设二元函数z=xex+y+(x+1)ln(1+y),则dz|(1,0)=__________.

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问题 设二元函数z=xex+y+(x+1)ln(1+y),则dz|(1,0)=__________.

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答案2edx+(e+2)dy

解析 因为dz=ex+ydx+xdex+y+ln(1+y)d(∞+1)+(x+1)dln(1+y)
    =ex+ydx+xex+yd(x+y)+ln(1+y)dx+ln(1+y)dx+(x+1)dy/1+y
    =ex+ydx+xex+y(dx+dy)+ln(1+y)dx+(x+1)dy /1+y,
所以dz|(1,0)=edx+e(dx+ay)+2dy=2edx+(e+2)dy.
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