高等数学求切线方程

Question

第九题

Answer 1

(0,1)
ln√(x^2+y^2) =arctan(x/y)
(1/2)ln(x^2+y^2) =arctan(x/y)
两边求导
(x+y.y')/(x^2+y^2) ={1/[1+(x/y)^2] } . [ 1/y -(x/y^2).y' ]
= [y^2/(x^2+y^2) ] . [ 1/y -(x/y^2).y' ]
x+y.y' = y^2 . [ 1/y -(x/y^2).y' ]
= y - xy'
(x+y)y' = y-x
y' =(y-x)/(x+y)
y'|(x,y)=(0,1)
= (1-0)/(0+1)
=1
切线方程 (0,1)
y-1 = x-0
x-y+1 =0