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The given problem asks for the differentiation of the function:  
\[
y = \sin^{-1}\left(\frac{x}{2}\right).
\]

### Solution:
To differentiate this, we use the **chain rule** and the known derivative of the inverse sine function. 

#### Step 1: Recall the derivative formula for \(\sin^{-1}(u)\):
\[
\frac{d}{dx} \left(\sin^{-1}(u)\right) = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx}.
\]
Here, \(u = \frac{x}{2}\).

#### Step 2: Apply the chain rule:
Now, differentiate the function step by step.

\[
y = \sin^{-1}\left(\frac{x}{2}\right) \quad \Rightarrow \quad \frac{dy}{dx} = \frac{1}{\sqrt{1 - \left(\frac{x}{2}\right)^2}} \cdot \frac{d}{dx}\left(\frac{x}{2}\right).
\]

#### Step 3: Simplify:
First, simplify the square inside the square root:
\[
\left(\frac{x}{2}\right)^2 = \frac{x^2}{4}.
\]
Thus:
\[
\frac{dy}{dx} = \frac{1}{\sqrt{1 - \frac{x^2}{4}}} \cdot \frac{1}{2}.
\]

#### Step 4: Combine terms:
The denominator simplifies to:
\[
\sqrt{1 - \frac{x^2}{4}} = \sqrt{\frac{4 - x^2}{4}} = \frac{\sqrt{4 - x^2}}{2}.
\]

Thus:
\[
\frac{dy}{dx} = \frac{1}{\frac{\sqrt{4 - x^2}}{2}} \cdot \frac{1}{2}.
\]

#### Step 5: Final simplification:
\[
\frac{dy}{dx} = \frac{2}{\sqrt{4 - x^2}} \cdot \frac{1}{2} = \frac{1}{\sqrt{4 - x^2}}.
\]

### Final Answer:
\[
y' = \frac{1}{\sqrt{4 - x^2}}.
\]

Would you like further explanation or another example?  

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Here are **5 related questions** you could explore:
1. What is the derivative of \(\sin^{-1}(x)\)?
2. How do you differentiate \(\cos^{-1}(u)\) using the chain rule?
3. What are the domain restrictions for the function \(\sin^{-1}(x)\)?
4. How does the chain rule apply to more complicated inverse trigonometric functions?
5. What is the integral of \(\frac{1}{\sqrt{4 - x^2}}\)?

**Tip:** Inverse trigonometric functions often require special care with domain and range when differentiating or integrating.

Learn how to differentiate y = sin^(-1)(x/2) using the chain rule and the formula for inverse trigonometric functions. Follow a detailed step-by-step explanation to get the derivative, suitable for students in Grades 11-12.

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