# Finding the derivative of $|x|^4$ using the chain rule.

I am presented with the following task:

Can you use the chain rule to find the derivatives of $|x|^4$ and $|x^4|$ in $x = 0$? Do the derivatives exist in $x = 0$? I solved the task in a rather straight-forward way, but I am worried that there's more to the task:

First of all, both functions is a variable to the power of an even number, so given that $x$ is a real number, we have that $|x^4| = |x|^4$. In order to force practical use of the chain rule, we write $|x|^4 = \sqrt{x^2}^4$. We are using the fact that taking a number to the power of an even number, and using the absolute value, gives us positive numbers exclusively. If we choose the chain $u = x^2$, thus $g(u) = \sqrt{u}^4$, we have that $u' = 2x$ og $g'(u) = (u^2)' = 2u$. Then we have that the derivative of the function, that I for pratical reasons will name $f(x)$, is $f'(x) = 2x^2 * 2x = 4x^3$. We see that the the general power rule applies here, seeing as we work with a variable to the power of an even number. The derivative in the point $x = 0$ is $4 * 0^3 = \underline{\underline{0}}$. Thus we can conclude that the derivative exists in $x = 0$.

Is this fairly logical? I'm having a hard time seeing that there is anything more to this task, but it feels like it went a bit too straightforward.

• Since $|x^4|=x^4$ for all real $x$, why bother keeping the bars? Or are you required to treat $x\mapsto |x|^4$ as composition of $x\mapsto|x|$ and of $y\mapsto y^4$? That would be problematic because the chain rule does not apply at $x=0$. Or maybe the expected answers are just "no and yes". Sep 14 '13 at 18:46

$${{\rm d}\left\vert x\right\vert^{4} \over {\rm d}x} = 4\left\vert x\right\vert^{3}\,{{\rm d}\left\vert x\right\vert \over {\rm d}x} = 4\left\vert x\right\vert^{3}\,{\rm sgn}\left(x\right) = 4\left\vert x\right\vert^{2} \left\lbrack\vphantom{\Large A}% \left\vert x\right\vert\,{\rm sgn}\left(x\right) \right\rbrack = 4x^{2}\left\lbrack\vphantom{\Large A}x\right\rbrack = 4x^{3}$$ We usually define $\quad{\rm sgn}:{\mathbb R} - \left\lbrace 0\right\rbrace \to {\mathbb R}$ such that $${\rm sgn}\left(x\right) = \left\lbrace% \begin{array}{rl} -1\,,\qquad & x < 0 \\[1mm] 1\,,\qquad & x > 0 \end{array}\right.$$ However, we usually see calculations where it is assumed "${\rm sgn}\left(0\right) = 0$" for practical purposes. The correct way is to perform the calculation for $x \not= 0$ and consider the case $x = 0$ as an independent one. It could be ( in particular cases ) that the "result $x = 0$" coincides with the "result $\not= 0$" in the limit $x \to 0^{\pm}$.

The above definition and "extremely care" are useful in 'practical calculations'. For example, let's solve ${\rm y}'\left(x\right) = 2\left\vert x\right\vert$ with ${\rm y}\left(-1\right) = -1$:

\begin{align} {\rm y}\left(x\right) - {\rm y}\left(-1\right) &= {\rm y}\left(x\right) + 1 = 2\int_{-1}^{x}{\rm sgn}\left(x'\right)x'\,{\rm d}x' = \left. \vphantom{\Huge A}x'^{2}{\rm sgn}\left(x'\right) \right\vert_{-1}^{x} - \int_{-1}^{x}x'^{2} \left\lbrack 2\delta\left(x'\right)\right\rbrack\,{\rm d}x' \\[3mm]&= x^{2}\,{\rm sgn}\left(x\right) + 1 \quad\Longrightarrow\quad {\rm y}\left(x\right) = x\,\left\vert x\right\vert\,,\quad x \not= 0 \end{align} We solved the differential equation without dividing the problem in two cases ( $x < 0$ and $x > 0$ ). Since ${\rm y}\left(0^{\pm}\right) = 0\left\vert 0\right\vert$, we 'adopt' as a solution ${\rm y}\left(x\right) = x\,\left\vert x\right\vert,\ {\large\forall\ x}$.

If you are in the Physics area, you'll find many situations like the one you addressed.

• To elaborate a bit, treating $|x|$ this way is an example of treating it as a distribution, or generalized function. For a distribution and a "test function" $\phi$ (which vanishes at infinity), $<|x|', \phi>$ = $-<|x|, \phi'> = -\int_{-\infty}^{\infty}|x|\phi(x)' = \int_{-\infty}^{0}x\phi(x)' - \int_{0}^{\infty}x\phi(x)' = \int_{-\infty}^{0}\phi(x) - \int_{0}^{\infty}\phi(x) = <sgn(x),\phi>$. Sep 15 '13 at 3:59
• @Bitrex You are quite right. It gives 'more room' for differential equations and, in addition, it becomes useful for practical purposes. Thanks. Sep 16 '13 at 0:57

It's worth noting that $|x^4|$ and $|x|^4$ equal $x^4$, but no matter. I'll assume that any simplifications like that are off-limits throughout.

One way to express the derivative of $|x|$ is $\frac{|x|}{x}$. So if we applied the chain rule to $|x^4|$ we have $\frac{|x^4|}{x^4}\cdot4x^3$, which is undefined at $0$. However this expression is defined in a neighborhood of $0$, and its limit exists, because $\frac{|x^4|}{x^4}$ is bounded and $4x^3$ approaches $0$.

Something similar could be done with $|x|^4$.

You can apply the chain rule to $(x^{2})^{2}$, which happens to equal $|x^{4}|=|x|^{4}$, since it is the composition of two functions differentiable at $0$ (i.e take $f(x)=x^{2}$ then this is $f\circ f$). You could actually proceed quicker since $|x^{4}|=|x|^{4}=x^{4}$ so we don't need chain rule. What you are using is that $f(x)=|x^{4}|=|x|^{4}$ agrees everywhere with a function that is differentiable everywhere and is hence differentiable everywhere itself. You can prove this using difference quotients.

Let $f$ be a function that agrees everywhere with a function $g$ whose derivative at $x_{0}$ exists.. Then is $f$ is differentiable at $x_{0}$ and $f'(x_{0})=g'(x_{0})$

$\lim_{h\to0}\frac{f(x_{0}+h)-f(x_{0})}{h}=\lim_{h\to0}\frac{g(x_{0}+h)-g(x_{0})}{h}=g'(x_{0})$. So $f'(x_{0})$ exists and equals $g'(x_{0})$.

• I think if it were OK to replace $|x^4|$ with $(x^2)^2$, then it would also be OK to just replace it with $x^4$. My guess is that OP needs to use the Chain Rule explicitly with one of the functions being the absolute value function. Sep 14 '13 at 18:02
• I believe my answer included representing $|x^{4}|$ as $x^{4}$ which allows us to avoid chain rule. Explicitly would be fine too since then $sgn(x^{4})$ is bounded and $4x^{3}$ goes to $0$ as $x$ goes to $0$. Sep 14 '13 at 18:11