If the function y = logax (a > 0, a ≠ 1) has | y | > 1 in the interval [2, + ∞), then the value range of real number a is______ ．

If the function y = logax (a > 0, a ≠ 1) has | y | > 1 in the interval [2, + ∞), then the value range of real number a is______ ．

If a > 1, the function y = logax is an increasing function, the inequality | logax | 1 & nbsp; is & nbsp; logax > 1, | loga2 > 1 = logaa, the solution is 1 < a < 2. If 1 > a > 0, the function y = logax is a decreasing function, and the function y = log1ax & nbsp; The inequality | logax | 1 is log1ax ﹥ 1. If log1a2 | 1 = log1a1a, the solution is 1 ﹤ 1a ﹤ 2, and the solution is 12 ﹤ a ﹤ 1. In conclusion, the value range of real number a is (12,1) ∪ (1,2), so the answer is (12,1) ∪ (1,2)

It is known that for any real number x, the function f (x) satisfies f (- x) = f (x). If the equation f (x) = 0 has 2009 real solutions, then the sum of these 2009 real solutions is______ ．

Let f (x) = 0 be the real solution of x1, X2 , X2009, let x1 ＜ x2 ＜ If there is x0 such that f (x0) = 0, then f (- x0) = 0, ｜ X1 + X2009 = 0, X2 + x2008 = 0 ，x1004+x1006=0，x1005=0，∴x1+x2+… +So the answer is: 0

The function f (x) satisfies f (2-x) = f (2 + x). If the equation f (x) = 0 has and only has 2009 real solutions, then the sum of these 2009 real solutions

0 ,
If y = 2-x, then 2 + x = 4-y,
So f (y) = f (4-y) = 0,
So this is a loop function

Y = ln (x / a) derivation, to process ah

y=lnx -lna
Y '= 1 / x, LNA is a constant and the derivative is 0

Derivation of [ln ^ 3 (x ^ 2) + 3] '

Let a = ln (x ^ 2)
Then we first derive a ^ 3, then LN, and finally x ^ 2
So [ln ^ 3 (x ^ 2) + 3] '
=[ln^3(x^2)]'
=3[ln(x^2)]^2*[ln(x^2)]'
=3[ln(x^2)]^2*(1/x^2)*(x^2)'
=6x[ln(x^2)]^2*(1/x^2)
=(6/x)*[ln(x^2)]^2

What is the derivative of F (x) = (LN (1 + x)) ^ 2,
How to analyze the derivative of this composite function?

f(x)=（ln（1+x））^2
f'(x)=2ln（1+x）*(ln（1+x）)'
=2ln（1+x）*1/(1+x)
=2ln（1+x）/(1+x)

Derivation of F (x) equal to 2A × ln (1 + x) - x
Please write down the detailed steps

f'(x)=2a×1/(1+x)-1

What is 1 / 2 [ln (1 + x) - ln (1-x)] derivation equal to
The teacher gave the result
1/2*[1/(1+x) + 1/(1-x)]
-Why does ln (1-x) become + 1 / (1-x)
F (x) = ln (SQR ((1 + x) / (1-x))) for f '(0)
zhang_ Heng: I see. So it is,

d( 1/2 * ( ln(1+x) - ln(1-x) )
=1/2*( dln(1+x) - dln(1-x) )
=1/2*( 1/(1+x)*d(1+x) - 1/(1-x)*d(1-x) )
=1/2* ( 1/(1+x) + 1/(1-x) )
=1/(1-x*x)
I just made a little mistake
W_ total：
After deriving LN, we should also derive the formula in LN
Do you know dlnx = 1 / X DX
Then replace x with 1 - x, and all x in the above formula should be replaced
So, DLN (1-x) = 1 / (1-x) * D (1-x)
The derivation (dlx-1) = (X-1-1)

Y = ln (1 / x)

y=-lnx
y'=-1/x

Derivation of Ln (x + 1) / 2

The original formula = ln (x + 1) - LN2
therefore
Its derivative = [ln (x + 1) - LN2] '
=1/(x+1)