Question

求隐函数导数或微分.求说明一下思路,

Answer 1

y = e^(xy) + tan(xy) y' = (y + xy')e^(xy) + (y+xy')sec^2(xy)

y' = ye^(xy) + xy'e^(xy) + ysec^2 (xy) +xy'sec^2 (xy)

y' [1-xe^(xy)-xsec^2(xy)] = y[e^(xy) + sec^2(xy)]

解出：

y' = y[e^(xy) + sec^2(xy)] / [1-xe^(xy)-xsec^2(xy)]

x - y +0.5siny = 0

dx - dy + 0.5cosy dy = 0 dy(0.5cosy - 1) +dx =0

dy = dx / (1 - 0.5cosy)

思路是：对等式两边同时对x求导数,遇到y 对x求导为：y‘

之后解出:y’ = f(x,y).要求dy ：则：dy = f(x,y) dx.