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

$y' = \frac{2\cos(x)}{1+4\sin^2(x)}.$

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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.