Question

e^(xy)+sinxy=y^2,对此函数求导

Answer 1

两边同时对x进行求导：
e^(xy)*(y+xy')+cosxy*(y+xy')=2yy'
∴[2y-xe^(xy)-xcosxy]y'=ye^(xy)+ycosxy
∴y'=[ye^(xy)+ycosxy]/[2y-xe^(xy)-xcosxy]