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The given equation is:

\[
-y = \sin^{-3}(x)
\]

We need to find \(\frac{dy}{dx}\). To do this, we can differentiate both sides of the equation with respect to \(x\).

Starting with the given equation:

\[
-y = \sin^{-3}(x)
\]

1. Differentiate both sides with respect to \(x\):

\[
\frac{d}{dx}(-y) = \frac{d}{dx}(\sin^{-3}(x))
\]

2. Applying the chain rule to the left side:

\[
-\frac{dy}{dx}
\]

3. Applying the chain rule to the right side:

Let \( u = \sin(x) \), then \( \sin^{-3}(x) = u^{-3} \). Differentiating \( u^{-3} \) with respect to \( u \) gives:

\[
\frac{d}{dx}(\sin^{-3}(x)) = -3\sin^{-4}(x) \cdot \cos(x)
\]

Putting these together, we get:

\[
-\frac{dy}{dx} = -3\sin^{-4}(x) \cdot \cos(x)
\]

4. Multiplying both sides by -1 to solve for \(\frac{dy}{dx}\):

\[
\frac{dy}{dx} = 3\sin^{-4}(x) \cdot \cos(x)
\]

Thus, the derivative \(\frac{dy}{dx}\) is:

\[
\frac{dy}{dx} = 3 \cos(x) \sin^{-4}(x)
\]

Would you like more details on any specific step or have any further questions?

Here are some related questions that might interest you:

1. How do we use the chain rule in differentiation?
2. What is the difference between implicit and explicit differentiation?
3. How can the derivative of a function with a trigonometric term be interpreted graphically?
4. How would the derivative change if the function was \(-y

Learn how to find the derivative of the function -y = sin^(-3)(x) using the chain rule. This problem involves understanding trigonometric functions and differentiation techniques. Suitable for high school calculus students in Grades 11-12.

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