方程e的x y次方 x xy平方=1能确定隐函数y=y(x),试求dy

如题所述

e^(xy) * (xy)^2 = 1
两边取ln:
(xy)+2ln(xy)=0
xy+2(lnx + lny) = 0
两边对x求导数:
y+xy'+2/x+y'/y = 0
y'(x+1/y)=-2/x - y
y' = -[(2+xy)/x]/[(xy+1)/y]
y' = -(xy+2)y/[x(xy+1)]
dy = -(xy+2)y/[x(xy+1)]dx
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