设z=arctan[(x+y)/(x-y)],则dz=?

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设z=arctan[(x+y)/(x-y)],则dz=?设z=arctan[(x+y)/(x-y)],则dz=?设z=arctan[(x+y)/(x-y)],则dz=?此题比较适合利用一阶微分形式不变

设z=arctan[(x+y)/(x-y)],则dz=?
设z=arctan[(x+y)/(x-y)],则dz=?

设z=arctan[(x+y)/(x-y)],则dz=?
此题比较适合利用一阶微分形式不变性,以及微分本身的运算法则求
令 x+y=u,x-y=v,t=(x+y)/(x-y)=u/v 则
dz=d(arctant)=[1/(1+t^2)] dt=[1/(1+t^2)]d[u/v]=[1/(1+t^2)][1/v^2](vdu-udv)
=[1/(1+u^2/v^2)][1/v^2](vdu-udv)=[1/(u^2+v^2)](vdu-udv)
={1/[(x+y)^2+(x-y)^2]}[(x-y)(dx+dy)-(x+y)(dx-dy)]
=[1/(2x^2+2y^2)](-2ydx+2xdy)
用这种方法解这类题目,条理清晰,不易算错,熟悉以后,不需要写出设置中间变量的过程.

dz={[-2y/(x-y)^2]dx +[2x/(x-y)^2]dy}][(x-y)^2/(2x^2+2y^2)]
刚才算错了特此道歉!