设z=z(x,y)由方程ez-xy2+sin(y+z)=0确定,求dz.

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问题 设z=z(x,y)由方程ez-xy2+sin(y+z)=0确定,求dz.

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答案对等式ez-xy2+sin(y+z)=0求微分得 d(ez)-d(xy2)+d[sin(y+z)]=0, ezdz-y2dx-2xydy+cos(y+z)(dy+dz)=0, [ez+cos(y+z)]dz=y2dx+[2xy-cos(y+z)]dy, 所以[*]

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