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?

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.