Single Variable Calculus, Chapter 3, 3.5, Section 3.5, Problem 72

Find $F'(1)$ if $F(x) = f(xf(xf(x)))$ where $f(1) = 2, f(2) = 3, f'(1) = 4, f'(2) = 5$ and $f'(3) = 6$

By using Chain Rule,


$
\begin{equation}
\begin{aligned}

F'(x) = \frac{d}{dx} [F(x)] =& \frac{d}{d(f(x f(x)))} [f(xf(xf(x)))] \left[ f(xf(x))) + x \frac{d}{d(xf(x))} [f(xf(x))] (f(x) + x f'(x)) \right]
\\
\\
F'(x) =& f'(xf(xf(x))) [f(xf(x)) + xf'(xf(x))(f(x) + xf'(x))]
\\
\\
F'(1) =& f'(1f(1f(1))) [f(1f(1)) + 1f'(1f(1))(f(1) + 1f'(1))]
\\
\\
F'(1) =& f'(1f(1 \cdot 2)) [f(1 \cdot 2) + 1f'(1 \cdot 2) (2 + 1 f'(1))]
\\
\\
F'(1) =& f'(f(2)) [f(2) + f'(2)(2 + f'(1))]
\\
\\
F'(1) =& f'(3) [3 + 5 (2 + 4)]
\\
\\
F'(1) =& 6 [3 + 30]
\\
\\
F'(1) =& 6[33]
\\
\\
F'(1) =& 198


\end{aligned}
\end{equation}
$