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xy+lnx+lny=1,求d^2y/dx^2
人气:194 ℃ 时间:2020-05-21 06:38:38
解答
两边对x求导
(xy)'+1/x+1/y*dy/dx=0
(x*dy/dx+y)+1/x+1/y*dy/dx=0
(x+1/y)dy/dx=-y-1/x
dy/dx=(-y-1/x)/(x+1/y)=y(-xy-1)/x(xy+1)=-y/x
d^2y/dx^2=(-y/x)'=-(y'x-y)/x^2=-(-y/x*x-y)/x^2=2y/x^2
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