7. Suppose \( F(x)=f(x f(x f(x)) \), where \( f(1)=2, f(2)=3 \) \( f^{\prime}(1)=4, f^{\prime}(2)=5 \), and \( f^{\prime}(3)=6 \). Find \( F^{\prime}(1) \).

To find \( F'(1) \), we start from the definition of \( F(x) \): \[ F(x) = f(x f(x f(x))) \] Next, we will apply the chain rule to differentiate \( F(x) \). 1. **Differentiation of \( F(x) \)**: \[ F'(x) = f'(x f(x f(x))) \cdot \frac{d}{dx}(x f(x f(x))) \] Now we need to differentiate \( x f(x f(x)) \). We will again use the product rule: \[ \frac{d}{dx}(x f(x f(x))) = f(x f(x)) + x \cdot \frac{d}{dx}(f(x f(x))) \] Now we apply the chain rule to differentiate \( f(x f(x)) \): \[ \frac{d}{dx}(f(x f(x))) = f'(x f(x)) \cdot \frac{d}{dx}(x f(x)) = f'(x f(x)) (f(x) + x f'(x)) \] Combining these parts, we have: \[ \frac{d}{dx}(x f(x f(x))) = f(x f(x)) + x(f'(x f(x))(f(x) + x f'(x))) \] 2. **Evaluate \( F'(1) \)**: Next, we will evaluate \( F'(1) \). First, we need to find \( f(1 f(1 f(1))) \). We know \( f(1) = 2 \). Therefore, we compute: - \( f(1 f(1)) = f(1 \cdot 2) = f(2) = 3 \) - Then, \( x f(x f(x)) \) at \( x=1 \): \[ 1 f(1 f(1)) = 1 \cdot 3 = 3 \] And: \[ f(1 f(1 f(1))) = f(3) = 4 \] Therefore: \[ F(1) = f(3) = 4 \] 3. **Calculating \( F'(1) \)**: Now we return to the derivative we calculated: \[ F'(1) = f'(3) \cdot (f(3) + 1(f'(1 f(1))(f(1) + 1 f'(1)))) \] We already know \( f'(3) = 6 \) and \( f(3) = 4 \). Now calculate \( f'(1) \): \[ f'(1) = 4 \] Therefore, we need \( 1 \cdot 4 + 1 \cdot 4 \cdot 4 = 4 + 16 = 20 \). Now we can multiply by \( f'(3) \): \[ F'(1) = 6 \cdot 20 = 120 \] Finally, the answer is: \[ F'(1) = 120 \]