正确答案及解析
正确答案
解析
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
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
