sin(xy)+ln(y-x)=x,求y`
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sin(xy)+ln(y-x)=x,求y`
sin(xy)+ln(y-x)=x,求y`
sin(xy)+ln(y-x)=x,求y`
两边对x求导得
(y+xy')cos(xy)+(y'-1)/(y-x)=1
解得
y'=[y-x+1-y(y-x)cos(xy)]/[x(y-x)cos(xy)+1]
自己可要亲自算啊.
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