Divide 406 into the sum of two prime numbers and find the minimum value of their product

Divide 406 into the sum of two prime numbers and find the minimum value of their product

5×401＝2005
If we want the minimum value of the product, we must make the difference as large as possible (and if it is a prime number). But 403 = 13 × 31, it is not a prime number, so the following is all wrong
And 406-401 = 5, which is also a prime number
Then the minimum value of their product is 5 × 401 = 2005

The sum of two primes is 50. What is the maximum product of these two primes?

50÷2=25
The two prime numbers closest to 25 are 19 and 32
19×31=589
So 589

The sum of two prime numbers is 50. What is the maximum product of the two prime numbers?

Max = 19 × 31 = 589

The equation XY ^ 2-e ^ x + e ^ y = 0 determines that y is the implicit function of X, and finds y '| x = 0

For X, note that y is a function of X
xy²-e^x+e^y=0
Derivation of X
y²+2xyy'-e^x+y'e^y=0－－(#)
When x = 0, 0-1 + e ^ y = 0, y = 0
Substituting x = 0, y = 0 into (?) formula, we get
0+0-1+y'=0,y'=1
That is, (y '| x = 0) = 1

Let e ^ y + xy-x ^ 2 = e ^ 2 determine the implicit function y = y (x)
Find the tangent equation and normal equation of the point on the curve whose abscissa is x = 0

When e ^ y + xy-x ^ 2 = e ^ 2, x = 0, e ^ y = e ^ 2, y = 2, take the derivative on both sides, e ^ y dy / DX + y + X * dy / dx-2x = 0, dy / DX = (2x-y) / (e ^ y + x), y '(0) = - 2 / e ^ 2, the tangent equation is: (Y-2) = (- 2 / e ^ 2) x, y = (- 2 / e ^ 2) x + 2, the slope of the normal equation is the negative reciprocal of the tangent, e ^ 2 / 2, and the normal equation is: (Y-2) /

Given the function y = cosx / X (1), find the derivative of the function (2) find the tangent equation of the function at x = π

Y = cosx / XY '= [(cosx)' × x-cosx × (x) '] / X & sup2; = [- SiNx × x-cosx × 1] / X & sup2; = [- SiNx × x-cosx] / X & sup2; when x = π, y' = [- SiNx × x-cosx] / X & sup2; = [0 + 1] / π & sup2; = 1 ／ π & sup2; so k = 1 ／ π & sup2; let the tangent be y = KX + B