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.