The problem asks you to find the derivative $y'$ given that $y = \sqrt{x}$.

Let's solve it step by step:

Step 1: Rewrite the function in exponential form

The square root of $x$ can be rewritten as: $y = \sqrt{x} = x^{\frac{1}{2}}$

Step 2: Differentiate with respect to $x$

To find $y'$, differentiate both sides with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx} \left( x^{\frac{1}{2}} \right)$

Step 3: Apply the power rule

The power rule for differentiation states that if $y = x^n$, then $\frac{dy}{dx} = n \cdot x^{n-1}$. Applying this rule: $y' = \frac{1}{2} \cdot x^{\frac{1}{2} - 1} = \frac{1}{2} \cdot x^{-\frac{1}{2}}$

Step 4: Simplify the expression

$y' = \frac{1}{2} \cdot \frac{1}{\sqrt{x}} = \frac{1}{2\sqrt{x}}$

This matches with one of the options provided. The correct answer is: $\boxed{\frac{1}{2\sqrt{x}}}$

This corresponds to the choice $\frac{1}{2} x^{-\frac{1}{2}}$.

Would you like more details or have any questions? Here are some related questions:

1. How does the power rule apply to derivatives?
2. What is the geometric interpretation of the derivative?
3. How do you differentiate other roots, like cube roots?
4. How do you find higher-order derivatives for the function $y = \sqrt{x}$?
5. What is the significance of negative exponents in derivatives?
6. Can you provide an example of applying the chain rule to a square root function?
7. How do you differentiate a product of functions, such as $y = x \sqrt{x}$?
8. How would the derivative change if $y = \sqrt{x^2 + 1}$?

Tip: Always try to express roots as exponents when differentiating, as it simplifies applying the power rule.