dz=d[xyP(z)]=yP(z)dx+xP(z)dy+xyP'(z)dz所以 dz=[ yP(z)dx+xP(z)dy] / [1- xyP'(z)] du=df(x,z) = f'x(x,z)dx+ f'z(x,z)dz= f'x(x,z)dx+ f'z(x,z)*{ [ yP(z)dx+xP(z)dy] })/ [1- xyP'(z)] ={ f'x(x,z) + f'z(x,z)*y...设u=f(x,z)而z(x,y)是由方程z=x +yP(z)所确定的函数,求du谢谢了不是求好了吗du由两部分组成{ f'x(x,z) +f'z(x,z)*yP(z)/ [1- xyP'(z)] } dx和{ f'z(x,z)*xP(z) / [1- xyP'(z)] }dyx 和y 是两个自变量呀