# Derivative of the absolute value $\vert x\vert^3$ with $\epsilon$-$\delta$-definition - check my proof

Consider $$f: \mathbb{R} \to \mathbb{R}, ~~ f(x) = \vert x \vert ^3$$ Show that the first derivative exists and it holds $$f'(x)=3x\vert x\vert ~~\forall x \in \mathbb{R}.$$

My approach:

edited

For a $$y > 0$$: I want to show that $$3y^2$$ is the limit of $$\frac{f(x)-f(y)}{x-y}$$. This means for any given $$\epsilon >0$$ I find a $$\delta >0$$ such that we have $$\forall x \in \mathbb{R}_{>0}~:\vert x - y\vert < \delta \Rightarrow \biggl|\frac{x ^3 - y^3}{x-y}- 3y^2\biggr| < \epsilon$$. After some manipulations and a polynomial long division I get the following: $$\biggl|\frac{x ^3 - y^3}{x-y}- 3y^2\biggr| =~ ... ~= \vert x^2-2y^2+xy\vert$$. But now I am stuck. How do I use this to show that the conition is satisfied?

I run into the same problem when I try the other case:

For $$y<0$$: I want to show that $$-3y^2$$ is the limit of $$\frac{f(x)-f(y)}{x-y}$$. This means for any given $$\epsilon >0$$ I find a $$\delta >0$$ such that we have $$\forall x \in \mathbb{R}_{<0}~:\vert x - y\vert < \delta \Rightarrow \biggl|\frac{y ^3 - x^3}{x-y}+ 3y^2\biggr| < \epsilon$$. After some manipulations and a polynomial long division I get the same as in the "$$y>0$$-case": $$\biggl|\frac{y ^3 - x^3}{x-y}+ 3y^2\biggr|=\biggl|\frac{x ^3 - y^3}{x-y}- 3y^2\biggr| =~ ... ~= \vert x^2-2y^2+xy\vert$$. How do I use this to show that the conition is satisfied?

However, the last case seems to be the easiest one...

If $$y=0$$ then I have to take a closer look at the limit. I will use the $$\epsilon$$ -$$\delta$$-definition of the limit. So if a value $$b$$ is the limit then for any given $$\epsilon >0$$ I find a $$\delta >0$$ such that we have $$\forall x \in \mathbb{R}~:\vert x - 0\vert < \delta \Rightarrow \biggl|\frac{\vert x \vert ^3 - 0}{x-0}-b \biggr| < \epsilon$$. I will now set $$b=0$$ and check if the condition is satisfied.

$$\biggl|\frac{\vert x \vert ^3 - 0}{x-0}-0 \biggr| = x^2$$. Setting $$\delta= \sqrt{\epsilon}$$ shows that the limit exists and hence $$f$$ is differentiable at $$0$$.

Did I use the $$\epsilon$$ -$$\delta$$-definition of the limit rigorously?

• They probably wanted you to handle all cases with the definition. Commented Aug 27, 2019 at 17:12
• @vonbrand, I have edited the post. Thanks for the remark. Commented Aug 27, 2019 at 17:19

No, since you only used the $$\varepsilon-\delta$$ definition at $$0$$ (correctly, by the way, although I would have taken $$\delta=\sqrt\varepsilon$$). In all other cases, what you did was to “apply the rules of differentiation”.
• Note that\begin{align}x^2-2y^2+xy&=x^2-y^2+xy-y^2\\&=(x-y)(x+y)+(x-y)y\\&=(x-y)(x+2y).\end{align}And, if you fix $y$ and $\varepsilon>0$, it is easy to show that there is a $\delta>0$ such that$$\lvert x-y\rvert<\delta\implies\bigl\lvert(x-y)(x+2y)\bigr\rvert<\varepsilon.$$ Commented Aug 28, 2019 at 14:55
• I would proceed as follows: $$\vert \delta (x+2y)\vert = \vert \delta (x+y-y+2y)\vert \leq \delta \vert x \vert + 2\delta \vert y \vert \leq \delta^2 + \delta \vert y \vert + 2 \vert y \vert \delta= 3 \vert y \vert \delta+ \delta^2.$$ Then I choose $\delta := min\{\delta_1, \delta_2\}$, where $\delta_1< \frac{\epsilon}{6\vert y\vert}$ and $\delta_2< \sqrt{\frac{\epsilon}{2}}$. This allows me to estimate: $$x^2-2y^2+xy < 3 \vert y \vert \delta+ \delta^2 < \frac{\epsilon}{2} + \frac{\epsilon}{2}=\epsilon.$$ So I have found an appropriate $\delta$. Commented Aug 28, 2019 at 17:37