Question

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

Answer 1

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)*yP(z)/ [1- xyP'(z)] } dx+{ f'z(x,z)*xP(z) / [1- xyP'(z)] }dy

追问

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

追答

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)] }dy
x 和y 是两个自变量呀