The sum of two prime numbers is 60. What is the maximum product of the two prime numbers?

31*29=899

The continuity of the derivative of xarctan (the square of 1 / x) at x = 0. When x is zero, the value of the function is zero
It's a piecewise function. That function has no definition at zero. The answer is continuous

f(x)=[xarctan(1/x²)]'=arctan(1/x²)-2x²/(x^4+1)limx→0f(x)=limx→0[arctan(1/x²)-2x²/(x^4+1)]=limx→0arctan(1/x²) - limx→02x²/(x^4+1)=π/2∵f(0)=0∴limx→0f(x)≠f(0)...

Is the derivative quotient of two functions at a certain point the limit of the quotient of two functions at that point?
In other words, G '(0) = a, f' (0) = B, then G (0) / F (0) = A / b?
When x approaches 0, G (x) / F (x) = A / B, where f (0) may be equal to zero

Of course not
g(x)=x^2+1,g'(x)=2x,g'(0)=0=a
f(x)=x+1 ,f'(x)=1,f'(0)=1=b
a/b=0
g(0)/f(0)=1/1=1

How to judge the maximum and minimum of function with derivative? What is stationary point?
How can we know whether it is the maximum or the minimum after finding out the function value?

The point where the first derivative is equal to 0 is the stationary point; the value of the derivative at the left positive and right negative points of the stationary point is the maximum, and the value at the left negative and right positive points is the minimum, then the maximum value is the maximum value, and the minimum value is the minimum value

Seeking maximum and minimum with derivative
Let the tangent slope of the tangent equation of the image with function y = x (cubic) + ax + 1 at point (0,1) be - 3, and find the maximum and minimum values of y = x (cubic) + ax + 1 on [0,2]

Let the tangent slope of the tangent equation at point (0,1) of the image with function y = x (cubic) + ax + 1 be - 3
==>The value of Y '= 3 * x ^ 2 + a at (0,1) is - 3
==>3*0^2+a=-3
==>a=-3
Then the original formula is y = x ^ 3-3x + 1, derivative y '= 3x ^ 2-3
Let y '= 0 get x = 1 or x = - 1, corresponding to the extremum of the whole domain, that is, when x = 1, the minimum value of y = - 1
On [0,1], y '

Find the implicit function y = y (x) determined by the equation x ^ 2Y ^ 2 + YX ^ 3 = 1, find its derivative and all limit values

Obviously, you can do this problem without the hidden function, and you can obviously do the hidden function, and you can obviously do the hidden function, and you can obviously do the hidden function, and you can obviously do the hidden function, and you can obviously you can obviously you can do the hidden function, and you can do. X & \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\(1) take the + sign Y & nbsp; = & nbsp; - X / 2 & nbsp; + & nbsp; [√ (X & # 178; & nbsp; + & nbsp; 1 / X & # 178;)] / 2Y & # 39; & nbsp; = -1/2 + （1/2)(1/2)(2x - 2/x³)/√(x² + 1/x²) = (1/2)[x - 1/x³-√(x² + 1/x²)/[√(x² + 1/x²) = 0x - 1/x³ = √(x² + 1/x²)3x^4 = 1x = ± 1 / 3 ^ (1 / 4) from the graph, here take X & nbsp; = & nbsp; - 1 / 3 ^ (1 / 4) & nbsp; (I don't know if there is a simpler way to determine) y & nbsp; = & nbsp; 3 / [2 * 3 ^ (1 / 4)] & nbsp; & nbsp; (minimum) (2) & nbsp; take - sign, which is similar to (1), X & nbsp; = & nbsp; ± & nbsp; 1 / 3 ^ (1 / 4) but here take & nbsp; X & nbsp; = & nbsp; 1 / 3 ^ (1 / 4) y & nbsp; = & nbsp; - 3 / [2 * 3 ^ (1 / 4)] & nbsp; (maximum)

The derivative of y when y ^ 2 + 2y-x = 3x ^ 7, x = 0

A:
Derivation
2yy'+2y'-1=21x^6
y'=(21x^6+1)/(2y+2)
When x = 0, y = 0 or - 2
We can get the result by substituting the above formula
The derivative of Y is 1 / 2 or - 1 / 2

Finding the second derivative of 3x ^ 3 + 5Y ^ 3 = 4

Y as a function, X as an independent variable
The derivation of X on both sides is: 9x & # 178; + 15y & # 178; y '= 0, that is: 3x & # 178; + 5Y & # 178; y' = 0 (1)
The solution is: y '= - 3x & # 178; / (5Y & # 178;)
(1) The derivation of X on both sides is: 6x + 10Y (y ') & #178; + 5Y & #178; y' '= 0
By substituting y '= - 3x & # 178; / (5Y & # 178;) into the above formula, we can get:
6x+10y[-3x²/(5y²)]²+5y²y''=0
The result is: 6x + 18x & # 8308; / (5Y & # 179;) + 5Y & # 178; y '' = 0
The solution is: y '' = - (30xy & # 179; + 18x & # 8308;) / (25y & # 8309;)
I hope I can help you. If you don't understand, you can ask. If you solve the problem, please click the "select as satisfactory answer" button below

How to determine who is the independent variable and who is the function about the independent variable

Generally, there will be explanations, such as Z = f (x, y), which means that X and y are independent variables and Z is a function of X and y
If there is no special explanation, take x as an independent variable

Let y (x) find y 'y' (0) from the implicit function determined by the equation E ^ y-e ^ x = XY

e^y-e^x=xy
It is necessary to find the derivative on both sides
e^y*y'-e^x=y+xy'
(e^y-x)y'=(e^x+y)
So y '= (e ^ x + y) / (e ^ Y-X)
When x = 0, e ^ y-e ^ 0 = 0, then e ^ y = 1, then y = 0
So y '(0) = (e ^ 0 + 0) / (e ^ 0-0) = 1 / 1 = 1

The sum of two prime numbers is 60. What is the maximum product of the two prime numbers?