First and second derivative of $|x|^3$

Question

I need to prove that $|x|^3$ is twice differentiable, by showing that the first and second derivatives exist using the definition. I've tried several ways, this is what I've got:

$$\lim_{h\to 0} \frac{|x+h|^3 - |x|^3}{h} = \lim_{h\to 0}\frac{\sqrt{(x+h)^2}^3 - \sqrt{x^2}^3}{h} = \lim_{h\to 0} \frac{\sqrt{(x+h)^6} - \sqrt{x^6}}{h}$$

then I rationalized the numerator:

$$= \lim_{h\to 0} \frac{(x+h)^6 - x^6}{h\left(\sqrt{(x+h)^6} + \sqrt{x^6}\right)}$$

and I'm stuck on what to do next, I'm skeptic that this is the right way, but I was not able to reach any answer using other ways either.

Any help would be great.

Thanks.

Answer 1

At $x\ne0$, you can take $h$ so small that $x$ and $x+h$ have the same sign. Then the function can be processed as two polynomial pieces which derive seamlessly:

$$|x^3|=\pm x^3$$ gives the derivatives $$\pm3x^2=3x|x|$$ and $$\pm6x=6|x|$$ (as the $\pm$ signs match that of $x$).

At $x=0$, the first derivative is

$$\lim_{h\to0}\frac{|h^3|}{h}=\lim_{h\to0}\pm h^2=0$$

(the sign doesn't matter as convergence is to zero). Note that the first derivative is continous as $3x|x|$ evaluates to $0$.

The second derivative is given by

$$\lim_{h\to0}\frac{3h|h|}{h}=\lim_{h\to0}3|h|=0.$$

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Note that for the third derivative we get

$$\pm6$$ where the sign is that of $x$, and for $x=0$,

$$\lim_{h\to0}\frac{6|h|}{h}$$ is not defined.

Answer 2

Just separate the last limit, i.e., \begin{aligned} L&=\lim_{h\to 0}\frac{(x+h)^6 - x^6}{h\left(\sqrt{(x+h)^6} + \sqrt{x^6}\right)}\\ &= \lim_{h\to 0}\frac{(x+h)^6 - x^6}{h}\cdot \lim_{h\to 0}{\frac1{\sqrt{(x+h)^6} + \sqrt{x^6}}} \\&= \frac{\mathrm{d}x^6}{\mathrm{d}x}\cdot\frac1{2\sqrt{x^6}} \\&= 6x^5\cdot\frac1{2|x^3|} \\ &=6x^5\cdot\frac{1}{2x^2|x|} \\ &=3\frac{x^3}{|x|} \end{aligned} and you're done.

If you need to prove $\frac{\mathrm{d}x^6}{\mathrm{d}x}=6x^5$, you can use this \begin{aligned} \frac{\mathrm{d}x^6}{\mathrm{d}x}&=\lim_{h\to0}\frac{(x+h)^6-x^6}h\\ &=\lim_{h\to0}\frac{(x+h-x)\left((x+h)^5+x(x+h)^4+x^2(x+h)^3+x^3(x+h)^2+x^4(x+h)+x^5\right)}h\\ &=\lim_{h\to0}\frac{h\left((x+h)^5+x(x+h)^4+x^2(x+h)^3+x^3(x+h)^2+x^4(x+h)+x^5\right)}h\\ &=\lim_{h\to0}\left((x+h)^5+x(x+h)^4+x^2(x+h)^3+x^3(x+h)^2+x^4(x+h)+x^5\right)\\ &=6x^5 \end{aligned}