It is known that the function f (x) is a differentiable function on (0, + ∞), if XF '(x) > F (x) is constant when x > 0 (1) It is proved that the function g (x) = f (x) / X is an increasing function on (0, + ∞) (2) When X1 > 0, X2 > 0, f (x1 + x2) > F (x1) + F (x2)

It is known that the function f (x) is a differentiable function on (0, + ∞), if XF '(x) > F (x) is constant when x > 0 (1) It is proved that the function g (x) = f (x) / X is an increasing function on (0, + ∞) (2) When X1 > 0, X2 > 0, f (x1 + x2) > F (x1) + F (x2)

1. Because G '(x) = (f (x) / x) = (XF' (x) - f (x)) / x ^ 2
And XF '(x) > F (x) is constant when x > 0, so XF' (x) - f (x) > 0
So G '(x) = (f (x) / x) = (XF' (x) - f (x)) / x ^ 2 > 0 is always true when x > 0
The function g (x) = f (x) / X is an increasing function on (0, + ∞)
2. It is known from 1 that the function g (x) = f (x) / X is an increasing function on (0, + ∞),
So when X1 > 0 and X2 > 0, there are X1 + x2 > X1 and G (x1 + x2) > G (x1)
There are f (x1 + x2) / (x1 + x2) > F (x1) / x1,
So X1 * f (x1 + x2) / (x1 + x2) > F (x1)
Similarly, X1 + x2 > x2 has g (x1 + x2) > G (x2)
If f (x1 + x2) / (x1 + x2) > F (x2) / x2 holds,
So x2 * f (x1 + x2) / (x1 + x2) > F (x2)
The sum of the two formulas gives X1 * f (x1 + x2) / (x1 + x2) + x2 * f (x1 + x2) / (x1 + x2) > F (x1) + F (x2)
f（x1+x2）＞f（x1）+f（x2）．

It is known that the function y = f (x) is differentiable on R and satisfies XF '(x) > - f (x). If a > b, then

xf'x+fx>0
Let f (x) = XF (x), then f '(x) > 0, f (x) is an increasing function
So f (a) > F (b)
AF (a) > BF (b)
This is a multiple-choice question. If you don't give me an option, you can only explain that. If you don't understand, you can ask

Given that the function f (x) is differentiable at x = 2 and lim △ x → 0, f (2 + △ x) - f (2) / Δ x = (# 189;), then f '(2) =?

That's one half