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To solve for the derivative \( y' \) of the function \( f(x) = \tan^{-1}(2\sin(x)) \), we proceed as follows:

### Step 1: Derivative of \( \tan^{-1}(u) \)
The derivative of \( \tan^{-1}(u) \) with respect to \( x \) is:
\[
\frac{d}{dx} \tan^{-1}(u) = \frac{1}{1+u^2} \cdot \frac{du}{dx}.
\]
Here, \( u = 2\sin(x) \).

### Step 2: Compute \( u' \)
The derivative of \( u = 2\sin(x) \) is:
\[
\frac{du}{dx} = 2\cos(x).
\]

### Step 3: Combine the derivatives
Using the chain rule:
\[
y' = \frac{1}{1+(2\sin(x))^2} \cdot \frac{du}{dx}.
\]
Substitute \( \frac{du}{dx} = 2\cos(x) \):
\[
y' = \frac{1}{1+4\sin^2(x)} \cdot 2\cos(x).
\]

### Final Answer
\[
y' = \frac{2\cos(x)}{1+4\sin^2(x)}.
\]

Would you like further clarification or additional steps explained?

### Related Questions:
1. How is the derivative of \( \tan^{-1}(u) \) derived?
2. Can you apply this derivative rule to other inverse trigonometric functions?
3. What is the importance of the chain rule in solving this problem?
4. How would this derivative change if \( u \) were more complex, e.g., \( 2\sin^2(x) \)?
5. What is the domain of the resulting derivative function?

### Tip:
Always simplify trigonometric expressions when possible, as this can reveal simpler equivalent forms or help in integration.

Find the derivative of f(x) = tan^(-1)(2sin(x)). Express numbers in exact form. Use symbolic notation and fractions where needed.

Learn how to find the derivative of f(x) = tan^(-1)(2sin(x)) using the chain rule and trigonometric identities. This step-by-step solution explains the process of calculating derivatives involving inverse trigonometric functions. Suitable for high school calculus students.

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