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设方程x^2+z^2=2ye^z确定z=z(x,y),求dz
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设方程x^2+z^2=2ye^z确定z=z(x,y),求dz
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答案和解析
两边对dx求导
2x+2zdz/dx=2ye^zdz/dx
dz/dx=x/(ye^z-z)
同理 对dy求导
2zdz/dy=2e^z+2ye^zdz/dy
dz/dy=e^z/(z-ye^z)
dz=xdx/(ye^z-z)+e^z dy/(z-ye^z)
2x+2zdz/dx=2ye^zdz/dx
dz/dx=x/(ye^z-z)
同理 对dy求导
2zdz/dy=2e^z+2ye^zdz/dy
dz/dy=e^z/(z-ye^z)
dz=xdx/(ye^z-z)+e^z dy/(z-ye^z)
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