Please Help me derive the derivative of the absolute value of x using the following limit definition. $$\lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x} $$ I have no idea as to how to get started.Please Help.

Thank You

  • $\begingroup$ Does $|x|$ differentiable at zero? $\endgroup$ – Hassan Muhammad Nov 20 '11 at 11:25

Since the absolute value is defined by cases, $$|x|=\left\{\begin{array}{ll} x & \text{if }x\geq 0;\\ -x & \text{if }x\lt 0, \end{array}\right.$$ it makes sense to deal separately with the cases of $x\gt 0$, $x\lt 0$, and $x=0$.

For $x\gt0$, for $\Delta x$ sufficiently close to $0$ we will have $x+\Delta x\gt 0$. So $f(x)= |x| = x$, and $f(x+\Delta x) = |x+\Delta x| = x+\Delta x$; plugging that into the limit, we have: $$\lim_{\Delta x\to 0}\frac{f(x+\Delta x) - f(x)}{\Delta x} = \lim_{\Delta x\to 0}\frac{|x+\Delta x|-|x|}{\Delta x} = \lim_{\Delta x\to 0}\frac{(x+\Delta x)-x}{\Delta x}.$$ You should be able to finish it now.

For $x\lt 0$, for $\Delta x$ sufficiently close to zero we will have $x+\Delta x\lt 0$; so $f(x) = -x$ and $f(x+\Delta x) = -(x+\Delta x)$. It should again be easy to finish it.

The tricky one is $x=0$. I suggest using one-sided limits. For the limit as $\Delta x\to 0^+$, $x+\Delta x = \Delta x\gt 0$; for $\Delta x \to 0^-$, $x+\Delta x = \Delta x\lt 0$; the (one-sided) limits should now be straightforward.

  • $\begingroup$ If i plug in the given function into the derivative formula i get $$\lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x) - x}{\Delta x}$$.Upon evaluating it i get the final answer as 1 $\endgroup$ – alok Nov 20 '11 at 7:22
  • 2
    $\begingroup$ @alok: Notice that you'll get different result for $x<0, x>0$, and $x=0$. $\endgroup$ – ofer Nov 20 '11 at 7:37
  • $\begingroup$ @alok: Be sure to keep track of what case you are in! You will not get $1$ except when $x\gt 0$, or when $x=0$ and $\Delta x\to 0^+$. If you are getting $1$ in all cases, you are doing it wrong. $\endgroup$ – Arturo Magidin Nov 20 '11 at 21:46
  • $\begingroup$ The derivative does not exist at 0! $\endgroup$ – The Great Duck Jun 12 '16 at 2:28


$=\lim\limits_{\Delta x\to 0}\dfrac{|x+\Delta x|-|x|}{\Delta x}$

$=\lim\limits_{\Delta x\to 0}\dfrac{(|x+\Delta x|-|x|)(|x+\Delta x|+|x|)}{\Delta x(|x+\Delta x|+|x|)}$

$=\lim\limits_{\Delta x\to 0}\dfrac{|x+\Delta x|^2-|x|^2}{\Delta x(|x+\Delta x|+|x|)}$

$=\lim\limits_{\Delta x\to 0}\dfrac{(x+\Delta x)^2-x^2}{\Delta x(|x+\Delta x|+|x|)}$

$=\lim\limits_{\Delta x\to 0}\dfrac{x^2+2x\Delta x+(\Delta x)^2-x^2}{\Delta x(|x+\Delta x|+|x|)}$

$=\lim\limits_{\Delta x\to 0}\dfrac{2x+\Delta x}{|x+\Delta x|+|x|}$





  • $\begingroup$ Can you show the further derivatives? $\endgroup$ – Math Lover Jan 9 at 15:16

We can write $|x| = \sqrt{x^2}$. Using the chain rule we then get $$|x|' = \frac{1}{2\sqrt{x^2}} \cdot 2x = \frac{x}{\sqrt{x^2}} = \frac{x}{|x|}$$

EDIT Using limit: $$\begin{align} \frac{\sqrt{(x+\Delta x)^2}-\sqrt{x^2}}{\Delta x} & = \frac{(\sqrt{(x+\Delta x)^2}-\sqrt{x^2})(\sqrt{(x+\Delta x)^2}+\sqrt{x^2})}{\Delta x (\sqrt{(x+\Delta x)^2}+\sqrt{x^2})} \\ & = \frac{(x+\Delta x)^2-x^2}{\Delta x (\sqrt{(x+\Delta x)^2}+\sqrt{x^2})} \\ & = \frac{2 x \Delta x + \Delta x^2}{\Delta x (\sqrt{(x+\Delta x)^2}+\sqrt{x^2})} \\ & = \frac{2x + \Delta x}{\sqrt{(x+\Delta x)^2}+\sqrt{x^2}} \\ & \to \frac{2x}{\sqrt{x^2}+\sqrt{x^2}} = \frac{2x}{2|x|} = \frac{x}{|x|} \end{align}$$

  • $\begingroup$ The OP wants to obtain the result using only the definition, nothing more. $\endgroup$ – Alex M. May 31 '17 at 7:11
  • $\begingroup$ I think there is a typo in the first method, the last squality should be $\dfrac {x}{\sqrt {x^2}} = \dfrac {x}{|x|}$? $\endgroup$ – Ovi May 31 '17 at 12:01
  • $\begingroup$ Oh, you're correct, I've missed ${}^2$. Will now fix it. $\endgroup$ – md2perpe May 31 '17 at 12:14

Well, the simple answer is if x < 0, it's obviously a linear line with a slope of -1, and when x > 0, it's a line with slope 1, and at x = 0, both formulas can be used and therefore we can't calculate the derivate.


when x > 0, |x|' = 1

when x < 0, |x|' = -1

when x = 0, |x|' is undefined

As doraemonpaul said, the implicit way to say this is |x|/x, because it returns the sign of x, except when x = 0, when dividing by 0 makes it undefined.

x/|x| could also work.

  • 2
    $\begingroup$ What is a linear line? :) $\endgroup$ – R_D Nov 8 '16 at 4:14

Cheap, non-rigorous, non-mathematical, engineering-type answer: sgn(x) ("signum x", the sign of x, being -1 for x<0 and +1 for x>0). Note that sgn(0) = 0, which is a practical compromise, being the average of -1 ("coming from the negatives") and +1 ("coming from the positives").

Of course we all know that d|x|/dx is not defined at x=0. Intuitively: the "tangent" of |x| at x=0 can be any line with slope -1 < s < +1 . So it is not unique. Hence: no derivative at that point.

Once again: these are not rigorous considerations (see @doraemonpaul 's answer for proper maths), but rather intuitive hints that help you grasp the issue.

Mathematica's answer (Version 11) is even more pragmatic: D[Abs[x], x] ==> Abs'[x] . I like it a lot :-)


Using alternative limit definition of the derivative of function:

$$\lim_\limits{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=\lim_\limits{x\to x_0}\frac{|x|-|x_0|}{x-x_0}=\lim_\limits{x\to x_0}\frac{(|x|-|x_0|)(|x|+|x_0|)}{(x-x_0)(|x|+|x_0|)}=$$ $$\lim_\limits{x\to x_0}\frac{x^2-x_0^2}{(x-x_0)(|x|+|x_0|)}=\lim_\limits{x\to x_0}\frac{(x-x_0)(x+x_0)}{(x-x_0)(|x|+|x_0|)}=\lim_\limits{x\to x_0}\frac{x+x_0}{|x|+|x_0|}=\frac{2x_0}{2|x_0|}=\frac{x_0}{|x_0|}.$$


protected by Alex M. May 31 '17 at 7:03

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.