# Thread: Derivativ eo asbolute value

1. ## Derivativ eo asbolute value

$y= \mid{x\mid}^2$
$y= \sqrt{x^2}^2$
$2 \sqrt{x^2}*\frac{1}{2 sqrt(x^2)}*2x$
$\frac{4x*sqrt(x^2)}{2*sqrt(x^2)}$ - how does the LaTex work with this?
where $\mid{x\mid}^2 = \sqrt{x^2}^2$
I got $2x.$
This seems too simplified. Does it really simplify that much? Thanks!

2. Originally Posted by Truthbetold
$y= \mid{x\mid}^2$
$y= \sqrt{x^2}^2$
$2 \sqrt{x^2}*\frac{1}{2 sqrt(x^2)}*2x$
$\frac{4x*sqrt(x^2)}{2*sqrt(x^2)}$ - how does the LaTex work with this?
where $\mid{x\mid}^2 = \sqrt{x^2}^2$
I got $2x.$
This seems too simplified. Does it really simplify that much? Thanks!
Hello,

surprising, isn't it? But you are right:

Use the definition of the absolute value:

$| x | =\left \{ \begin{array}{lr}x, & x > 0 \\0, & x = 0\\-x, &x < 0\end{array}\right.$

Now take your function:

$y = | x |^2 = \left \{ \begin{array}{lr}x^2, & x > 0 \\0, & x = 0\\(-x)^2 = x^2, &x < 0\end{array}\right.$ which will give the derivation $y' = 2 \cdot | x | = \left \{ \begin{array}{lr}2x, & x > 0 \\0, & x = 0\\2 \cdot (-x) \cdot (-1), &x < 0\end{array}\right.$

3. Hmm.. Isn't ${|x|}^2 = x^2$ anyway?

4. Originally Posted by wingless
Hmm.. Isn't ${|x|}^2 = x^2$ anyway?
in fact, you are correct..

5. Originally Posted by Truthbetold
$\frac{4x*sqrt(x^2)}{2*sqrt(x^2)}$ - how does the LaTex work with this?
just put a \ before sqrt..