题目

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

提问时间：2021-07-01

答案

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