases{ (a+1)x^a&$x>0$\cr (a+1)(-x)^a&$x<0$}$$ i.e., $$f'(x) = (a+1)|x|^a,\qquad x\ne 0$$ and $f'$ is continuous iff $a\ge0$. The study of the differentiabily of $f'$ is analogous and $f''$ is much like $f$, namely $$f''(x)=\cdots$$ More details required?

First of all note that if $g(x)=|x|^{\alpha+2}$, then $g'(x)=(\alpha+2)|x|^{\alpha}x$ and thus it is enough (and more convient) to study the differentiability of $f(x)=|x|^\alpha$ to answer your question. Juste note that for every $D\subset \Bbb R$ and $k\geq 1$ we have $g \in C^k(D) \iff g'\in C^{k-1}(D)$. My conclusions are summarized in the following list and are direct consequences of the Theorem below.

For $\alpha >0$, we write $[\alpha]$ the integer part of $\alpha$ (i.e. $[\cdot]$ is the floor function).

Let $\alpha \in \Bbb R$ and $f\colon \Bbb R\to \Bbb R, \ x \mapsto |x|^{\alpha}$, then

• $f \in C^{\infty}(\Bbb R\setminus\{0\})$,
• If $\alpha <0$, then $f\notin C^{0}(\Bbb R)$,
• If $\alpha \geq 0$ and $\alpha = 2n$ for some $n\in \Bbb N$, then $f\in C^{\infty}(\Bbb R)$
• If $\alpha>0$ and $\alpha \neq 2n$ for every $n\in \Bbb N$, then $f\in C^{[\alpha]}(\Bbb R)$ but $f\notin C^{[\alpha]+1}(\Bbb R)$

Remember that for every $A \subset B \subset \Bbb R$ and $k\in \Bbb N$, we have $C^{k}(B)\subset C^{k}(A)\subset C^{k+1}(A)$.

For any function $f\colon \Bbb R \to \Bbb R$ and $n\in \Bbb N$, we write $f^{(n)}$ the $n$-th derivative of $f$.

Theorem. Let $\alpha \in \Bbb R, \alpha \neq 1$ and $f\colon \Bbb R\to \Bbb R, \ x \mapsto |x|^{\alpha}$, then for every $n\in \Bbb N$ and every $x\neq 0$, we have $$f^{(2n)}(x)=\frac{\alpha !}{(\alpha-2n)!}|x|^{\alpha-2n},$$ and $$f^{(2n+1)}(x)=\frac{\alpha !}{(\alpha-2n-1)!}|x|^{\alpha-2(n+1)}x,$$ Moreover, $f^{(2n)}$ and $f^{(2n+1)}$ are continuous on $\Bbb R\setminus\{0\}$ and

• $f^{(2n)}(0)=0 \quad \text{ if } \alpha >2n,$
• $f^{(2n)}(0)=0 \quad \text{ if } \alpha=2n,$
• $f^{(2n+1)}(0)=0 \quad \text{ if } \alpha >2n+1,$
• $\lim_{x\to 0}f^{(2n+1)}(x)$ does not exists if $\alpha=2n+1, $
• $\lim_{x\to 0}f^{(2n)}(x)$ does not exists if $2n< \alpha$,
• $\lim_{x\to 0}f^{(2n+1)}(x)$ does not exist if $2n+1<\alpha$.

Proof: See appendix.

Appendix:

Proposition 1. Let $f\colon \Bbb R\to \Bbb R,\ x\mapsto|x|$, then $f'(x)$ exists for every $x\neq 0$ and $f'$ is continuous on $(0,\infty)$ and $(-\infty,0)$. Moreover, $$f'(x)=\operatorname{sign}(x)=\frac{x}{|x|}=\frac{|x|}{x} \qquad \forall x \neq 0,$$ where $\operatorname{sign}$ is the sign function.

Proof:

• Let $x>0$, then for every $|h|<x$ we have $$\frac{|x+h|-|x|}{h}=\frac{x+h-x}{h}=1 \implies f'(x)=\lim_{h\to 0}\frac{|x+h|-|x|}{h}=1=\operatorname{sign}(x).$$
• If $x<0$, for every $|h|<x$ we have $$\frac{|x+h|-|x|}{h}=\frac{-x-h-x}{h}=-1 \implies f'(x)=\lim_{h\to 0}\frac{|x+h|-|x|}{h}=-1=\operatorname{sign}(x).$$
• If $x=0$, then for every $h_1<0<h_2$ we have $$\frac{|x+h_1|-|x|}{h_1}=\frac{-h_1}{h_1}=-1 \quad \text{ and } \quad \frac{|x+h_2|-|x|}{h_2}=\frac{h_2}{h_2}=1,$$ thus $$\lim_{h_1\to 0^-}\frac{|x+h_1|-|x|}{h_1}=-1 \neq 1 = \lim_{h_2\to 0^-}\frac{|x+h_2|-|x|}{h_2},$$ i.e. $f'(0)$ does not exists.

Finally, clearly for every $x\neq 0$, we have $\operatorname{sign}(x)=\frac{x}{|x|}=\frac{|x|}{x}$ and since this function is constant on $(0,\infty)$ (resp. $(-\infty,0)$) it is continuous on $(0,\infty)$ (resp. $(-\infty,0)$).$\qquad \square$

The next Proposition, Theorem as well as Corollary are classical results of real-analysis and a proof can be found in any analysis introductory book.

Proposition 2. Let $f:D\subset \Bbb R\to \Bbb R,g:E \subset f(D)\to \Bbb R$ and $x_0\in D$ such that $f(x_0)\in E$. If $f'(x_0)$ and $g'(f(x_0))$ exist, then $$\frac{d}{dx}g\big(f(x_0)\big)=g'\big(f(x_0)\big)f'(x_0).$$

Theorem 3. Let $a<b$ and $f[a,b]\to \Bbb R$ such that $f'$ exists and is continuous on $]a,b[$. If $\lim_{x\to a^+}f'(x)$ exists, then $$\lim_{x\to a^+}f'(x)=\lim_{h\to 0^+}\frac{f(a+h)-f(a)}{h}=:f'(a).$$ Corollary 4. Let $I\subset \Bbb R$ open, $x_0\in I$ and $f\colon I \to \Bbb R$ continuous. Suppose that $f'(x)$ exists for every $x\in I\setminus\{0\}$. If $\lim_{x\to x_0^-}f'(x)=\lim_{x\to x_0^+}f'(x)=l$, then $f'(x_0)$ exists and $f'(x_0)=l$.

Proposition 5. Let $\alpha \in \Bbb R$ and $f\colon [0,\infty)\to \Bbb R,\ x\mapsto x^\alpha$, then $f'(x)$ exists for every $x\neq 0$ and is continuous on $(0,\infty)$. Moreover, if $\alpha\geq 1$, then $f'(0)$ exists and $f'$ is continuous on $[0,\infty)$. Moreover, $$f'(x)=\alpha x^{\alpha-1} \quad \forall x>0, \qquad f'(0)=0 \quad \text{ if }\alpha> 1 \qquad\text{ and } \qquad f'(0)=1 \quad \text{ if }\alpha = 1.$$

Proof: Since $\alpha\in \Bbb R$, we have $f(x)=x^\alpha=e^{\alpha \ln(x)}$ by definition. Now, by proposition 2 we have $$f'(x)=\frac{\alpha}{x}\ e^{\alpha \ln(x)} = \alpha x^{\alpha-1} \qquad \forall x >0.$$ It follows that $f'$ is continuous on $(0,\infty)$. Now, suppose that $\alpha > 1$, then, from Theorem 3, we get $$f'(a)=\lim_{x\to a^+}f'(x)=\alpha \ 0^{\alpha-1}=0.$$ Finally, if $\alpha = 1$, then, from Theorem 3, we get $$f'(a)=\lim_{x\to a^+}f'(x)=1.$$ Continuity on $[0,\infty)$ when $\alpha \geq 1$ should now be clear. $\qquad \square$

Proof of the main Theorem: We proceed by induction over $n \in \Bbb N$, so let $x\neq 0$, then (by Proposition 1 and 4) $$f^{(0)}(x)=\frac{\alpha !}{\alpha !}|x|^{\alpha}=f(x) \quad \text{and} \quad f^{(1)}= \alpha |x|^{\alpha-1}\frac{x}{|x|}=\alpha |x|^{\alpha-2}x.$$ Now suppose that the formula is true for $2n$, then (by Propositions 1 and 5) $$f^{(2n+1)}(x)=\frac{d}{dx}\left(\frac{\alpha !}{(\alpha-2n)!}|x|^{\alpha-2n}\right)=\frac{\alpha !(\alpha-2n)}{(\alpha-2n)!}|x|^{\alpha-2n-1}\frac{x}{|x|}=\frac{\alpha !}{(\alpha-(2n+1))!}|x|^{\alpha-2n-2}x.$$ If the formula is true for $2n+1$, then (by Propositions 1 and 5) $$f^{(2n+2)}(x)=\frac{d}{dx}\left(\frac{\alpha !}{(\alpha-2n-1)!}|x|^{\alpha-2(n+1)}x\right)\\ =\frac{\alpha !}{(\alpha-2n-1)!}\Big(\big(\alpha-2(n+1)\big)|x|^{\alpha-2(n+1)-1}x\frac{x}{|x|}+|x|^{\alpha-2(n+1)}\Big)=\frac{\alpha !}{(\alpha-2n-1)!}\Big((\alpha-2n-1)|x|^{\alpha-2(n+1)}\Big)=\frac{\alpha !}{(\alpha-2n-2)!}|x|^{\alpha-2(n+1)}.$$ Now, if $\alpha>2n$, then, by Corollary 4, clearly $f^{(2n)}(0)=\lim_{x\to 0}f^{(2n)}(x)=0$ and if $\alpha >2n+1$, then, by Corollary 4, $\lim_{x\to 0}f^{(2n+1)}(x)=0$. Now, if $\alpha =2n$, then $f(x)=|x|^{2n}=(x^2)^n=x^{2n}$ and thus $f^{(k)}(0)=0$ for every $k\in \Bbb N$. If $\alpha = 2n+1$, then $$f^{(2n+1)}(x)=\alpha!\frac{x}{|x|}=\alpha!\operatorname{sign}(x)$$ and thus $f^{(2n+1)}(0)$ does not exists. Finally, if $\alpha < 2n$, then $\lim_{x\to 0^+}f^{(2n)}(x)=\infty\neq-\infty=\lim_{x\to 0^-}f^{(2n)}(x)$, i.e. $\lim_{x\to 0}f^{(2n)}(x)$ does not exists. The same argument shows that if $\alpha < 2n+1$, then $\lim_{x\to 0}f^{(2n)}(x)$ does not exists. $\qquad \square$

PS Thank you for this interesting question.

HINT: Using one-side derivatives it is easy to show differentiability in 0 for $a>0$.