数学隐函数求导，求二阶导数
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arctan(y/x) = ln√(x^2+y^2)
[1/( 1+ (y/x)^2 )] .( -y/x^2 + (1/x) y' ) = (x +y.y')/(x^2+y^2)
[x^2/( x^2+ y^2 )] .( -y/x^2 + (1/x) y' ) = (x +y.y')/(x^2+y^2)
x^2 .( -y/x^2 + (1/x) y' ) = x +y.y'
-y + xy' = x +y.y'
(x-y)y' = x+y
y' = (x+y)/(x-y)
y''
=[(x-y)(1+y') - (x+y)(1-y') ] /(x-y)^2
=[(x-y)(1+y') - (x+y)(1-y') ] /(x-y)^2
=2(xy'-y)/(x-y)^2
=2(x[(x+y)/(x-y)]-y)/(x-y)^2
=2[x(x+y)-y(x-y) ]/(x-y)^3
=2(x^2-y^2)/(x-y)^3
=2(x+y)/(x-y)^2
[1/( 1+ (y/x)^2 )] .( -y/x^2 + (1/x) y' ) = (x +y.y')/(x^2+y^2)
[x^2/( x^2+ y^2 )] .( -y/x^2 + (1/x) y' ) = (x +y.y')/(x^2+y^2)
x^2 .( -y/x^2 + (1/x) y' ) = x +y.y'
-y + xy' = x +y.y'
(x-y)y' = x+y
y' = (x+y)/(x-y)
y''
=[(x-y)(1+y') - (x+y)(1-y') ] /(x-y)^2
=[(x-y)(1+y') - (x+y)(1-y') ] /(x-y)^2
=2(xy'-y)/(x-y)^2
=2(x[(x+y)/(x-y)]-y)/(x-y)^2
=2[x(x+y)-y(x-y) ]/(x-y)^3
=2(x^2-y^2)/(x-y)^3
=2(x+y)/(x-y)^2
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