Absolute Value Derivatives Let's say I have the function $f(x) = 10|x-2| - x^2 $ and I want to find the derivative. Since I can rewrite $ |x| $ as $\sqrt(x^2) $, can I also rewrite $ |x-2| $ as $ \sqrt{(x-2)^2}$ ?   (sorry for weird formatting Im new here)
Generally we introduce the sign function $\operatorname{sgn(x)}=\dfrac x{|x|}=\pm 1$ for $x\neq 0$ and $\operatorname{sgn}(0)=0$ 

[ The value in zero is a convention but we can also choose to let it undefined. ]

This way we can derivate $f(x)=|x-2|$

$f(x):\begin{cases}x-2 &amp; \text{for } x\ge 2\\ 2-x &amp; \text{for } x\le 2\end{cases}\quad$ thus $\quad f'(x):\begin{cases}+1 &amp; \text{for } x&gt; 2\\ -1 &amp; \text{for } x&lt; 2\end{cases}$

Or similarly $f'(x)=\operatorname{sgn}(x-2)$ for $x\neq 2$

Your remark about using the square root, leads indeed to that same result:

$f(x)=|x-2|=\sqrt{(x-2)^2}$ 

$f'(x)=\dfrac{\frac 12\times2\times(x-2)}{\sqrt{(x-2)^2}}=\dfrac{x-2}{|x-2|}=\operatorname{sgn}(x-2)\quad$ as defined above for $x\neq 2$.

[ Again, the derivative in $x=2$ does not exists, in the strict functional sense since there is a discontinuity there, it only exists as a distribution, but this is another subject. ]
Absolute Value can be viewed as finding the magnitude, or radius of a vector.
So using Pythagorean Theorem:$|x+y|=\sqrt{x^2+y^2}$ for $x+y\in \Bbb R^{2}$. Since your term is in $\Bbb R^{1}$, one of the vector components is necessarily $0$.

Letting $y=0$ and $x=u-2$:

You get $|u-2|=\sqrt{(u-2)^2}$

You can also view the Absolute Value as a piece-wise function $|x-a|=x-a$, for $x&gt;a$ and $|x-a|=a-x$, for $x&lt;a$.

Differentiating:$$\frac{d|x-a|}{dx}=1,x&gt;a$$
and $$\frac{d|x-a|}{dx}=-1, x&lt;a$$