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.