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Let's solve the two parts of this problem step by step:

### Part (i): Find the derivative of \( \sin^{-1}(2x) \)

We know that the derivative of the inverse sine function is given by:

\[
\frac{d}{dx} \left( \sin^{-1}(u) \right) = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx}
\]

Here, \( u = 2x \), so:

\[
\frac{d}{dx} \left( \sin^{-1}(2x) \right) = \frac{1}{\sqrt{1 - (2x)^2}} \cdot \frac{d}{dx}(2x)
\]

Now, \( \frac{d}{dx}(2x) = 2 \), so we have:

\[
\frac{d}{dx} \left( \sin^{-1}(2x) \right) = \frac{2}{\sqrt{1 - 4x^2}}
\]

### Part (ii): Find the integral \( \int \frac{\sin^{-1}(2x)}{\sqrt{1 - 4x^2}} \, dx \)

We already know the derivative of \( \sin^{-1}(2x) \), which helps us in the second part.

Let \( I = \int \frac{\sin^{-1}(2x)}{\sqrt{1 - 4x^2}} \, dx \).

At this point, solving this integral involves using integration by parts, where we can take:

- \( u = \sin^{-1}(2x) \)
- \( dv = \frac{1}{\sqrt{1 - 4x^2}} \, dx \)

Then, the integration process will follow the rule:

\[
\int u \, dv = uv - \int v \, du
\]

Let's perform this computation next.

1. \( u = \sin^{-1}(2x) \) implies \( du = \frac{2}{\sqrt{1 - 4x^2}} \, dx \).
2. \( dv = \frac{1}{\sqrt{1 - 4x^2}} \, dx \) implies \( v = \sin^{-1}(2x) \).

So, applying integration by parts leads to:

\[
I = \sin^{-1}(2x) \cdot \sin^{-1}(2x) - \int \sin^{-1}(2x) \cdot \frac{2}{\sqrt{1 - 4x^2}} \, dx
\]

At this stage, solving the integral becomes more complex and may require advanced techniques, including substitution or reduction formulas.

Would you like a further breakdown of the steps or assistance in completing this part of the solution?

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Here are some related questions that extend the above concepts:
1. What is the derivative of \( \cos^{-1}(3x) \)?
2. How can we use substitution to solve integrals of trigonometric inverse functions?
3. What are some techniques for solving integrals involving square roots of quadratic expressions?
4. How does integration by parts work for more complex functions?
5. What are common applications of inverse trigonometric functions in calculus?

**Tip:** When working with inverse trigonometric functions, always remember their derivatives and how they relate to square root expressions in integrals.

Find the derivative of sin^(-1)(2x). Hence, find the integral of (sin^(-1)(2x)) / sqrt(1 - 4x^2) dx.

Learn how to find the derivative of sin^(-1)(2x) and solve the integral of (sin^(-1)(2x)) / sqrt(1 - 4x^2) using integration by parts. This problem covers key calculus concepts like inverse trigonometric derivatives and advanced integration techniques.

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