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The problem asks to find \( F'(1) \), given that \( F(x) = f(xf(x)) \), with specific values for \( f(1) = 5 \), \( f(5) = 6 \), \( f'(1) = 3 \), \( f'(5) = 4 \), and \( f'(6) = 5 \).

### Solution:

We will apply the chain rule to differentiate \( F(x) \).

1. \( F(x) = f(x f(x)) \)
2. Differentiating both sides with respect to \( x \), applying the chain rule:
   \[
   F'(x) = f'(x f(x)) \cdot \frac{d}{dx}[x f(x)]
   \]

3. Now differentiate the inner expression \( x f(x) \):
   \[
   \frac{d}{dx}[x f(x)] = f(x) + x f'(x)
   \]

4. Therefore:
   \[
   F'(x) = f'(x f(x)) \cdot (f(x) + x f'(x))
   \]

5. Now substitute \( x = 1 \):
   \[
   F'(1) = f'(1 f(1)) \cdot (f(1) + 1 \cdot f'(1))
   \]
   
6. From the given values, we know:
   \[
   f(1) = 5, \quad f'(1) = 3, \quad f'(5) = 4
   \]
   
   Substituting these into the equation:
   \[
   F'(1) = f'(5) \cdot (5 + 1 \cdot 3)
   \]
   \[
   F'(1) = 4 \cdot (5 + 3) = 4 \cdot 8 = 32
   \]

So, \( F'(1) = 32 \).

Would you like further details or clarifications?

### Related Questions:
1. How does the chain rule work in composite functions like this one?
2. What are the specific steps for differentiating products of functions?
3. Why do we substitute \( x = 1 \) in this problem?
4. What other rules of differentiation are useful for similar problems?
5. How can we use implicit differentiation in composite functions?

### Tip:
Always remember to carefully apply the chain rule when differentiating nested functions, especially with products inside.

If F(x) = f(x*f(x)), where f(1) = 5, f(5) = 6, f'(1) = 3, f'(5) = 4, and f'(6) = 5, find F'(1).

Learn how to differentiate composite functions using the chain rule. This example explores the function F(x) = f(x*f(x)), where f(1) = 5, f(5) = 6, f'(1) = 3, and f'(5) = 4, to find F'(1). Suitable for students learning advanced calculus differentiation techniques.

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