Finding the Derivative of |x| using the Limit Definition 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
 A: 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:
$$\begin{align}
\frac{\sqrt{(x+\Delta x)^2}-\sqrt{x^2}}{\Delta x} 
& = \frac{(\sqrt{(x+\Delta x)^2}-\sqrt{x^2})(\sqrt{(x+\Delta x)^2}+\sqrt{x^2})}{\Delta x (\sqrt{(x+\Delta x)^2}+\sqrt{x^2})} \\
& = \frac{(x+\Delta x)^2-x^2}{\Delta x (\sqrt{(x+\Delta x)^2}+\sqrt{x^2})} \\
& = \frac{2 x \Delta x + \Delta x^2}{\Delta x (\sqrt{(x+\Delta x)^2}+\sqrt{x^2})} \\
& = \frac{2x + \Delta x}{\sqrt{(x+\Delta x)^2}+\sqrt{x^2}} \\
& \to \frac{2x}{\sqrt{x^2}+\sqrt{x^2}} = \frac{2x}{2|x|} = \frac{x}{|x|}
\end{align}$$
A: $\dfrac{d}{dx}|x|$
$=\lim\limits_{\Delta x\to 0}\dfrac{|x+\Delta x|-|x|}{\Delta x}$
$=\lim\limits_{\Delta x\to 0}\dfrac{(|x+\Delta x|-|x|)(|x+\Delta x|+|x|)}{\Delta x(|x+\Delta x|+|x|)}$
$=\lim\limits_{\Delta x\to 0}\dfrac{|x+\Delta x|^2-|x|^2}{\Delta x(|x+\Delta x|+|x|)}$
$=\lim\limits_{\Delta x\to 0}\dfrac{(x+\Delta x)^2-x^2}{\Delta x(|x+\Delta x|+|x|)}$
$=\lim\limits_{\Delta x\to 0}\dfrac{x^2+2x\Delta x+(\Delta x)^2-x^2}{\Delta x(|x+\Delta x|+|x|)}$
$=\lim\limits_{\Delta x\to 0}\dfrac{2x+\Delta x}{|x+\Delta x|+|x|}$
$=\dfrac{x}{|x|}$
$=\dfrac{x|x|}{|x|^2}$
$=\dfrac{x|x|}{x^2}$
$=\dfrac{|x|}{x}$
A: 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 :-)
A: 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.
A: 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. 
So: 
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. 
A: One can cast this question about the differentiation of absolute value operator into the more general setting of (partial) differentiation of the Euclidean norm. This article gives the formula in $\mathbb{R^n}$ with coordiantes $x=(x_1,x_2,\cdots x_n)$ :
$$\dfrac{\partial}{\partial x_k}\|x\|=\dfrac{x_k}{\|x\|}$$
Considering the case $n=1$, one gets :
$$\dfrac{ \mathrm{d}}{ \mathrm{d}x}\|x\|=\dfrac{x}{|x|}$$
A: Using alternative limit definition of the derivative of function:
$$\lim_\limits{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=\lim_\limits{x\to x_0}\frac{|x|-|x_0|}{x-x_0}=\lim_\limits{x\to x_0}\frac{(|x|-|x_0|)(|x|+|x_0|)}{(x-x_0)(|x|+|x_0|)}=$$
$$\lim_\limits{x\to x_0}\frac{x^2-x_0^2}{(x-x_0)(|x|+|x_0|)}=\lim_\limits{x\to x_0}\frac{(x-x_0)(x+x_0)}{(x-x_0)(|x|+|x_0|)}=\lim_\limits{x\to x_0}\frac{x+x_0}{|x|+|x_0|}=\frac{2x_0}{2|x_0|}=\frac{x_0}{|x_0|}.$$
