In differential, why can the increment of dependent variable of function be expressed as independent variable multiplied by a plus higher-order infinitesimal Function is unknown. There may be many cases. Why does dy = a when the argument has an increment Δ x. And Δ y=dy+o( Δ x),o( Δ x) Is infinitesimal, that is, for any function, the increment of the dependent variable is a multiple of the increment of the independent variable?

In differential, why can the increment of dependent variable of function be expressed as independent variable multiplied by a plus higher-order infinitesimal Function is unknown. There may be many cases. Why does dy = a when the argument has an increment Δ x. And Δ y=dy+o( Δ x),o( Δ x) Is infinitesimal, that is, for any function, the increment of the dependent variable is a multiple of the increment of the independent variable?

That's right. This multiple is the derivative of the independent variable requiring differentiation. The derivative can be changed, so the multiple is not fixed. Just like you look for two numbers at random, there is always a multiple relationship between them

Judgment of the sum of two functions on the infinitesimal of independent variables X → a, f (x) and G (x) are n-order and m-order infinitesimals of x-a respectively. Why are the two functions n-order infinitesimals of x-a when n < m? Missing a word, the sum of the two functions is the n-order infinitesimal of x-a?

Divide by (x-a) n times. That's the definition of infinitesimal of order K

Let the function z = ln (x + y ^ 2), then find the total differential DZ =? What is total differential and how to find it?

Definition of total differential
The sum of the products of two full differential partial derivatives f'x (x, y) and f'y (x, y) of function z = f (x, y) and the increments △ X and △ y of independent variables respectively
　　f'x(x,y)△x + f'y(x,y)△y
If the difference between the expression and the full increment △ Z of the function,
When ρ → 0, yes ρ ( )
Higher order infinitesimal,
Then the expression is called the total differential of the function z = f (x, y) at (x, y) (about △ x, △ y)
Recorded as: DZ = f'x (x, y) △ x + F'y (x, y) △ y
According to the definition of total differential, the partial derivatives of X and y are obtained respectively
f‘x（x,y）=(1/x+y^2)*1=1/x+y^2
f'y (x,y) =(1/x+y^2)*2y=2y/x+y^2
By substituting into the total differential expression, we can get: DZ = (1 / x + y ^ 2) △ x + (2Y / x + y ^ 2) △ y
(the key to this problem is to understand the definition of total differential and be able to find two partial derivatives of Z)

Find the differential of function y = ln (x + radical (1 + x ^ 2))

y＝ln[x+√(1+x ²)]
∴y'＝[x+√(1+x ²)]'/ [x+√(1+x ²)]
＝[1+x/√(1+x ²)]/ [x+√(1+x ²)]
＝[x+√(1+x ²)]/ [1+x ²+ x√(1+x ²)]

Solution equation: dy / DX = x / Y Add another question dy / DX = Y / X

dy/dx=x/y
ydy=xdx
xdx=∫ydy
x^2/2=y^2/2+C1
Therefore, y ^ 2-x ^ 2 = C is the result, and C is any real number
dy/dx=y/x
dy/y=dx/x
Two side integral
ln|y|=ln|x|+C1
ln|y/x|=C1
Y / x = C is the desired value, and C is any real number

Y = Tan (x + y), find dy / DX

dy/dx=sec ² (x + y) * (1 + dy / DX) then [1-sec ² (x+y)]dy/dx=sec ² (x+y)
Then dy / DX = sec ² (x+y)/[1-sec ² (x+y)]=1/cos ² (x+y)÷[1-1/cos ² (x+y)]=1/(cos ² (x+y)-1)=-1/sin ² (x+y)