设y=f(x)是由方程xy+lnx+y=1所确定的函数,求dy.
问题描述:
设y=f(x)是由方程xy+lnx+y=1所确定的函数,求dy.
答
方程两边同时求x对y的导:y+xdy/dx+1/x+2ydy/dx=0,dy/dx=-(y+1/x)/(x+2y),dy=-(y+1/x)dx/(x+2y)
设y=f(x)是由方程xy+lnx+y=1所确定的函数,求dy.
方程两边同时求x对y的导:y+xdy/dx+1/x+2ydy/dx=0,dy/dx=-(y+1/x)/(x+2y),dy=-(y+1/x)dx/(x+2y)