已知xy+lnx+lny=1，求dy/dx，d2y/dx2．
xy+lnx+1ny=1，求dy/dx，d2y/dx2，移项，等式成为xy+lnx+lny-1=0．令F(x，y)=xy+lnx+lny-1，则有dy/dx=-[(∂F/∂x)/(∂F/∂y)=-(y+1/x)/(x/1/y)=-(y/x)，d2y/dx2=(d/dx)[-(y/x)]=-[(dy/dx)x-y]/x2=-[-(y/x)x-y]/x2=2y/x2
已知xy+lnx+lny=1，求dy/dx，d2y/dx2．
xy+lnx+1ny=1，求dy/dx，d2y/dx2，移项，等式成为xy+lnx+lny-1=0．令F(x，y)=xy+lnx+lny-1，则有dy/dx=-[(∂F/∂x)/(∂F/∂y)=-(y+1/x)/(x/1/y)=-(y/x)，d2y/dx2=(d/dx)[-(y/x)]=-[(dy/dx)x-y]/x2=-[-(y/x)x-y]/x2=2y/x2
