https://file.aiquickdraw.com/mathbot/file/17299264794396dtff87e.jpg

To find the derivative \( y' \) of the function \( y = \sin^{-1}(f(\sqrt{x})) \), we'll proceed as follows:

1. **Identify the function and apply the chain rule.**

Given:
\[
y = \sin^{-1}(f(\sqrt{x}))
\]

Let \( u = f(\sqrt{x}) \), so the function becomes \( y = \sin^{-1}(u) \).

2. **Differentiate using the chain rule.**

The derivative of \( y = \sin^{-1}(u) \) is:
\[
\frac{dy}{du} = \frac{1}{\sqrt{1 - u^2}}
\]

Now, differentiate \( u = f(\sqrt{x}) \) with respect to \( x \):
\[
\frac{du}{dx} = f'(\sqrt{x}) \cdot \frac{d}{dx}(\sqrt{x}) = \frac{f'(\sqrt{x})}{2\sqrt{x}}
\]

3. **Combine the results:**

\[
y' = \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{\sqrt{1 - (f(\sqrt{x}))^2}} \cdot \frac{f'(\sqrt{x})}{2\sqrt{x}}
\]

\[
y' = \frac{f'(\sqrt{x})}{2\sqrt{x} \sqrt{1 - (f(\sqrt{x}))^2}}
\]

4. **Compare with the answer choices:**

The correct derivative expression matches:
\[
y' = \frac{f'(\sqrt{x})}{2\sqrt{x} \sqrt{1 - (f(\sqrt{x}))^2}}
\]

So, the correct answer is:
\[
\boxed{y' = \frac{f'(\sqrt{x})}{2\sqrt{x} \sqrt{1 - (f(\sqrt{x}))^2}}}
\]

This matches the second option from the top in the image.

Would you like a more detailed explanation of any step, or do you have further questions?

Here are five related questions:
1. How would the derivative change if the function were \( \cos^{-1}(f(\sqrt{x})) \) instead?
2. What is the general formula for the derivative of the inverse sine function \( \sin^{-1}(x) \)?
3. Can you derive the formula using implicit differentiation instead of the chain rule?
4. What happens if \( f(x) \) is a constant function? How does that simplify the derivative?
5. How would you approach finding the second derivative of \( y = \sin^{-1}(f(\sqrt{x})) \)?

**Tip:** When differentiating composite functions, always remember to use the chain rule step by step to ensure accuracy.

This page explains how to find the derivative of the function y = sin^(-1)(f(√x)) using the chain rule and the properties of inverse trigonometric functions. It includes step-by-step differentiation and matches the solution with answer choices.

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