Z = (1 + x ^ 2 + y ^ 2), then DZ (1,1) is equal to (DX + dy) Such as the title Missed a sign. It should be Z = (1 + x ^ 2 + y ^ 2) open root, then DZ (1,1) is equal to (DX + dy)

∵dz=(z/x)dx+(z/y)dy =[x/√（1+x²+y²）]dx+[y/√（1+x²+y²）]dy ∴dz(1,1)=(1/√3)dx+(1/√3)dy

RELATED INFORMATIONS

1. （2x^2)dy/dx+y=4y^3 The original equation is reduced to (1 / 2x ^ 2) DX = (1 / 4Y ^ 3-y) dy Integral on both sides is - 1 / 6x ^ 3 + C = (1 / 12Y ^ 2-1) ln (4Y ^ 3-y)
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19. The general solution of the first order linear differential equation (x ^ 3 + y ^ 3) dx-3xy ^ 2dy = 0 Such as the title
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