Differentiate root x exponent of 2

"Root x" is written as $\sqrt{x}$.
"Exponent of 2" means raising the expression to the power of 2.
So, the function is $f(x) = (\sqrt{x})^2$.
For $\sqrt{x}$ to be defined, $x$ must be greater than or equal to 0 ($x \ge 0$).
When we square a square root, they cancel each other out:

$$f(x) = (\sqrt{x})^2 = x \quad \text{for } x \ge 0$$

Now we need to find the derivative of $f(x) = x$ with respect to $x$.
Using the power rule for differentiation, which states that $\frac{d}{dx}(x^n) = nx^{n-1}$:

$$\frac{d}{dx}(x) = \frac{d}{dx}(x^1) = 1 \cdot x^{1-1} = 1 \cdot x^0 = 1 \cdot 1 = 1$$

This derivative is valid for $x > 0$.

The derivative of "root x exponent of 2" is:

$$1$$