Calculus Help! Differentiation / Chain Rule

If F(x) = f(xf(xf(x))), where f(1) = 2, f(2) = 3, f'(1) = 4, f'(2) = 5, and f'(3) = 6, find F'(1).

Using the Chain Rule (and occasional Product Rule):
F'(x) = f '(x f (x f (x))) * (d/dx) x f (x f (x))
.......= f '(x f (x f (x))) * [1 * f (x f (x)) + x * (d/dx) f (x f (x))]
.......= f '(x f (x f (x))) * [f (x f (x)) + x * f '(x f (x)) * (d/dx) x f(x)]
.......= f '(x f (x f (x))) * [f (x f (x)) + x f '(x f (x)) * (f(x) + x f '(x))].

Let x = 1:
F'(1) = f '(f(f(1))) * [f(f(1)) + f '(f(1)) * (f(1) + f '(1))]
........= f '(f(2)) * [f(2) + f '(2) * (2 + 4)]
........= f '(3) * [3 + 5 * 6]
........= 6 * 33
........= 198.

I hope this helps!