F (x) = 3x2-2cosx + 3x (square) - lne, find f '(x)

F (x) = 3x2-2cosx + 3x (square) - lne, find f '(x)

f(x)=6x²-2cosx-1
So f '(x) = 12x + 2sinx

Is LNX an increasing function under (0, + infinity)? Then (lne) ^ 2 should be greater than LNX. How can LNX > (LNX) ^ 2? The domain is [1,2]

LNX is indeed an increasing function in its domain of definition, and (lne) ^ 2 is a constant, which can only be compared with a variable function in a certain region

1) Lgx = - 2 (2) lgx = 21ga LGB (3) LNX = 0.7 (4) LNX = 1-ln4 process

（1）lgx=-2
x=10^(-2)=1/100
（2）lgx=21ga-lgb
lgx=lg(a²/b)
x=a²/b
（3）lnx=0.7
x=e^0.7
（4）lnx=1-ln4
lnx=ln(e/4)
x=e/4