Excuse me: 1. The relationship between derivative and differential integral 2. The relationship between differential and integral 3. The relationship between derivative and probability 4. The relationship between probability density and distribution function Probability of the relevant knowledge! The best meaning of white point, I am very stupid~

The rate of change of a function relative to a white variable is usually called the rate of change of a function
Derivative is a concept arising from the study of the rate of change. Therefore, let's discuss change first
Rate problem, which leads to the concept of derivative
1、 Example of rate of change
2. Instantaneous velocity of moving object
Speed is a concept that we often encounter. For example, when walking, it is 5 meters per hour
The speed of walking is 5km / J, h; another example is the speed of a car in three hours
Driving 120 km, its speed is half = 40 (km / h), these are average
Speed, which to some extent reflects the movement of the object. However, it can not explain this car
When the car goes fast, when it goes slow, and how fast and how slow
The development of science, not only need to calculate the average speed, but also calculate any moment
How to calculate the instantaneous velocity?
Now let's suppose that an object moves in a variable speed linear motion, and its equation of motion is 5; 5 (*),
Where 5 is the distance, C is the time, and the instantaneous velocity of the object at "O" is calculated
p(cD)—
We know that when the time changes from ". To". * 4f, the object changes in this period of time
The distance is
45 = 5 (CO + AF) - 5 (20),
Therefore, in this period of time, the average rate of change of object moving distance to time is the average speed
by
- A5 5 (o? 2) - 5 (6.)
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When AC is very small, we can use; to approximate the velocity of the object at 'o'
The smaller the db-o is, the closer it is to "(" O ")
Instantaneous velocity at 'O, i.e
U (C.,; cauldron = lllj, ljllj
2. Change rate of total cost
Suppose that the total cost C of a product is a function of the output x, denoted as C = C (s), and then calculate the output
When the quantity is 2 = "O", the change rate of total cost to output
We know that when the output changes from ". To". TDx ", the total cost will change accordingly
The quantity is
DC = C (O + DK) - C (o),
Therefore, when the output changes from "O" to 50 * DK, the average change rate of total cost to output is 0
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d6— dw ’
When dk-o, the limit of the above formula is the change of total cost to output when the total output is "
Rate of change (called marginal cost in economics, and recorded as fertilizer), i.e
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The practical significance of the above two examples is different, but the solution to the problem is the same
The mathematical form of the rate of change of a number at a certain point is attributed to the change of the calculation function
The limit of the ratio of (Ay) to the change of the independent variable (DX) (when d6-o),
We call this particular form of limit the derivative of the function y = 8 "
Definition of derivative
Definition 3.1 let function 2 = / (z) be defined in a neighborhood of point. O
When the variable gets the change DW in KO, the function gets the corresponding change
A7 = / (XD? A5) - eight "
If the ratio of ay to a 2 is extreme when Congli is used
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If it exists, then function 8 (2) is said to be differentiable at point X. and this limit value is called function 8 (2)
The derivative at point "O" is denoted as / '(XO), i.e
/ ''ko) = Central cup; weaving l14j, making ljll4
Derivative / '(o) can also be written as
JK -. 5vd2,5t cutting JL
7m-xo alum dZi ':, alum DK
If we let "=" O "ten this, then when this one is 0, X-X
The definition of can be expressed as
4v1—f／*.1
/ '(XO) = block Dan (:. Z)
Function 8 (x) is derivable at point "O", sometimes it is also called 8 (x) at point: o has derivative or
If the limit (3.1) does not exist, then eight (x) is at the point
XO is not differentiable; if it is not differentiable, the reason is because of this
For convenience, we also say that the derivative of function 8 (2) at point "O" is infinite and denoted as / '("O) =
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Solving the equation (x2 + Y2) dy / DX = 2XY

Maybe x2 in the above differential equation is x ^ 2 and Y2 is y ^ 2. If so, change the variables X and Y into the functions of P and T in the following way:
Let x = PCOS (T), y = PSIN (T)
The above differential equation can be transformed into the following form of differential equation:
dp/p=f(cos(t),sin(t))dt
Because f (COS (T), sin (T)) is a bit complicated, it is not written out. But f (COS (T), sin (T)) DT is an integrable function. After integrating both sides of the equation at the same time, P, cos (T) and sin (t) are replaced by X and y to get the final solution
Maybe there are other simpler ways

Finding the special solution of dy / DX = x + x ^ 2 / y + y ^ 2, y (0) = 1

dy/dx=(x+x^2)/(y+y^2)
(y+y^2)dy=(x+x^2)dx
y^2/2+y^3/3=x^3/3+x^2/2+C
x=0,y=1
1/2+1/3=C
5/6=C

General solution of dy / DX - (1 / x) y = 2x ^ 2

∵ dy / DX - (1 / x) y = 2x ^ 2 = = > dy YDX / x = 2x ^ 2DX = = > dy / x-ydx / x ^ 2 = 2xdx (both ends of the equation divide by the same x) = = > d (Y / x) = D (x ^ 2) = = > y / x = x ^ 2 + C (C is a constant) = = = > y = x ^

The general solution of dy / DX + (1-2x / x ^ 2) * y = 1

Y = (1-x) (1-2x) = 2x ^ 2-3x + 1dy / DX = y '= 2 * 2x + 3 = 4x + 3DX, the derivative of Y is dy

General solution of dy / DX + (1-2x / x ^ 2) y = 1

Using undetermined coefficient method
dy/dx+(1-2x/x^2)y=1
The solution of homogeneous equation is
y*=Cx^2*e^(1/x)
The general explanation is
y=[e^(-1/x)+C]*x^2*e^(1/x)

Dy / DX = ycosx / 1 + Y2, y (0) = 1

dy/dx=ycosx/1+y2
therefore
(1+y2)dy/y=cosxdx
(1/y+y)dy=cosxdx
So y ^ 2 / 2 + LNY = SiNx + C
By introducing y (0) = 1, we get
c=1/2
So get
y^2+2lny=2sinx+1

Solving a linear differential equation (dy / DX) - 2XY = Xe ^ (- x ^ 2)

The general solution is y = e ^ (x ^ 2) ((- 1 / 4) e ^ (- 2x ^ 2) + C)

Dy / DX = [Xe ^ (x ^ 2)] / [(1 / 2) e ^ y], how to solve this differential equation

Separation coefficient e ^ YDY = e ^ (x ^ 2) DX ^ 2
e^y=e^(x^2)+c

U = arctan (X-Y) ^ Z partial derivative
U / Z is only the derivative of Z,

This is simple. You can take X and y as constants and take derivatives of Z
Partial (U / z) = (X-Y) ^ Z * ln (X-Y) / (1 + (X-Y) ^ (2Z))