To solve the problem $\frac{dy}{dx}$ for $y = (\sin x)^{-3}$, we'll use the chain rule and some differentiation techniques.

Given: $y = (\sin x)^{-3}$

Step 1: Rewrite the function for easier differentiation.

$y = \sin^{-3} x$

Step 2: Apply the chain rule.

The chain rule states: [ \frac{dy}{dx} = \frac{d}{dx} \left[\sin^{-3} x\right] = -3 \cdot \sin^{-4} x \cdot \frac{d}{dx}[\sin x] ]

Step 3: Differentiate $\sin x$ with respect to $x$.

$\frac{d}{dx}[\sin x] = \cos x$

Step 4: Substitute the derivative of $\sin x$ back into the equation.

$\frac{dy}{dx} = -3 \cdot \sin^{-4} x \cdot \cos x$

Step 5: Simplify the expression.

$\frac{dy}{dx} = -3 \cdot \frac{\cos x}{\sin^4 x}$

So the derivative is: $\boxed{\frac{dy}{dx} = -\frac{3\cos x}{\sin^4 x}}$

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

Here are some related questions to expand your understanding:

1. How does the chain rule apply to more complex functions?
2. What is the derivative of $\sin^3 x$ using the chain rule?
3. Can you explain the product rule and when it would be applicable?
4. How would you find the second derivative of $y = (\sin x)^{-3}$?
5. What happens if we differentiate $y = (\cos x)^{-3}$ instead?

Tip: When differentiating functions with negative exponents, rewriting them can simplify the application of differentiation rules.