The given function is:

$y = \frac{1}{\sin^2(x)}$

To find the derivative of $y$, we can rewrite it as:

$y = (\sin(x))^{-2}$

Now, use the chain rule for differentiation:

$\frac{dy}{dx} = -2 \cdot (\sin(x))^{-3} \cdot \frac{d}{dx}(\sin(x))$

The derivative of $\sin(x)$ is $\cos(x)$, so:

$\frac{dy}{dx} = -2 \cdot (\sin(x))^{-3} \cdot \cos(x)$

This simplifies to:

$\frac{dy}{dx} = -\frac{2\cos(x)}{\sin^3(x)}$

Thus, the derivative of $y = \frac{1}{\sin^2(x)}$ is:

$\frac{dy}{dx} = -\frac{2\cos(x)}{\sin^3(x)}$

Would you like more details on the chain rule, or any other step explained in detail?

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Here are 5 related questions you might explore:

1. What is the derivative of $y = \frac{1}{\cos^2(x)}$?
2. How do you find the second derivative of $y = \frac{1}{\sin^2(x)}$?
3. What happens to the derivative when using the product rule in similar trigonometric functions?
4. How does this derivative relate to the cotangent and cosecant functions?
5. How do you differentiate $y = \sin^{-2}(x)$ using implicit differentiation?

Tip: When dealing with trigonometric functions, rewriting them in alternative forms can often make differentiation easier.