设函数y=y()x由方程cos(xy)-ln[(x+y)/y]=y确定：求dy/dx|x=0
对cosxy-ln(x+y)+lny=y两边对x求导得 -siny[y+x(dy/dx)-1/(x+y)(1+dy/dx)+(1/y)y′=y′ 所以 dy/dx=[-ysinxy-1/(x+y)]/[1-1/y+1/(x+y)+xsinxy] 又x=0时，y=cosO-lnl=1 所以 dy/dx|x=0={[-(1/y)]/(1-1/y+1/y)}|y=1=-1
设函数y=y()x由方程cos(xy)-ln[(x+y)/y]=y确定：求dy/dx|x=0
对cosxy-ln(x+y)+lny=y两边对x求导得 -siny[y+x(dy/dx)-1/(x+y)(1+dy/dx)+(1/y)y′=y′ 所以 dy/dx=[-ysinxy-1/(x+y)]/[1-1/y+1/(x+y)+xsinxy] 又x=0时，y=cosO-lnl=1 所以 dy/dx|x=0={[-(1/y)]/(1-1/y+1/y)}|y=1=-1
