Finding the second derivative d ^ 2 * y / DX ^ 2 established by parametric equation x = t - 2arctant ; y = t^3/3 -t I calculate DX / dt and dy / DT, and then divide the second by the first to get dy / DX. Is this correct? Then I differentiate the result of dy / DX to t to get the second derivative. Is this correct? But the calculation is far from the answer. The first step should be right, mainly how to deal with the second step

Finding the second derivative d ^ 2 * y / DX ^ 2 established by parametric equation x = t - 2arctant ; y = t^3/3 -t I calculate DX / dt and dy / DT, and then divide the second by the first to get dy / DX. Is this correct? Then I differentiate the result of dy / DX to t to get the second derivative. Is this correct? But the calculation is far from the answer. The first step should be right, mainly how to deal with the second step

DX / dt = - 2 [1 / (1 + T ^ 2)], dy / dt = T ^ 2-1, then y '= dy / DX = (dy / DT) / (DX / DT) = [T ^ 2-1] / {- 2 [1 / (1 + T ^ 2)]} = (1 / 2) (1-T ^ 4)
Then dy '/ dt = - 2T ^ 3
y"=dy'/dx=(dy'/dt)/(dx/dt)=(-2t^3)/{-2[1/(1+t^2)]}=t^3(1+t^2)

Finding the second derivative (d ^ 2Y) / (DX ^ 2) of the implicit function y = y (x) determined by the equation Ye ^ x + LNY = 1

The derivation of X on both sides is obtained
y'e^x+ye^x+y'/y=0
y'=-ye^x/(e^x+1/y)=-y^2e^x/(ye^x+1)
y''=[(-2yy'e^x-y^2e^x)(ye^x+1)+y^2e^x(y'e^x+ye^x)]/(ye^x+1)^2

All integer solutions of the system of equations: {x ^ 3 + y ^ 3 + Z ^ 3 = x + y + Z x ^ 2 + y ^ 2 + y ^ 2 = XYZ

From the basic inequality:
x+y+z=x^3+y^3+z^3≥3xyz=3(x^2+y^2+z^2)≥(x+y+z)^2
If x + y + Z (1) x + y + Z > 0, x + y + Z ≤ 1 (2) may be x + y + Z = 0
(1) x^2+y^2+z^2=xyz≤1/27*(x+y+z)^3≤1/27
So 0

Given a group: x ^ 3 + y ^ 3 + Z ^ 3 = x + y + Z x ^ 2 + y ^ 2 + Z ^ 2 = XYZ, find the positive integer solution of X, y, Z

If one of X, y, Z is not zero, then according to x ^ 2 + y ^ 2 + Z ^ 2 = XYZ, it will not be zero. And because XYZ = x ^ 2 + y ^ 2 + Z ^ 2 ＞ x ^ 2 + y ^ 2 ≥ 2XY, if x, y are positive integers, Z ≥ 2 is the same, X ≥ 2; y ≥ 2 obviously, at this time, x ^ 3 + y ^ 3 + Z ^ 3 > 4x + 4Y + 4Z > x + y + Z, the equation does not hold

It is known that f (x) = (AX + b) / (x2 + 1) is an odd function defined on (- 1,1), and f (1 / 2) = - 2 / 5
(2) Judge the monotonicity of F (x) and prove your conclusion
(3) Solving inequality f (t-1) + F (T)

Because it is an odd function, it is defined on 0, so f (0) = 0, substituting into the function, we get b = 0, f (1 / 2) = - 2 / 5, we get a = - 1
F (x) = - x (x ^ 2 + 1) = x ^ 3 + X monotone increasing function in (- 1,1)
Proof: using the general method, let x1, X2 be very simple
The function f (T) ＜ - f (t-1) is odd
F (T) < f (1-T) because the function is an increasing function on (- 1,1), so - 1 < T < 1-T < 1 is 0 < T < 1 / 2

Can the solution of the set composed of function values of quadratic function y = x ^ 2-4 be: {x ∈ r ｜ y = x ^ 2-4}? Why?

No, your way of writing is to represent a set of function domains
It should be written as {y | y = x ^ 2-4} = {y | y ≥ - 4}
If you don't understand, please hi me, I wish you a happy study!

|+What is 3 | + | - 7 |?

The absolute value of 3 is equal to 3, and the absolute value of - 7 is equal to 7, so the absolute value of 3 plus negative 7 is equal to 10; the absolute value of positive number remains unchanged, and the absolute value of negative number is equal to the value itself without negative sign

The positions of the numbers a, B and C on the number axis are shown in the figure, and / A-B / + / B-C / = / C-A / - / - / - / - / - / -, b 0 a C are simplified
The first slash is B, the second slash is zero, and so on

a>b
a-b>0
|a-b|=a-b
b0
|c-a|=c-a
So the original formula = (a-b) + (C-B) + (C-A) = 2c-2b

If one root of the quadratic equation AX square + BX + C = 0 (AC is not equal to 0) about X is 2, then there must be one root of the quadratic equation CX square + BX + a = 0 about X
Online and so on ~!

Let (X-2) (ax-c / 2) = 0 - 2a-c / 2 = B. let (x-n) (CX-A / N) = 0-nc-a / N = BN = 1 / 2. For the quadratic equation CX square + BX + a = 0 of X, then (x-n) (CX-A / N) = 0-nc-a / n = BN = 1 / 2

The difference between a good / great many, a large number of, a great deal (of)

1. Modify countable nouns: many, fee, a fee, a couple of, several, a large / small number of = large / small numbers of, a good / great many = many, a + noun singular, a large / small quantity of = large / small quantities of