Let f (x) = ln (1 / x) - LN2, then what is the derivative of F (x), please help me

Let f (x) = ln (1 / x) - LN2, then what is the derivative of F (x), please help me

[ln(1/x)-ln2]′
=[ln(1/x)]′
=1/(1/x)*(1/x)′
=x*(-1/x^2))
=-1/x

F (x) is differentiable. On (0, + ∞), there is f (x) > F '(x) ln (x ^ x)

Because f (x) is differentiable, in (0, + ∞), there are: F (x) > F '(x) ln (x ^ x) = x * f' (x) * ln (x), and y = f (x) / ln (x) (x > 1) differentiable. So f (x) / x 1) f '(x) * ln (x) - f (x) / x > 0

Parabola y = a (x-1) square opening upward, a (radical 2, Y1) B (2, Y2) C (- radical 3, Y3), judge the size of Y1, Y2, Y3

y1

If there is (0, Y1) (5, Y2) (4, Y3) (5, Y4) on the image of parabola y = (x-4) 2, then the size relationship of Y1, Y2, Y3 is?

When x = 0, Y1 = 16
x=5 y2=1
x=4 y3=0
y1>y2>y3