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For which $a\in {\bf R}$ is $|x|^a$ differentiable at $x=0$ and what is the derivative there?

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What do you have so far? –  GEdgar Sep 23 '12 at 20:57

3 Answers 3

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Hint: Let $a\gt 1$. Calculate $\displaystyle\lim_{x\to 0} \frac{|x|^a-0}{x-0}$. It is useful to work separately with the limit from the right and the limit from the left.

If $x\gt 0$, then $\dfrac{|x|^a-0}{x-0}=\dfrac{x^a}{x}=x^{a-1}$.

If $x\lt 0$, then $\dfrac{|x|^a-0}{x-0}=\dfrac{|x|^a}{-|x|}=-|x|^{a-1}$.

Now we need to show that the limit does not exist when $a\le 1$. For $a=1$, we are working with the familiar absolute value function, and the limit of the differential quotient from the right is $1$, while the limit from the left is $-1$.

If $a\lt 1$, the differential quotient blows up as $x$ approaches $0$ from the right.

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Hint: take the definition of differential and separate it to left and right limits.

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By definition, $f$ is called differentiable at the point $x_0$, if its increment $Δf=f(x_0+h)−f(x_0)$ may be represented as $$f(x_0+h)−f(x_0)=f'(x_0)h+α(x_0,h),$$ where $\alpha$ satisfies $|\alpha(x_0,h)|=o(|h|),\quad h\rightarrow 0$. Thus is sufficient to write increment for $f(x)=|x|^\alpha$ at point $x_0=0$ for $h>0$ and $h<0$

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