The second derivative of y = 1 + 1 / X

The second derivative of y = 1 + 1 / X

y'=[1/(1+x)]'
=-1 / (1 + x) square
Y '' = 2 / (1 + x) cube

Y = 1 / 1 - √ x second derivative

Draw the graph of the following functions: (1) y = x2-2, X ∈ Z and | x | ≤ 2; (2) y = - 2x2 + 3x, X ∈ (0,2]; (3) y = x | 2-x|
(4）y={3,x

(1) Y = x2-2, X ∈ Z and | x | ≤ 2, that is, the value of X is [- 2,2]
Other similar, their own brains

If the distance between two points a and B representing X and - 1 on the number axis is equal to 2, find the value of X

According to the meaning of the title:
x-(-1)=±2
Ψ x = 1, or x = - 3

Let the probability density of (x, y) be
f(x、y)=｛8xy,0≤x≤y,0≤y≤1,
0, others
Find the edge probability density of X and y
Need to replace ideas and processes~

Let f (x) be the marginal probability density of X and G (y) be the marginal probability density of Y. the upper and lower limits of F (x) = ∫ f (x, y) dy integral are positive and negative infinity. From the definition domain of joint function, we know that f (x) = ∫ 8xydy integral is 0 and XF (x) = 4x ^ 3. Similarly, G (y) = ∫ 8xydx integral is y and 1g (y)

Why is the second derivative of higher number y '' = D ^ 2Y / DX ^ 2? What is D / DX
Finding the second derivative of the implicit function determined by the equation XY + sinx-1 = 0

d^2y=d(dy)
dx^2=dxdx

Three fourths of a is equal to four fifths of B. A: B = (): ()

16:15

The parallelogram is divided into three triangles: A, B and C. A is 15 square centimeters more than B, and the ratio of B to C is 2:3. What is the area of this parallelogram?

Let B be 2x,
A parallelogram is divided into three triangles. If the area of one triangle is equal to the sum of the areas of the other two triangles, then 2x + 15 = 2x + 3x
The solution is x = 5
The area of this parallelogram is 2x + 15 + 2x + 3x = 7x + 15 = 50cm & sup2;

The inverse scale function y = 2x of K and the first-order function y = 2x-1, where the first-order function image passes through points (a, b), (a + 1, B + k)
(1) Finding the analytic expression of inverse proportion function
(2) Point a is the intersection of two functions, and in the first quadrant, find the coordinates of point a

1 2a-1=b
2(a+1)-1=b+k
So k = 2
We can know the analytic formula
2. Wait for the analytic expressions of two functions
If you know the value of X, you can take it to y
Let X and y be greater than 0

1. Find the analytic expression of F, the minimum value of quadratic function f is 1, and F = f = 3
If f is monotone in the interval [2A. A + 1], find the direct range of A

1. By substituting f = f = 3 into quadratic function f (x) = ax ^ 2 + BX + C, C = 3B = - 2A
And because the function has a minimum value, the opening of the function is upward, and the axis of symmetry x = - (B / 2a)
The minimum value of x = 1 is 1
So f (1) = a + B + 3 = 1
The solution is a = 2, B = - 4
So f (x) = 2x ^ 2-4x + 3
2. The derivative of F (x) is 4x-4
If the derivative is greater than 0, it will increase and if it is less than 0, it will decrease
The solution shows that the f (x) function decreases first and increases later with x = 1 as the symmetry axis
So if f is monotone in the interval [2A. A + 1], then 2A < 1 < A + 1
The solution is 0 < a < 0.5