Finding F'(1) for F(x) = f(xf(xf(x))) using chain rule

[aslyons]

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, find F'(1).

I feel I have a decent grasp on the chain rule, product rule, etc, but when faced with a problem like this I just blank out. I don't even really know where to begin.

Unfortunately I haven't found anything in my class notes that would be of help, and I haven't found any explanation online that's intuitive, or even close to the magnitude of this composite.

Could someone please explain to me in plain english the strategy to solve this problem? This isn't for a grade; I'm just studying, so I'm not as interested in the final answer as I am interested in the method to solving this.

Thanks in advance. Sorry my first post here is a question; I've lurked for years now, and always been able to find some help w/o the need to post :)

 

[HallsofIvy]

Let u= xf(x) so that f(xf(x))= f(u). let v= xf(u) so that f(xf(xf(x)))= f(xf(u))= f(v).

Now, use the chain rule:
What is df/dv? What is dv/du? What is du/x?