Z = f (x, y) is the function given by equation E ^ (- XY) - 2Z + e ^ Z, find the total differential DZ

Z = f (x, y) is the function given by equation E ^ (- XY) - 2Z + e ^ Z, find the total differential DZ

e^(-xy)-2z+e^z=0-ye^(-xy)-2z'(x)+e^z z'(x)=0z'(x)=ye^(-xy)/(e^z-2)-xe^(-xy)-2z'(y)+e^z z'(y)=0z'(y)=xe^(-xy)/(e^z-2)dz=ye^(-xy)/(e^z-2) *dx + xe^(-xy)/(e^z-2) *dy=[ye^(-xy)dx+xe^(-xy)dy]/(e^z-2)

∫ xdx + YDY + (x + Y-1) DZ, where the integral is a straight line from point (1,1,1) to point (2,3,4) Thank you, brothers and sisters. Thank you very much

The linear equation is X-1 = (Y-1) / 2 = (Z-1) / 3
The solution is x = (Z + 2) / 3, y = (2Z + 1) / 3, so x + Y-1 = Z
So ∫ xdx + YDY + (x + Y-1) DZ = ∫ (1,2) xdx + ∫ (1,3) YDY + ∫ (1,4) ZDZ = 13

In total differential, DZ = a △ x + B △ y= э z/ э* △x+ э z/ э y*△y there э z/ э* Is delta x the partial derivative of X? э z/ э Is y * △ y the partial derivative of Y?

Yes, the form of total differential is DZ = f'x (x, y) △ x + f'y (x, y) △ y
a. B represents the partial derivative

It is known that the total differential of function z = f (x, y) is DZ = 2xdx-2ydy, and f (1,1) = 2, when f (x, y) is in the region D = {(x, y) |x ^ 2 + y ^ 2 / 4 ≤ 1} Find the maximum and minimum values of F (x, y) (the more detailed the process, the better,

F (x, y) = x ^ 2 - y ^ 2 + C, f (1,1) = 2 = > C = 2F (x, y) = x ^ 2 - y ^ 2 + 2, region D = {(x, y) | x ^ 2 + y ^ 2 / 4 ≤ 1}, (1) in the interior of region D, from 2x = 0, 2Y = 0: stagnation point (0,0), f (0,0) = 2 (2) on the boundary of region D, x ^ 2 = 1

(x + ay) DX + YDY is total differential, then a =?

a=0
(x+0*y)dx+ydy=d(1/2*x^2+1/2*y^2)

[(x + ay) DX + 4ydy] / (x + 2Y) is the total differential of a function U (x, y). Find a

du=[（x+ay）dx+4ydy]/(x+2y)
u'y=4y/(x+2y)=(4y+2x-2x)/(x+2y)=2-2x/(x+2y)
Integral to Y: u = 2Y XLN (x + 2Y) + G (x)
u'x=-ln(x+2y)-x/(x+2y)+g'(x)
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