# What is the derivative of sqrtx^2?

Jun 22, 2015

The root of a square is the number itself.

#### Explanation:

${\sqrt{x}}^{2} = x$

The derivative of this is $1$

Jun 22, 2015

If $f \left(x\right) = {\sqrt{x}}^{2}$, the $f \left(x\right) = x$, so $f ' \left(x\right) = 1$, but

#### Explanation:

Be careful:

$g \left(x\right) = \sqrt{{x}^{2}} = \left\mid x \right\mid = \left\{\begin{matrix}x & \text{if" & x >= 0 \\ -x & "if} & x < 0\end{matrix}\right.$

So $g ' \left(x\right) = \left\{\begin{matrix}1 & \text{if" & x > 0 \\ -1 & "if} & x < 0\end{matrix}\right.$

$g ' \left(0\right)$ does not exist.