The domain of y = ln (2-lnx)

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Firstly, ① x > 0 and ② LNX > 0 are required
From (2) we know that x > 1
So the final answer is: x > 1
The domain of definition is: (1, + infinity)
(sorry, I don't know how to input the infinite symbol.)

What is the definition of LNY = LNX?

Ln is a special case of log function, which is based on 10. The definition domain of y = LNX is x > 0

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dy=4sinxdsinx-3xdx
=4sinx·cosxdx-3xdx
=(4sinxcosx-3x)dx

Let y be equal to the x power of X, then dy is equal to

X times (1 + LNX)

Let f (x) be derivative and y = f (the third power of x), then dy / DX is?

3*x^2*f`(x^3)

Let the x power of the equation xy-e + the Y power of E = 0, determine the function y = y (x), and find the fraction dy of DX

How to get ln (square of t-1)

Square on both sides
e^x-1=t²
e^x=t²+1
Natural logarithm of both sides
x=ln(t²+1)

Y = ln (x + the square of x minus one to find the nth power of Y As the title~

First of all, according to the meaning of the problem, the domain of definition is solved

It is proved that y = (x power of 2 + x power of 2) ln (under x + radical sign (1 + x square)) is an odd function. Try to be more detailed, my foundation is not very good

Because (x + √ (1 + x ^ 2)) (- x + √ (1 + x ^ 2))
=（√（1+x^2）+ x）（√（1+x^2）-x）
Using the formula of square difference
=（1+x^2）-x2=1,
So (- x + √ (1 + x ^ 2)) = 1 / (√ (1 + x ^ 2) + x) （*）
f(x)=(2^x+2^(-x)）ln(x+√（1+x^2））
f(-x)= (2^(-x) +2^x）ln(-x+√（1+x^2））…… Replace (*) with
=(2^x+2^(-x)）ln[1/（√（1+x^2）+ x）]
=(2^x+2^(-x)）[-ln(x+√（1+x^2））]
=-(2^x+2^(-x)）ln(x+√（1+x^2））
=- f(x),
The domain of function definition is r,
So the function is odd,

The domain of y = ln (2-lnx)