x^2+y^2-z = f(x+y+z),兩邊對x求偏導數(shù),得2x-z'=(1+z')f'(x+y+z),解得 z'=[2x-f'(x+y+z)]/[1+f'(x+y+z)];兩邊對y求偏導數(shù),得2y-z'=(1+z')f'(x+y+z),解得 z'=[2y-f'(x+y+z)]/[1+f'(x+y+z)].則 dz = {...（【請用“兩邊同時取微分”的方法做這道題】）x^2+y^2-z = f(x+y+z),
兩邊對x求微分，得
2xdx-z'dx=(dx+z'dx)f'(x+y+z),
解得 z'dx = [2x-f'(x+y+z)]dx/[1+f'(x+y+z)];
兩邊對y求微分，得
2ydy-z'dy=(dy+z'dy)f'(x+y+z),
解得 z'dy = [2y-f'(x+y+z)]dy/[1+f'(x+y+z)].
則 dz={[2x-f'(x+y+z)]dx+[2y-f'(x+y+z)]dy}/[1+f'(x+y+z)].