To find the derivative of $y = \sqrt{\arcsin(x)}$, we can apply the chain rule. Let's break this down step by step:

Step 1: Rewrite the function

We have: $y = \sqrt{\arcsin(x)}$ This is the same as: $y = (\arcsin(x))^{\frac{1}{2}}$

Step 2: Apply the chain rule

We differentiate using the chain rule. The derivative of $(\arcsin(x))^{\frac{1}{2}}$ is: $\frac{dy}{dx} = \frac{1}{2} (\arcsin(x))^{-\frac{1}{2}} \cdot \frac{d}{dx}(\arcsin(x))$

Step 3: Derivative of $\arcsin(x)$

The derivative of $\arcsin(x)$ with respect to $x$ is: $\frac{d}{dx}(\arcsin(x)) = \frac{1}{\sqrt{1 - x^2}}$

Step 4: Combine the results

Now, combining the derivatives, we get: $\frac{dy}{dx} = \frac{1}{2} (\arcsin(x))^{-\frac{1}{2}} \cdot \frac{1}{\sqrt{1 - x^2}}$

Final Answer:

$\frac{dy}{dx} = \frac{1}{2\sqrt{\arcsin(x)}} \cdot \frac{1}{\sqrt{1 - x^2}}$

Let me know if you'd like more details or if you have any questions!

Related Questions:

1. How do we differentiate more complex compositions like $y = \sqrt{\arccos(x)}$?
2. What is the geometric interpretation of the derivative of $\arcsin(x)$?
3. How does the derivative of inverse trigonometric functions behave as $x$ approaches the boundaries of their domain?
4. What happens if we apply the product rule to functions involving inverse trig functions?
5. Can you derive the second derivative of $y = \sqrt{\arcsin(x)}$?

Tip: When applying the chain rule, always make sure to differentiate the inner function as well as the outer function!