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
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dz=d[xyP(z)]=yP(z)dx+xP(z)dy+xyP'(z)dz
所老察以 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'侍袭茄x(x,z) + f'z(x,z)*yP(z)/ [1- xyP'(z)] } dx+{ f'z(x,z)*xP(z) / [1- xyP'禅差(z)] }dy
所老察以 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'侍袭茄x(x,z) + f'z(x,z)*yP(z)/ [1- xyP'(z)] } dx+{ f'z(x,z)*xP(z) / [1- xyP'禅差(z)] }dy
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