To find the derivative \( y' \) of the function \( y = \sin^{-1}(4x + 1) \), we can use the chain rule. The derivative of \( \sin^{-1}(u) \) with respect to \( u \) is given by:

\[
\frac{dy}{du} = \frac{1}{\sqrt{1 - u^2}}
\]

where \( u = 4x + 1 \). Then, we also need to find \( \frac{du}{dx} \):

\[
\frac{du}{dx} = 4
\]

Now applying the chain rule:

\[
y' = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{\sqrt{1 - (4x + 1)^2}} \cdot 4
\]

So, we get:

\[
y' = \frac{4}{\sqrt{1 - (4x + 1)^2}}
\]

### Summary
The derivative \( y' \) is:

\[
y' = \frac{4}{\sqrt{1 - (4x + 1)^2}}
\]

Would you like more details on any part of this solution? Here are some related questions you might consider:

1. What are the restrictions on the domain of \( y = \sin^{-1}(4x + 1) \)?
2. How do you find the second derivative \( y'' \)?
3. Can you solve for \( x \) given a specific value of \( y \)?
4. How does the graph of \( y = \sin^{-1}(4x + 1) \) look?
5. What is the relationship between inverse trigonometric functions and their derivatives?

**Tip:** Remember to consider the domain of the function when working with inverse trigonometric functions, as they have specific input restrictions.

Explore how to find the derivative of the function y = sin^{-1}(4x + 1) using the chain rule. This solution breaks down the differentiation process, including necessary formulas and theorems, ideal for high school students in Grades 11-12.

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