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The given problem asks to find \( h'(1) \), where \( h(x) = f(x)f(x) \), given the following conditions:

- \( f(1) = 7 \)
- \( f'(1) = -8 \)

### Step-by-step solution:

1. First, express \( h(x) \) as:
   \[
   h(x) = f(x) \cdot f(x) = f(x)^2
   \]
   
2. Now, differentiate \( h(x) \) using the chain rule. The derivative of \( f(x)^2 \) is:
   \[
   h'(x) = 2f(x)f'(x)
   \]
   
3. Substitute \( x = 1 \) into the derivative:
   \[
   h'(1) = 2f(1)f'(1)
   \]
   
4. Use the given values \( f(1) = 7 \) and \( f'(1) = -8 \):
   \[
   h'(1) = 2(7)(-8) = 2 \times 7 \times (-8) = -112
   \]

Thus, the value of \( h'(1) \) is:
\[
h'(1) = -112
\]

Would you like a detailed explanation of any part of this? 

### Related questions:
1. How do you apply the product rule to derivatives?
2. Can you explain the chain rule in more detail?
3. How do you find the derivative of composite functions?
4. What are other examples where the chain rule is used?
5. How would the derivative change if \( h(x) = f(x)^3 \)?

### Tip:
Always check if you can simplify the function before applying the rules of differentiation—this often makes the problem easier to solve.

Let f be a function such that f(1) = 7 and f'(1) = -8. Find h'(1) for the function h(x) = f(x)f(x).

Learn how to find the derivative of the function h(x) = f(x)^2 using the chain rule. In this problem, f(1) = 7 and f'(1) = -8, and we find h'(1).

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