Given the function y = x ㏑ x, find the derivative of the function and the tangent equation of the function at point x = 1

Given the function y = x ㏑ x, find the derivative of the function and the tangent equation of the function at point x = 1

y'=lnx+1
Where x = 1, y '= 1
y=0
So the tangent equation is
y=x-1
Namely
x-y-1=0

How to find the first derivative dy / DX of the implicit function y = y (x) determined by the equation Ye ^ x + LNY = 2?

Both sides of X at the same time to obtain the derivative
y'e^x+ye^x+y'/y=0
(e^x+1/y)y'=-ye^x
y'=-ye^x/(e^x+1/y)

How about the solution of this system? XYZ = 1, x + y + Z = 2, x + y + Z = 3

We can use the following substitution method. First, square x + y + Z to get x ^ 2 + y ^ 2 + Z ^ 2 + 2XY + 2XZ + 2yz = 4. We know that x ^ 2 + y ^ 2 + Z ^ 2 = 3, so 2XY + 2XZ + 2yz = 2 (XY + XZ + YZ) = 1. We know that swallow = 1, we can know that x = 1 / YZ, YZ = 1 / X. substitution, 2 (XY + XZ + YZ) = 1 = 2 (1 / Z + 1 / y + 1 / x)

It is known that three positive integers x, y, Z satisfy x + y + Z = XYZ, and X

xyz＝x＋y＋z＜3z
∴xy＜3
Because x < y, xy = 2, x = 1, y = 2
∴z＝3

Given that x, y and Z are three positive integers less than 100, and that x ＞ y ＞ Z & nbsp; and (X-Y), (x-z) and (Y-Z) are prime numbers, we can find the maximum value of (x-z)

Because 100 ＞ x ＞ y ＞ Z ＞ 0, the maximum possible value of X is 99. If X-Y & nbsp; is a prime number, the maximum possible value of Y is 97, and X-Y = 2 is the maximum possible value of prime number (x-z). If Z is the minimum and (x-z) and (Y-Z) are prime numbers, let z = 3, 4, 5 . 9x-z is not prime number, let z = 10x-z = 89 & nbsp; be prime number, Y-Z = 87, not prime number, let z = 16x-z = 83 & nbsp; be prime number, Y-Z = 81, not prime number, let z = 20x-z = 79 & nbsp; be prime number, Y-Z = 77, not prime number, let z = 26x-z = 73 & nbsp; be prime number, Y-Z = 71, be prime number, so the maximum possible value of (x-z) is 73

Find out that all positive integer solutions XYZ = 4 (x + y + Z) satisfy z > y > X
Both answers are very good. I don't know which one to choose

If z > y > x, we might as well use enumeration
The minimum value of X is 1
4+4(y+z)=yz
Then Y > 4 can satisfy z > y > 0
When y = 5, z = 24; y = 6, z = 14; y = 8, z = 9; there is no more
Similarly, when x = 2, there is a
y=3,z=10;y=4,z=6.
When x = 3
The minimum of Y is 4 and the minimum of Z is 5. The equation will never hold
So there are five sets of integer solutions

Let the positive real solution of the system of equations X & # 179; - XYZ = - 5, Y & # 179; - XYZ = 2, Z & # 179; - XYZ = 21 be (x, y, z), then x + y + Z=

First, I change the problem to a simple point. In the range of positive integers, I solve the system of equations X & # 179; - XYZ = - 5, Y & # 179; - XYZ = 2. By subtracting the two equations, I get (Y-X) (x ^ 2 + XY + y ^ 2) = 7, so I get Y-X = 1 or Y-X = 7 (rounding off), so I get Y-X = 1, then x ^ 2 + XY + y ^ 2 = 7. Combining the two equations, I get y = 2, x = 1, and bring in X & # 179; - XYZ =

Given x + y + Z = 0, find the value of (x + y) (y + Z) (Z + x) + XYZ

x+y+z=0
=>x+y=-z
=>y+z=-x
=>z+x=-y
So (x + y) (y + Z) (Z + x) + XYZ = - XYZ + XYZ = 0

Given x + y + Z = 0, find the value of (x + y) (y + Z) (x + Z) + XYZ

Because x + y + Z = 0, so
x+y=-z
y+z=-x
x+z=-y
(x+y)(y+z)(x+z)+xyz
=-z*(-x)*(-y)+xyz
=-xyz+xyz
=0

Given x + Y: z = y + Z: x = Z + X: y, find the value of (x + y) (y + Z) (Z + x): XYZ
Given that x ^ 2-3xy-4y ^ 2 = 0 and XY is not equal to 0, find the value of X: y

1. Let k = (x + y) / z = (y + Z) / x = (Z + x) / YX + y = KZY + Z = KXz + x = KY add 2 (x + y + Z) = K (x + y + Z) (K-2) (x + y + z) = 0 K = 2 or x + y + Z = 0, then x + y = - ZK = (x + y) / z = - 1, so k = 2, k = - 1, so the original formula = [(x + y) / Z] * [(y + z) / x] * [(Z + x) / y] = K & sup3; = 8 or - 1, 2, (x-4y) (x + y)