Question

1,求高等数学Z=arctan[（x+y）/（x-y）]的全微分 2,求z=arctan[(x+y)]/(1-xy)]的全微分

Answer 1

你好！

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Answer 2

1
(x+y)/(x-y)=1+2y/(x-y)
[(x+y)/(x-y)]'x=2y'/(x-y) -2y(1-y')/(x-y)^2
[(x+y)/(x-y)]'y=2/(x-y)+2y(-1)/(x-y)^2 1+(x+y)^2/(x-y)^2=2x^2+2y^2/(x-y)^2
dz=[(dy/dx)(x-y)-y(1-dy/dx)]/(x^2+y^2) *dx + (x-2y)/(x^2+y^2) dy

2
[(x+y)/(1-xy)]'x=(1+y')/(1-xy)-(x+y)(-y-xy')/(1-xy)^2
[(x+y)/(1-xy)]'y=(1+x')/(1-xy)-(x+y)(-x-x'y)/(1-xy)^2
1+(x+y)^2/(1-xy)^2=[(x+y)^2+(1-xy)^2]/(1-xy)^2

dz=[(1+dy/dx)(1-xy)-(x+y)(-y-xdy/dx)]/[(1-xy)^2+(x+y)^2] dx
+ [(1+dx/dy)(1-xy)-(x+y)(-x-ydx/dy)]/[(1-xy)^2+(x+y)^2] dy

追问

诶，答案不对。
第一个答案是，dz=(xdy-ydx)/(x²+y²)
第二个答案是，dz=dx/(1+x^2)+dy/(1+y^2)

Answer 3

1、dz=(xdy-ydx)/(x²+y²)
2、dz=[(1+y²)dx+(1+x²)dy]/[(1-xy)²+(x+y)²]

追问

第一个答案是对的，可有详解？
第二个答案不对，答案是dx/(1+x^2)+dy/(1+y^2)

追答

dz=[(1+y²)dx+(1+x²)dy]/[(1-xy)²+(x+y)²]=dx/(1+x^2)+dy/(1+y^2)

Answer 4