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
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Sign up to join this communityPlease 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
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.
$\dfrac{d}{dx}|x|$
$=\lim\limits_{\Delta x\to 0}\dfrac{|x+\Delta x|-|x|}{\Delta x}$
$=\lim\limits_{\Delta x\to 0}\dfrac{(|x+\Delta x|-|x|)(|x+\Delta x|+|x|)}{\Delta x(|x+\Delta x|+|x|)}$
$=\lim\limits_{\Delta x\to 0}\dfrac{|x+\Delta x|^2-|x|^2}{\Delta x(|x+\Delta x|+|x|)}$
$=\lim\limits_{\Delta x\to 0}\dfrac{(x+\Delta x)^2-x^2}{\Delta x(|x+\Delta x|+|x|)}$
$=\lim\limits_{\Delta x\to 0}\dfrac{x^2+2x\Delta x+(\Delta x)^2-x^2}{\Delta x(|x+\Delta x|+|x|)}$
$=\lim\limits_{\Delta x\to 0}\dfrac{2x+\Delta x}{|x+\Delta x|+|x|}$
$=\dfrac{x}{|x|}$
$=\dfrac{x|x|}{|x|^2}$
$=\dfrac{x|x|}{x^2}$
$=\dfrac{|x|}{x}$
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: $$\begin{align} \frac{\sqrt{(x+\Delta x)^2}-\sqrt{x^2}}{\Delta x} & = \frac{(\sqrt{(x+\Delta x)^2}-\sqrt{x^2})(\sqrt{(x+\Delta x)^2}+\sqrt{x^2})}{\Delta x (\sqrt{(x+\Delta x)^2}+\sqrt{x^2})} \\ & = \frac{(x+\Delta x)^2-x^2}{\Delta x (\sqrt{(x+\Delta x)^2}+\sqrt{x^2})} \\ & = \frac{2 x \Delta x + \Delta x^2}{\Delta x (\sqrt{(x+\Delta x)^2}+\sqrt{x^2})} \\ & = \frac{2x + \Delta x}{\sqrt{(x+\Delta x)^2}+\sqrt{x^2}} \\ & \to \frac{2x}{\sqrt{x^2}+\sqrt{x^2}} = \frac{2x}{2|x|} = \frac{x}{|x|} \end{align}$$
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 :-)
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.
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 :
$$\dfrac{ \mathrm{d}}{ \mathrm{d}x}\|x\|=\dfrac{x}{|x|}$$
Using alternative limit definition of the derivative of function:
$$\lim_\limits{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=\lim_\limits{x\to x_0}\frac{|x|-|x_0|}{x-x_0}=\lim_\limits{x\to x_0}\frac{(|x|-|x_0|)(|x|+|x_0|)}{(x-x_0)(|x|+|x_0|)}=$$ $$\lim_\limits{x\to x_0}\frac{x^2-x_0^2}{(x-x_0)(|x|+|x_0|)}=\lim_\limits{x\to x_0}\frac{(x-x_0)(x+x_0)}{(x-x_0)(|x|+|x_0|)}=\lim_\limits{x\to x_0}\frac{x+x_0}{|x|+|x_0|}=\frac{2x_0}{2|x_0|}=\frac{x_0}{|x_0|}.$$