əz/əx = (-1/x²) / [ 1+ (1/x+y)²]
əz/əy = 1/ [ 1+ (1/x+y)²]
dz = əz/əx dx + əz/əy dy = ...(-1/x²)怎么來的？為什么下面對(duì)Y偏導(dǎo)就只有1啦？əz/əx = (-1/x²) / [ 1+ (1/x+y)²]əz/əy = 1/[ 1+ (1/x+y)²]dz = əz/əx dx + əz/əy dy =z=arctan(1/x + y) 還是 z=arctan[ 1/(x+y) ] ??是 z=arctan[ 1/(x+y) ]不好意思，弄錯(cuò)啦。əz/əx = ﹣1/(x+y)² / [ 1+ 1/ (x+y)²]= ﹣1 / [ (x+y)² + 1 ]同理，əz/əy = ﹣1/(x+y)² / [ 1+ 1/ (x+y)²]= ﹣1 / [ (x+y)² + 1 ]dz = ﹣(dx + dy) / [ (x+y)² + 1 ]﹣1/(x+y)²是不是函數(shù)1/(X+Y)的連鎖求導(dǎo)？是的。我知道，謝謝啦。