<s>[INST] How is it, that $\sqrt{x^2}$ is not $ x$, but $|x|$? As far as I see, $\sqrt{x^2}$ is not $x$, but $|x|$, meaning the "absolute". I totally get this, because $x^2$ is positive, if $x$ is negative, so $\sqrt{y}$, whether $y = 10^2$ or $y = -10^2$: $y$ is positive.
But then I remember that $\sqrt{x}$ is the same as $x^{1/2}$ and thus, $\sqrt{x^2}$ is the same as $x^{1}$.
So, as far as I get it, $\sqrt{x^2} = x^{2/2}$.
But then, I can cancel $\frac{2}{2}$ to $1$. So: $\sqrt{x^2} = x^1 = x$.
Where does the absolute value in this derivation of the calculation come from?
I get it why its there: Because $(-x)^2 = x^2$, so the radicand is always positive. But from $\sqrt{x^n} = x^{n/2}$ I dont understand how it gets there.
 [/INST] It is true that $\sqrt{x} = x^{\frac{1}{2}}$ and $(\sqrt{x})^2 = (x^{\frac{1}{2}})^2 = x$. This is true for all $x$ in the domain of $\sqrt{x}$, namely $x \in [0, \infty)$.
The absolute value comes from composing the square root and square in the opposite order; that is $\sqrt{x^2} = (x^2)^{\frac{1}{2}}$. For any $x \in \mathbb{R}$, $x^2 \geq 0$ so $x^2$ is in the domain of the square root function. The square root function however returns only non-negative values, so if $x < 0$, $(x^2)^{\frac{1}{2}} \neq x$ (but if $x \geq 0$, $(x^2)^{\frac{1}{2}} = x$). In fact, as you have noted, if $x < 0$ write $x = -a$ for some $a \in (0, \infty)$, then $x^2 = (-a)^2 = a^2$, so $(x^2)^{\frac{1}{2}} = (a^2)^{\frac{1}{2}}$. Now $a > 0$ so we obtain $(a^2)^{\frac{1}{2}} = a$, but $a$ is just the absolute value of $x$, i.e. $a = |x|$.
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