一道大一高数题,设方程xtany=cos(xy)确定了隐函数y=y(x),求dy/dx.

两边同时对x求偏导
而且注意y=y(x)
所以由链式法则,如果有关于y的函数对x求导,最后会多出一个因子dy/dx
左边积法则+链式,右边链式+积法则
tany+sec^2 y *dy/dx = -sin(xy)*d(xy)/dx
tany+sec^2 y *dy/dx = -sin(xy)*(y+x*dy/dx)
把dy/dx的整理到一起
(sec^2 y+xsin(xy))(dy/dx)=-ysin(xy)-tany
dy/dx=[-ysin(xy)-tany]/[sec^2 y+xsin(xy)]