The problem you provided is asking to find $f'(x)$, the derivative of the function $f(x)$, given that:

$f(x + h) - f(x) = -8h x^2 + 1h x - 8h^2 x - 3h^2 + 3h^3.$

This expression can be interpreted as the difference quotient, $\frac{f(x+h) - f(x)}{h}$, from which we can derive $f'(x)$.

Step-by-step Solution:

1. Start with the definition of the derivative: $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}.$

2. Divide the given expression by $h$: $\frac{f(x+h) - f(x)}{h} = \frac{-8h x^2 + 1h x - 8h^2 x - 3h^2 + 3h^3}{h}.$

3. Simplify the expression:

   • The first term simplifies to $-8x^2$.
   • The second term simplifies to $1x = x$.
   • The third term becomes $-8h x$.
   • The fourth term simplifies to $-3h$.
   • The fifth term becomes $3h^2$.

   Therefore, the expression simplifies to: $\frac{f(x+h) - f(x)}{h} = -8x^2 + x - 8h x - 3h + 3h^2.$

4. Take the limit as $h \to 0$: As $h \to 0$, any terms involving $h$ will vanish. Specifically:

   • The term $-8h x$ goes to 0.
   • The term $-3h$ goes to 0.
   • The term $3h^2$ goes to 0.

   Therefore, the limit is: $f'(x) = -8x^2 + x.$

Final Answer:

$f'(x) = -8x^2 + x.$

Would you like further details on any step or have any questions? Here are five related questions to deepen your understanding:

1. How does the difference quotient help in finding derivatives?
2. What happens to higher-order terms of $h$ when taking the limit as $h \to 0$?
3. Can you apply the same process to find $f'(x)$ for a different function $f(x)$?
4. What would the second derivative $f''(x)$ be for this function?
5. How can you use this method to find derivatives for more complex functions?

Tip: When dealing with limits in derivative calculations, remember that any term involving $h$ will vanish as $h \to 0$.