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To find the derivative of \( y = (8 - 3x^3)^3 \) at \( x = 2 \), we first find the derivative of \( y \) with respect to \( x \) using the chain rule:

1. **Differentiate the outer function**:
   If \( u = 8 - 3x^3 \), then \( y = u^3 \). The derivative of \( y \) with respect to \( u \) is \( \frac{dy}{du} = 3u^2 \).

2. **Differentiate the inner function**:
   The derivative of \( u = 8 - 3x^3 \) with respect to \( x \) is \( \frac{du}{dx} = -9x^2 \).

3. **Apply the chain rule**:
   Combine these derivatives: \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 3u^2 \cdot (-9x^2) = -27u^2x^2 \).

Now, substitute \( x = 2 \) into \( u \):
\[ u = 8 - 3(2)^3 = 8 - 3 \cdot 8 = 8 - 24 = -16 \]

Thus, \( \frac{dy}{dx} \) at \( x = 2 \) is:
\[ \frac{dy}{dx} = -27(-16)^2 \cdot (2)^2 = -27 \cdot 256 \cdot 4 = -27648 \]

Therefore, the derivative of \( y \) at \( x = 2 \) is \( -27648 \).

Do you need a detailed breakdown or have any other questions related to this? Here are some additional questions you might consider:

1. How would you find the second derivative of the function?
2. How does changing the exponent from 3 to another value affect the derivative?
3. What happens to the derivative as \( x \) approaches infinity?
4. How can you confirm this result using a numerical approximation?
5. Would the derivative at \( x = 0 \) be different, and how would you calculate it?

**Tip:** When applying the chain rule, always keep track of your inner and outer functions to correctly apply their derivatives.

Find the derivative of y = (8 - 3x^3)^3 at x = 2.

Find the derivative of the function y = (8 - 3x^3)^3 at x = 2 using the chain rule. This detailed solution covers key calculus concepts and is suitable for high school students in Grades 11-12.

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