Is the total differential of a binary function the partial derivative?

Is the total differential of a binary function the partial derivative?

For example, the definition of Z = Z (x, y) total differential is the sum of the products of the two partial derivatives f'x (x, y) and f'y (x, y) of the function z = f (x, y) and the increment △ X and △ y of the independent variable, f'x (x, y) △ x + F'y (x, y) △ y. if the difference between the expression and the total increment △ Z of the function, when ρ → 0, yes ρ Higher order infinitesimal of ()

The definition field of function y = (x ^ 2-1) / (x ^ 2 + 1) is? The value range is?

Y = (x ^ 2 + 1-2) / (x ^ 2 + 1) = 1 - ((2 / (x ^ 2 + 1)) therefore, the definition field is {x ‖ x is not equal to - 1 and 1} because the minimum value of x ^ 2 + 1 is 1 and there is no maximum value, the value field of 2 / (x ^ 2 + 1) is (0,2), so the value field of 1 - ((2 / (x ^ 2 + 1)) is [- 1,1). Therefore, the definition field of y = (x ^ 2-1) / (x ^ 2 + 1) is {x ‖ x ≠ - 1 and

What does the value range of a function mean? Which two parts determine? What is the difference between it and a domain?

The value range of a function refers to the range of the function! For example, y = 3x + 6, X is the independent variable and Y is the dependent variable! That is, y changes with the change of X. the value range of this function refers to the value range of Y, and the definition field refers to the value range of X. the value range is determined by the nature of the definition field and the function!

In the function of senior one mathematics, in the definition field of F (x + 1) obtained by knowing the definition field of F (x) [1,0], are their value fields the same?

Same as 0

Given that the domain of F (x + 1) is [- 1,1], then the domain of f| (x + 1) | is____ The answer is (- 3,

The definition field of F (x + 1) is: [- 1,1]
Then:
In F (x + 1), X ∈ [- 1,1]
Then:
In F (x + 1), x + 1 ∈ [0,2]
Get:
In F (T), t ∈ [0,2]
So:
In F (|x + 1|), |x + 1 ∈ [0,2], get: X ∈ [- 3,1]
Namely:
In F (|x + 1|), X ∈ [- 3,1]
Then: the definition field of F (x + 1) is: X ∈ [- 3,1]

By using the definition of function monotonicity, it is proved that the function f (x) = negative (root sign x) is a subtractive function in its definition domain

The definition field is {x | x ≥ 0.}
For any x1 ＞ x2 ≥ 0
F (x1) - f (x2) = [- (root x1)] - [- (root x2)] = (root x2) - (root x1)
=[[(root x2) - (root x1)] [(root x2) + (root x1)]] / [(root x2) + (root x1)]
=(x2-x1) / [(root x2) + (root x1)]
Because x2-x1 < 0, (root x2) + (root x1) > 0
Therefore, f (x1) - f (x2) < 0
f（x1)＜f(x2)
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