Question

由方程x^y-2x+y=0确定了隐函数y=y(x),求微分dy

Answer 1

x^y-2x+y=0

e^(ylnx)-2x+y=0

对x求导,得e^(ylnx)*(dy/dx*lnx+y/x)-2+dy/dx=0

即x^y(dy/dx*lnx+y/x)-2+dy/dx=0

dy/dx=(2-y/x)/(x^y+1)=(2x-y)/[x(2x-y+1)]

dy=(2x-y)dx/[x(2x-y+1)]