I am a physics student and would apologize if in advance if you find my way of asking the question inapproperiate. I have the following function $F(x)=2f(x) - f(2x).$ Given that $\frac{d}{dx}f(x) > 1$, what can one say about $\frac{d}{dx}F(x)$ ?
Since
$\frac{d}{dx}F(x) = 2\frac{d}{dx}f(x) - \frac{d}{dx}f(2x)$.
The first term, $2\frac{d}{dx}f(x)$, is greater than 2. How about the second term and hence the overall derivative $\frac{d}{dx}F(x)$?
By chain rule
$$F'(x)=2f'(x)-2f'(2x)=2(f'(x)-f'(2x))$$
but you can't conclude anything else because might be
$$f'(x)\geq f'(2x) \quad \lor \quad f'(x)\leq f'(2x)$$
we need more information on $f'(x)$ or $f''(x)$ to can predict something else on $F'(x)$.
I will here give a couple of examples of functions $f$ which both satisfy the preconditions but for which the corresponding $F$:s have opposite behaviors.
First note that by chain rule, $$F'(x) = 2 f'(x) - 2 f'(2x) = 2 \left( f'(x) - f'(2x) \right).$$
For $f(x) = x - e^{-x}$ you have $f'(x) = 1 + e^{-x} > 1$ and $$F'(x) = 2 \left( (1+e^{-x}) - (1+e^{-2x}) \right) = 2 e^{-x} (1 - e^{-x}) \begin{cases}>0 & (x>0), \\ <0 & (x<0).\end{cases}$$
However, for $f(x) = x + e^{x}$ you have $f'(x) = 1 + e^{x} > 1$ and $$F'(x) = 2 \left( (1+e^{x}) - (1+e^{2x}) \right) = 2 e^{x} (1 - e^{x}) \begin{cases}<0 & (x>0), \\ >0 & (x<0).\end{cases}$$
Thus it's difficult to say anything about $F'.$
