Question

2.求曲线x2+xy+y2=1上点(1,-1)处的切线方程与法线

Answer 1

x^2+xy+y^2=1

d/dx(x^2+xy+y^2)=0

2x + xy'+y +2y.y'=0

(x+2y)y' = -(2x+y)

y' = -(2x+y)/(x+2y)

y'|(x,y)=(1,-1) 

= -(2-1)/(1-2)

=1

点(1,-1)处的切线方程

y+1= x-1

x-y-2=0

点(1,-1)处的法线方程

y+1 =-(x-1)

x+y=0

Answer 2

求隐函数的导数。

详情如图所示:

供参考，请笑纳。

Answer 3

求隐函数的导数。