Why is $|x|'=\frac{x}{|x|}$? Why is $|x|'=\frac{x}{|x|}$ ?
May someone explain this clearly? If $x>0$ then it should be 1, and if $x<0$ then it's $-1$...

You can derive this equation by writing the absolute value as $|x| =\sqrt{x^2}$. From here we see that
$$
\frac{d}{dx} |x| = \frac{d}{dx} \sqrt{x^2} = \frac{1}{2 \sqrt{x^2} } \left( \frac{d}{dx} x^2 \right) =  \frac{2x}{2 \sqrt{x^2} }  =  \frac{x}{\sqrt{x^2} } =     \frac{x}{|x|}
$$
where we use the chain rule. Notice that the function $|x|$ is differentiable on $\mathbb{R}\setminus\{0\}$, which means that in our derivation above $x \neq 0$, which is why we can have an $x$ in the denominator.
You can see that this is consistent with the equation
$$
|x|'=\begin{cases}1 & \text{ if } x > 0\\ -1 & \text{ if } x <0\end{cases}
$$
Since if $x>0$, then $|x| =x$, this means that
$$
\frac{x}{|x|} = \frac{x}{x}  = 1
$$
And similarly, for $x<0$ we have $|x| = -x$, so we get
$$
\frac{x}{|x|} = \frac{x}{-x}  = -\frac{x}{x} = -1
$$
