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How do I find the the derivative of $|x|^p$ for $x\in \mathbb{R}$ and $p\in [1,\infty)$ via the definition of the derivative? I know the derivative is equal to $px|x|^{p-2}$. If I use the answer to this question, I don't get to anywhere useful:

$$f(x)'=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\frac{|x+h|^p-|x|^p}{h}=\frac{(|x+h|^p-|x|^p)(|x+h|^p+|x|^p)}{h(|x+h|^p+|x|^p)}=\frac{|x+h|^{2p}-|x|^{2p}}{h(|x+h|^p+|x|^p)}$$

So I don't think myself that adopting the method from the question is correct. But then I can't think of any other way to get to the desired answer. Any help is appreciated.

2 Answers2

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Hint: Break it up into cases. $x>0$ and $x<0$ are trivial as once $h$ is sufficiently close to 0, so is $x+h$.

The $x=0$ case break into two one sided limits, one as $h\to 0^-$, the other $h\to 0^+$. This lets you get rid of the absolute value in your calculations.

Alan
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  • I still get stuck when I factor $\frac{(x+h)^p-x^p}{h}$. Is there any technique that I have to know, or some trick? I need to factor $h$ out, right? –  Jun 21 '21 at 09:34
  • You can use the fact that that's the definition of the derivative of $x^p$ – Alan Jun 21 '21 at 09:37
  • I'm not sure I expressed myself clearly, sorry if that's the case. My struggle is to factor $\frac{(x+h)^p-x^p}{h}$, so that I can get $px|x|^{p-2}$. In particular, I'm not sure how to get rid of the $h$ at the bottom, so the limit is not $\infty$ or undefined. So, my question is, what technique do I use to do this? –  Jun 21 '21 at 09:43
  • The exponential answer here shows how to calculate that limit as a derivative. https://math.stackexchange.com/questions/1335500/proof-of-the-derivative-of-xn – Alan Jun 21 '21 at 09:48
  • thank you very much –  Jun 21 '21 at 10:29
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If $x>0$, you can evaluate the limit with $|h|<x$ so that $x+h>0$ and the limit is just the derivative of $x^p$ (because $|x|^p=x^p,|x+h|^p=(x+h)^p$). The function is even, so that its derivative is odd.

The only remaining case is for $x=0$,

$$\lim_{h\to 0}\frac{|h|^p}{h}=\lim_{h\to 0}|h|^{p-1}=0$$ for $p>1$. Finally, if $x=0,p=1$ the derivative does not exist.