请问这道全微分题怎么做,谢谢

如题所述

z=arctan (x-y)/(x+y)
那么Z'x=1/[1+(x-y)²/(x+y)²] *[(x-y)/(x+y)]'
=1/[1+(x-y)²/(x+y)²] * 2y/(x+y)²
=2y/(2x²+2y²)
同理Z'y=1/[1+(x-y)²/(x+y)²] *[(x-y)/(x+y)]'
=1/[1+(x-y)²/(x+y)²] * -2x/(x+y)²
= -2x/(2x²+2y²)
即dz=2y/(2x²+2y²) dx -2x/(2x²+2y²) dy
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