Question

设函数y=y(x)由方程ln(x^2+y^2)=arctany/x所确定,求dy|x=1,y=0

Answer 1

简单计算一下即可，答案如图所示

Answer 2

ln(x^2+y^2)=arctany/x
(2x+2yy')/(x^2+y^2)=[y'/(1+y^2)·x-arctany]/x^2
将x=1，y=0代入上式：
(2×1+2×0y')/(1^2+0^2)=[y'/(1+0^2)×1-arctan0]/1^2
2=[y'-0]/1
y'=2
dy/dx=2
dy=2dx