设z=z(x,y)由x-yz+yez-x-y=0确定,求

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问题 设z=z(x,y)由x-yz+yez-x-y=0确定,求

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答案方程x-yz+yez-x-y=0两边对x求偏导得1-ydz/dx+yez-x-y(dz/dx-1)=0,解得dz/dx=(yez-x-y-1)/y(ez-x-y-1),方程x-yz+yez-x-y=0两边对y求偏导得-ydz/dx-z+ez-x-y+yez-x-y(dz/dx-1)=0,解得dz/dx=[(y-1)ez-x-y+z]/y(ez-x-y-1),则dz=(yez-x-y-1)/y(ez-x-y-1)dx+[(y-1)ez-x-y+z]/y(ez-x-y-1)dy.

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