【正确答案】：x=0代入等式，得e-e^y=sin(0·y) y=0 等式两边同时求导 e^x-(e^y)·y'=cos(xy)·(y+xy') [xcos(xy)+e^y]·y'=e^x-ycos(xy) y'=[e^x-ycos(xy)]/[xcos(xy)+e^y] y'|x=0 =[e-ycos(0·y)]/[0·cos(0·y)+e] =(1-y)/1 =(1-0)/1 =1