The given function is: $y = \tan(\sin^2(x))$

To find the derivative $\frac{dy}{dx}$, let's go through the steps.

Step 1: Differentiate $y$ with respect to $x$

Since $y = \tan(\sin^2(x))$, we need to use the chain rule. Let: $u = \sin^2(x)$ and $y = \tan(u)$

Then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

Step 2: Differentiate $y = \tan(u)$ with respect to $u$

The derivative of $\tan(u)$ with respect to $u$ is: $\frac{dy}{du} = \sec^2(u)$

Step 3: Differentiate $u = \sin^2(x)$ with respect to $x$

Since $u = \sin^2(x)$, use the chain rule again: $\frac{du}{dx} = 2\sin(x) \cdot \cos(x) = \sin(2x)$ (using the double-angle identity for sine: $\sin(2x) = 2\sin(x)\cos(x)$).

Step 4: Combine the results

Now we can combine these derivatives: $\frac{dy}{dx} = \sec^2(\sin^2(x)) \cdot \sin(2x)$

Final Answer

$\frac{dy}{dx} = \sec^2(\sin^2(x)) \cdot \sin(2x)$

Would you like more details on any of the steps? Here are some related questions that could expand on this topic:

1. How would you find the second derivative of $y = \tan(\sin^2(x))$?
2. What is the behavior of $\sec^2(\sin^2(x)) \cdot \sin(2x)$ for large values of $x$?
3. How does the chain rule apply to functions with multiple nested layers?
4. What would be the derivative if the function were $y = \tan(\cos^2(x))$ instead?
5. Can you explain why the double-angle identity for sine is useful here?

Tip: When taking derivatives involving trigonometric functions, keep trigonometric identities in mind as they often simplify the expression.