X ^ 2 + 3xy = 28, XY + 4Y ^ 2 = 8 to solve special type equations

X ^ 2 + 3xy = 28, XY + 4Y ^ 2 = 8 to solve special type equations

Add
x²+4xy+4y²=36
(x+2y)²=36
x+2y=±6
If x = 6-2y
Substituting X & sup2; + 3xy = 28
4y²-24y+36+18y-6y²=28
y²+3y-4=0
(y+4)(y-1)=0
y=1,y=-4
x=6-2y
If x = - 6-2y
Substituting X & sup2; + 3xy = 28
4y²+24y+36-18y-6y²=28
y²-3y-4=0
(y-4)(y+1)=0
y=-1,y=4
x=-6-2y
therefore
x=4,y=1
x=14,y=-4
x=-4,y=-1
x=-14,y=4

The square of x plus 3xy equals 28, and the square of XY plus 4Y equals 8

This is the result of the two-way addition of the sum of X \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\y

The partial derivative of Z = Ye ^ XY,

The partial derivative of Z = Ye ^ (XY),
∂z/∂x=y²e^(xy)；∂z/∂y=e^(xy)+xye^(xy)=(1+xy)e^(xy)；

The partial derivative of Z = e ^ XY / (e ^ x + e ^ y)

zx=[xe^xy(e^x+e^y)-e^x*e^xy]/(e^x+e^y)²
zy=[ye^xy(e^x+e^y)-e^y*e^xy]/(e^x+e^y)²

2Sin (3x-2y + Z) = 3x-2y + Z, find the partial derivative

Let f (U, n) have continuous second order partial derivatives, z = f (3x + 2Y, Y & sup2;), and find 2Z / 2x + 2 & sup2; Z / 2x2y

∂z/∂x = ∂f /∂u * ∂u/∂x = 3 fu'∂²z / ∂x∂y = 3 fuu'' * ∂u/∂y + 3 fuv'' * ∂v/∂y= 3 fuu'' * 2 + 3 fuv'' * 2y= 6 fuu'' + 6...

Finding mixed partial derivatives of Z = ln (3x + 2Y)

The partial derivative of X is equal to = 3 / (3x + 2Y)
Then we use the above formula to derive Y: equal to = - 6 / (3x + 2Y) ^ 2, which is the mixed partial derivative
If the partial derivatives of X and y are both continuous functions, then the results of the two methods are equal

Let V ^ 2V = lnu-2 and y = LNU
Is this correct? There is another question that I don't understand,

There is no problem found, the idea is completely correct. However, the chain rule of composite function is generally used. The applicability of chain rule is good

Find u = x ^ y ^ Z partial derivative, find u to x, u to y, u to Z respectively

u=x^y^z
au/ax=a(x^(y^z))/ax=(y^z) * x^[(y^z)-1]
au/ay=a(x^(y^z))/ay=lnx * x^(y^z) * z*y^(z-1)
au/az=a((x^y)^z)/az=ln(x^y) * (x^y)^z
If you don't understand, please ask

Z = e ^ (u-2v), u = SiNx, v = y ^ 2, respectively

u=sinx
∂u/∂x = cosx
v=y^2
∂v/∂y= 2y
z=e^(u-2v)
∂z/∂x = e^(u-2v) .∂u/∂x
= cosx.e^(u-2v)
∂z/∂y= e^(u-2v) .(-2(∂v/∂y))
=-4y.e^(u-2v)