设z=x^2+y^2,其中y=y(x)是由方程x^2-xy+y^2=1所确定的隐函数，求dy/dx|x=1,y=0

Answer 1

x^2-xy+y^2=1
2x-y-xy'+2yy'=0
y'=(2x-y)/(x-2y)
dy/dx|(x=1,y=0) =2

dz/dx=2x+2ydy/dx=2x+2y(2x-y)/(x-2y)=2(x^2-y^2)/(x-2y)
dz/dx (1,0) =2

Answer 2

由隐函数求导法可得
dy/dx=-(2x-y)/(2y-x)
故dy/dx|x=1,y=0=2