Z = (1 + XY) ^ (x + 2Y) find AZ / ax (1,1) AZ / ay (1,1) thank you

Z = (1 + XY) ^ (x + 2Y) find AZ / ax (1,1) AZ / ay (1,1) thank you

z(1,1)=8
lnz=(x+2y)ln(1+xy)
dz/dx(1,1)=zln(1+xy)+zy(x+2y)/(1+xy)=8ln2+12
dz/dy(1,1)=2zln(1+xy)+zx(x+2y)/(1+xy)=16ln2+12

Z = f (u.v), u = x-y.v = e, Z / ax. And AZ / ay
Find the partial derivative of compound function

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Z = e ^ (XY) sin (x + y), then AZ / ay=

z=e^(xy)sin(x+y),
az/ay=xe^(xy)sin(x+y)+e^(xy)cos(x+y)

Given ax / (a-b) = ay / (B-C) = AZ / (C-A) (a must not be 0), try to find the value of X + y + Z

Let ax / (a-b) = ay / (B-C) = AZ / (C-A) = K
Then AX = K (a-b); ay = K (B-C); AZ = K (C-A)
By adding the three formulas, we can get a (x + y + Z) = 0; because a cannot be 0, so x + y + Z = 0

Given z = arctany / x, find AZ / ax, AZ / ay

∂z/∂x
=1/(1+y^2/x^2)*(y/x)'
=x^2/(x^2+y^2)*y*(-1/x^2)
=-y/(x^2+y^2)
∂z/∂y
=1/(1+y^2/x^2)*(y/x)'
=x^2/(x^2+y^2)*(1/x)
=x/(x^2+y^2)

The differential equation x * AZ / ax + y * AZ / ay = Z can be transformed into a variable U = x, v = Y / X?
The answer is Z = u * AZ / Au

AZ / AX = AZ / Au * Au / ax + AZ / AV * AV / AX = AZ / Au + AZ / AV (- Y / x ^ 2) AZ / ay = AZ / Au * Au / ay + AZ / AV * AV / ay = AZ / AV * (1 / x) x * AZ / ax + y * AZ / ay = x * AZ / Au + AZ / AV (- Y / x) + AZ / AV (Y / x) = x * AZ / Au = u * AZ / aux * AZ / ax + y * AZ / ay = Z, i.e. z = u * AZ / Au

The definition field of function f (x) is {x | x ∈ R, X is not equal to 0}. For all x.y ∈ R, f (XY) = f (x) + F (y)
The definition field of function f (x) is {x | x ∈ R, X is not equal to 0}. For all x.y ∈ R, f (XY) = f (x) + F (y)
In the first step, it is found that f (x) is an even function
(2) If f (4) = 1 and f (x) is an increasing function on (0, positive infinity), then the solution set of inequality f (3x + 1) + F (2x-6) ≤ 3 is?

f(3x+1)+f(2x-6)≤3
3f(4)=3;
f(3x+1)+f(2x-6)≤3f(4);
f((3x+1)(2x-6))≤f(4*4*4)
And f (x) is an even function,
f(|(3x+1)(2x-6)|)≤f(|64|)
F (x) is an increasing function on (0, positive infinity),
|(3x+1)(2x-6)|≤|64|
-64

The function f (x) defined on (0, positive infinity) satisfies f (XY) = f (x) + F (y) and if x > 1, f (x) 1, f (x) X1 > 0
f(x2/x1)=f(x2)+f(1/x1)=f(x2)-f(x1)

Let x = y = 1 and substitute it into the original formula to get f (1) = 0
Let y = 1 / x, then f (x) = - f (1 / x)

For any x, y ∈ R, f (x + y) = f (x) + F (y), f (XY) = f (x) f (y) holds. When x is not equal to y, ()
The definition field of function y = f (x) is R. for any x, y ∈ R, f (x + y) = f (x) + F (y), f (XY) = f (x) f (y) is constant. When x is not equal to y, f (x) is not equal to f (y). It is proved that 1; if x > 0, then f (x) > 0; 2: F (x) is a monotone increasing function on R

1. F (x + y) = f (x) + F (y) let y = 0f (x) = f (x) + F (0) f (0) = 0x ≠ 0, f (x) ≠ 0 is a monotone increasing function on R for any x > 0f (x) = f (√ x * √ x) = f (√ x) * f (√ x) > 02

If z = (x + y) ^ 3, then the second order partial derivative is ᦉ 8706; ^ 2 Z / ᦉ 8706; X ᦉ 8706; y=

6（x+y）