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To solve for \( f'(-4) \) given that \( f(x) = \frac{x^3}{5} - 10x \), we need to follow these steps:

### Step 1: Differentiate the function \( f(x) \)
The derivative of \( f(x) \) with respect to \( x \) is:
\[
f'(x) = \frac{d}{dx} \left( \frac{x^3}{5} - 10x \right)
\]

We can differentiate each term separately:

- The derivative of \( \frac{x^3}{5} \) is:
\[
\frac{d}{dx} \left( \frac{x^3}{5} \right) = \frac{1}{5} \cdot 3x^2 = \frac{3x^2}{5}
\]

- The derivative of \( -10x \) is:
\[
\frac{d}{dx} \left( -10x \right) = -10
\]

So, the derivative function is:
\[
f'(x) = \frac{3x^2}{5} - 10
\]

### Step 2: Substitute \( x = -4 \) into \( f'(x) \)
We now substitute \( x = -4 \) into the derivative function to find \( f'(-4) \):

\[
f'(-4) = \frac{3(-4)^2}{5} - 10 = \frac{3(16)}{5} - 10 = \frac{48}{5} - 10
\]

Converting 10 to a fraction:
\[
f'(-4) = \frac{48}{5} - \frac{50}{5} = \frac{-2}{5}
\]

Thus, \( f'(-4) = \frac{-2}{5} \).

### Final Answer:
\[
f'(-4) = \frac{-2}{5}
\]

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Would you like further details or clarification on any of the steps? 

Here are five related questions to explore:
1. How do we apply the power rule of differentiation to more complex functions?
2. What happens when we differentiate a product of functions instead of a sum?
3. How do we handle higher-order derivatives, such as \( f''(x) \)?
4. What are the implications of finding critical points using \( f'(x) \)?
5. How can we use derivatives to determine the concavity of a function?

**Tip:** When differentiating powers of \( x \), always reduce the exponent by 1 and multiply by the original exponent.

Learn how to find the derivative of f(x) = (x^3)/5 - 10x and evaluate f'(-4). This problem covers differentiation using the power rule and is suitable for students in Grades 11-12.

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