求隐函数导数或微分.求说明一下思路,

1个回答

  • 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.