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The given function is:

\[
f(x) = 7x^3(x^2 - 7)
\]

To evaluate \( f'(-4) \), we first need to differentiate the function \( f(x) \) using the product rule. Let's proceed step by step.

### 1. Differentiate \( f(x) = 7x^3(x^2 - 7) \):

We use the **product rule** for differentiation:

\[
\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)
\]

Here, \( u(x) = 7x^3 \) and \( v(x) = x^2 - 7 \).

#### Differentiate \( u(x) = 7x^3 \):
\[
u'(x) = 7 \cdot 3x^2 = 21x^2
\]

#### Differentiate \( v(x) = x^2 - 7 \):
\[
v'(x) = 2x
\]

Now, applying the product rule:

\[
f'(x) = u'(x)v(x) + u(x)v'(x)
\]
\[
f'(x) = (21x^2)(x^2 - 7) + (7x^3)(2x)
\]

Simplifying each term:

\[
f'(x) = 21x^2(x^2 - 7) + 14x^4
\]
\[
f'(x) = 21x^4 - 147x^2 + 14x^4
\]
\[
f'(x) = (21x^4 + 14x^4) - 147x^2
\]
\[
f'(x) = 35x^4 - 147x^2
\]

### 2. Evaluate \( f'(-4) \):

Now substitute \( x = -4 \) into \( f'(x) = 35x^4 - 147x^2 \):

\[
f'(-4) = 35(-4)^4 - 147(-4)^2
\]
\[
f'(-4) = 35(256) - 147(16)
\]
\[
f'(-4) = 8960 - 2352
\]
\[
f'(-4) = 6608
\]

Thus, \( f'(-4) = 6608 \).

Would you like more details or have any questions?

### Related Questions:
1. How does the product rule apply in other derivative problems?
2. What is the significance of the fourth and second powers of \( x \) in this differentiation?
3. How can this method be applied to higher-degree polynomials?
4. How does the chain rule differ from the product rule?
5. What are some real-world applications of such polynomial derivative calculations?

**Tip:** For complex functions, always break them into simpler parts before applying derivative rules like the product or chain rule.

Let f(x) = 7x^3(x^2 - 7). Evaluate f'(x) at the following points, specifically f'(-4).

Learn how to differentiate the function f(x) = 7x^3(x^2 - 7) using the product rule. The solution also includes evaluating f'(-4) with detailed steps. This problem focuses on key calculus concepts like differentiation and polynomial functions, ideal for students in Grades 11-12.

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