微积分求大神 2.(4)
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方程两边对x求导
cos(xy)*(y+xy')-2(y'-1)/(y-x)=2
cos(xy)*y+cos(xy)*xy'+2y'/(x-y)-2/(x-y)=2
y'=[2+2/(x-y)-cos(xy)*y]/[2/(x-y)+cos(xy)*x]
=[2x-2y+2-cos(xy)*y(x-y)]/[2+cos(xy)*x(x-y)]
cos(xy)*(y+xy')-2(y'-1)/(y-x)=2
cos(xy)*y+cos(xy)*xy'+2y'/(x-y)-2/(x-y)=2
y'=[2+2/(x-y)-cos(xy)*y]/[2/(x-y)+cos(xy)*x]
=[2x-2y+2-cos(xy)*y(x-y)]/[2+cos(xy)*x(x-y)]
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