Question

xy+lnx+lny=1,求d^2y/dx^2

Answer 1

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