設u=f(x,z)而z(x,y)是由方程z=x yP(z)所確定的函數,求du

設u=f(x,z)而z(x,y)是由方程z=x yP(z)所確定的函數,求du

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…