To find the derivative $f'(x)$ of the function

$f(x) = \left( -2x^8 + (-2)x^4 + 1 \right)^7,$

we will use the chain rule.

Step 1: Set the inner function and outer function

Let

$g(x) = -2x^8 - 2x^4 + 1 \quad \text{and} \quad h(x) = x^7.$

Then, $f(x) = (g(x))^7$, so we can apply the chain rule:

$f'(x) = 7(g(x))^6 \cdot g'(x).$

Step 2: Find $g'(x)$

Now, we differentiate $g(x) = -2x^8 - 2x^4 + 1$:

$g'(x) = \frac{d}{dx} \left( -2x^8 - 2x^4 + 1 \right).$

Using the power rule:

$g'(x) = -16x^7 - 8x^3.$

Step 3: Put everything together

Now, substitute $g(x)$ and $g'(x)$ into the derivative formula:

$f'(x) = 7 \left( -2x^8 - 2x^4 + 1 \right)^6 \cdot (-16x^7 - 8x^3).$

Step 4: Factor the result

Let's factor the expression:

$f'(x) = -8 \left( -2x^8 - 2x^4 + 1 \right)^6 \cdot (2x^7 + x^3).$

Thus, the derivative is:

$f'(x) = -8 \left( -2x^8 - 2x^4 + 1 \right)^6 \cdot x^3 (2x^4 + 1).$

This is the factored form of the derivative.

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Would you like a more detailed explanation or have any questions about this solution? Here are some follow-up questions you might find useful:

1. How does the chain rule apply in this problem?
2. What happens if we apply the product rule to the outer and inner functions instead of the chain rule?
3. Can you find the second derivative of the function?
4. What is the behavior of $f'(x)$ as $x \to \infty$?
5. How would you approach differentiating a more complex function with multiple compositions?

Tip: Always check your factored forms to ensure that all common terms are accounted for, especially when factoring polynomials.