2
$\begingroup$

Why is $|x|'=\frac{x}{|x|}$ ?

May someone explain this clearly? If $x>0$ then it should be 1, and if $x<0$ then it's $-1$...

$\endgroup$
6
  • 4
    $\begingroup$ ... which is exactly what $\frac{x}{\lvert x\rvert}$ is. $\endgroup$
    – user239203
    Jun 19, 2020 at 21:21
  • 1
    $\begingroup$ By definition of absolute value, $|x|=x$ if $x\ge0$ and $|x|=-x$ otherwise. So $\frac{|x|}x=\frac x{|x|}=\pm1$ depending on the sign of $x$, and undefined if $x=0$. $\endgroup$
    – user170231
    Jun 19, 2020 at 21:21
  • $\begingroup$ And it is, use the definition of absolute value $\endgroup$
    – Presage
    Jun 19, 2020 at 21:21
  • 1
    $\begingroup$ Yes because if x<0, |x|=-x is decreasing. $\endgroup$
    – Naj Kamp
    Jun 19, 2020 at 21:21
  • 1
    $\begingroup$ Just think about what the function $x/|x|$ does, and consider the derivative of $|x|$ when $x>0$ and when $x<0$. Out of curiosity, why did you include "induction" among the tags? $\endgroup$
    – D Ford
    Jun 19, 2020 at 21:29

5 Answers 5

6
$\begingroup$

Well, for a full answer, let's deal with this problem in $n$ dimensions. Let $\mathbf{x}=(x_1,...,x_n)=\sum_{i=1}^{n} x_i \widehat{\mathbf{e}_i}$ and $\Vert\mathbf{x}\Vert = \sqrt{\sum_{i=1}^{n} {x_i}^2}$ Then $$\nabla \Vert\mathbf{x}\Vert = \sum_{i=1}^{n} \frac{\partial \Vert\mathbf{x}\Vert}{\partial x_i} \widehat{\mathbf{e}_i}$$ Using the chain rule, $$\frac{\partial \Vert\mathbf{x}\Vert}{\partial x_i} = \frac{1}{2\sqrt{\sum_{i=1}^{n} {x_i}^2}} {2 x_i}$$ $$=\frac{x_i}{\Vert\mathbf{x}\Vert}$$ Thus $$\nabla \Vert\mathbf{x}\Vert = \sum_{i=1}^{n} \frac{x_i}{\Vert\mathbf{x}\Vert} \widehat{\mathbf{e}_i}$$ $$= \frac{\mathbf{x}}{\Vert\mathbf{x}\Vert}.$$

$\endgroup$
1
  • 1
    $\begingroup$ I was typing my answer from mobile while you posted your answer. I didn't see it, otherwise I would have not posted mine (since it is just the 1-dimensional case) +1 :) $\endgroup$
    – Quillo
    Jun 19, 2020 at 21:48
4
$\begingroup$

You may also want to consider this alternative derivation. Write the modulus as $|x| =\sqrt{x^2}$. Therefore, using the chain rule: $$ \frac{d}{dx}|x| = \frac{2x}{2\sqrt{x^2}} = \frac{x}{|x|} $$

$\endgroup$
1
  • 2
    $\begingroup$ Interesting, but asking for trouble, I think, insofar as it changes the issues to some other implicit/concealed issues... $\endgroup$ Jun 19, 2020 at 21:39
2
$\begingroup$

You can derive this equation by writing the absolute value as $|x| =\sqrt{x^2}$. From here we see that $$ \frac{d}{dx} |x| = \frac{d}{dx} \sqrt{x^2} = \frac{1}{2 \sqrt{x^2} } \left( \frac{d}{dx} x^2 \right) = \frac{2x}{2 \sqrt{x^2} } = \frac{x}{\sqrt{x^2} } = \frac{x}{|x|} $$ where we use the chain rule. Notice that the function $|x|$ is differentiable on $\mathbb{R}\setminus\{0\}$, which means that in our derivation above $x \neq 0$, which is why we can have an $x$ in the denominator.

You can see that this is consistent with the equation $$ |x|'=\begin{cases}1 & \text{ if } x > 0\\ -1 & \text{ if } x <0\end{cases} $$ Since if $x>0$, then $|x| =x$, this means that $$ \frac{x}{|x|} = \frac{x}{x} = 1 $$ And similarly, for $x<0$ we have $|x| = -x$, so we get $$ \frac{x}{|x|} = \frac{x}{-x} = -\frac{x}{x} = -1 $$

$\endgroup$
1
$\begingroup$

Note that $$|x|=\begin{cases}x & \text{ if } x \geq 0\\ -x & \text{ if } x <0\end{cases}$$ The function has a derivative on $(-\infty, 0) \cup (0,\infty)$ and is given by $$\frac{d}{dx}|x|=\begin{cases}1 & \text{ if } x > 0\\ -1 & \text{ if } x <0\end{cases}.$$ This is just the function $\frac{|x|}{x}$ (see the function definition of $|x|$ given above) for all $x \neq 0$.

$\endgroup$
0
$\begingroup$

Since $|x|^2=x^2\ $ we get $\ 2x=(x^2)'=(|x|^2)'=2|x||x|'\ $ thus $\ |x|'=\dfrac{x}{|x|}$ for $x\neq 0$.

$\endgroup$

You must log in to answer this question.