见图.理论方面说不清楚,我把答案写出来了.
两边对x求导，y看成常数σz/σx=e^(xy)[(y+0)]+cos(yz)[0+yσz/σx)+zx^(z-1)σz/σx得[1-ycos(xy)-zx^(z-1)]σz/σx=ye^xy∴σz/σx=ye^(xy)/[1-ycos(xy)-zx^(z-1)]
（ye^xy+2x)/-ycosyz
