To find the derivative of the function \( f(x) = \arcsin(x) \), we'll use the chain rule and the derivative formula for the inverse sine function.

The derivative of \( \arcsin(x) \) with respect to \( x \) is given by:

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

Thus, the derivative of the function \( f(x) = \arcsin(x) \) is:

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

This result is valid for \( -1 < x < 1 \), where the function \( \arcsin(x) \) is defined and differentiable.

If you have any further questions or need additional details, please let me know!

Here are some related questions to consider:
1. What is the domain of the function \( f(x) = \arcsin(x) \)?
2. How do you apply the chain rule in differentiation?
3. What is the range of the function \( f(x) = \arcsin(x) \)?
4. How do you find the derivative of inverse trigonometric functions?
5. What are the applications of the derivative of \( \arcsin(x) \)?
6. How does the derivative of \( \arcsin(x) \) behave as \( x \) approaches 1 or -1?
7. Can you differentiate \( \arcsin(x) \) with respect to another variable?
8. How do you interpret the derivative \( f'(x) = \frac{1}{\sqrt{1 - x^2}} \) geometrically?

**Tip:** Remember that the derivative of an inverse function gives the rate of change of the function's output with respect to its input.

Understand how to find the derivative of arcsin(x) with a step-by-step explanation using the chain rule and the derivative formula for inverse sine. Suitable for advanced high school students and anyone interested in calculus and trigonometry.

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