Math problem thread

Find all differentiable functions f: R -> R+ for which the following is true: f'(x) = f(f(x))

Such a function does not exist. 

Since f(x)>0 for all x, f'(x)=f(f(x))>0 so f must be strictly increasing.  Then f(x)>0 implies f(f(x))>f(0), or f'(x)>f(0). So f(0) is a lower bound on the derivative.

Now consider the slope between f(0) and f(x) for x<0.  We have
f(0)<(f(0)-f(x))/(0-x), or f(x)<f(0)+xf(0)=(1+x)f(0).  But then f(x)<0 for any x<-1, which is a contradiction.