What is the meaning of calculus and definite integral in high school mathematics? Definite integral is to find the area of a curve trapezoid, and calculus is to find that the calculus of the original derivative is equal to the two differences of the original function of the derivative What is the meaning of calculus itself? Is the meaning of calculus in high school mathematics convenient to find the value of definite integral? In addition, in the derivation of definite integral and calculus, the algebraic property of derivative is that △ x tends to 0, right? The main application in calculus derivation is the geometric meaning of derivative, that is, slope,

What is the meaning of calculus and definite integral in high school mathematics? Definite integral is to find the area of a curve trapezoid, and calculus is to find that the calculus of the original derivative is equal to the two differences of the original function of the derivative What is the meaning of calculus itself? Is the meaning of calculus in high school mathematics convenient to find the value of definite integral? In addition, in the derivation of definite integral and calculus, the algebraic property of derivative is that △ x tends to 0, right? The main application in calculus derivation is the geometric meaning of derivative, that is, slope,

Calculus is the general term of differential and integral. In a broad sense, calculus also refers to the whole mathematical analysis
The integral in high school is basically used to calculate the geometric area
Derivative is the quotient of differential. In high school, derivative basically refers to slope

Determination of the existence of limit and derivative in freshman calculus

phillipster ,
To determine whether a function exists at a certain point, it is equivalent to the left limit, the right limit exists, and the left and right limits are equal. Whether the derivative exists at a certain point is equivalent to the left derivative, the right derivative exists, and the left and right derivatives are equal

On the continuity of limit derivatives When x → 1, the left and right limits of the function (x squared - 1) / (x-1) exist and are equal to 2, that is, the limit exists, but according to the limit, so the function f (x) is differentiable at x0, that is, at x = 1, and continuous according to the differentiable ratio, so I think it is continuous at x = 1, but I find that it is discontinuous at x = 1. Why? I really can't understand it,

According to the definition field, X-1 cannot be equal to 0, so x = 1 is meaningless. In addition, continuity must be differentiable, and differentiable is not necessarily continuous. The derivative is that the left limit is equal to the right limit, and continuity also needs to be equal to the function value at that point. It must meet its definition field

What is the relationship between derivative and limit Such as title Can we say that there are many kinds of limits For example, X - > 0, X - > infinity, X - > a specific number But there is only one derivative, that is δ x->0 Nothing else What I'm talking about is that in terms of function, the limit of the sequence will not be considered for the time being The point of my question is the relationship between limits and functions, not each

There is some truth in your statement. Indeed, from the perspective of trend, the trend of derivative is only δ X - > 0 (in addition, the unilateral derivative has δ When x approaches to 0 from the left or right, correspondingly, the limit also has a one-sided limit), and the function limit has x - > infinity, X - > a specific number. The X - > 0 you say is also X - > a specific number

Let f (x) = (A-2) ln (- x) + 1 / x + 2aX, where a is a real number A = 0 for extremum Should the derivative of F (x) = (A-2) ln (- x) + 1 / x + 2aX a = 0 not be 2 / X-1 / x ^ 2 Let the derivative be zero, x = 1 / 2, which is in contradiction with the definition domain Please explain

When a = 0, f (x) = - 2ln (- x) + 1 / X
f'(x)=-2*(-1/x)*(-1)-1/x^2=-2/x-1/x^2=-(2x+1)/x^2=0
x=-1/2
Your -2ln (- x) derivative is wrong. It should be: - 2 * (- 1 / x) * (- x) '= - 2 / X

Find the derivative of F (x) = (A-2) ln (- x) + 1 / x + 2aX,

A set of commonly used derivation formulas, there is really nothing to say. Ln'x = 1 / X; ( 1/x)'=-1/(x^2) ;
Then f '(x) = - (A-2) / X-1 / (x ^ 2) + 2A = - 1 / (x ^ 2) - (A-2) x + 2A