Question

y=arcsin根号1+x分之1-x求导数

Answer 1

y=arcsin√[(1-x)/(1+x) ]
siny =√[(1-x)/(1+x) ]
(siny)^2 = (1-x)/(1+x)
2(siny).(cosy) . dy/dx = -1/(1+x)^2
2√[(1-x)/(1+x) ] . √ [ 1 - (1-x)/(1+x) ] .dy/dx = -1/(1+x)^2
2√[(1-x)/(1+x) ] . √ [ 2x/(1+x) ] .dy/dx = -1/(1+x)^2
{ 2√[2x(1-x)]/(1+x) } .dy/dx = -1/(1+x)^2
dy/dx = -1/{ 2(1+x) .√[2x(1-x)] }

追问

有跟号的

追答

改了
y=arcsin√[(1-x)/(1+x) ]
siny =√[(1-x)/(1+x) ]
(siny)^2 = (1-x)/(1+x)
2(siny).(cosy) . dy/dx = -1/(1+x)^2
2√[(1-x)/(1+x) ] . √ [ 1 - (1-x)/(1+x) ] .dy/dx = -1/(1+x)^2
2√[(1-x)/(1+x) ] . √ [ 2x/(1+x) ] .dy/dx = -1/(1+x)^2
{ 2√[2x(1-x)]/(1+x) } .dy/dx = -1/(1+x)^2
dy/dx = -1/{ 2(1+x) .√[2x(1-x)] }

追问

y=arcsin√[(1-x)/(1+x) ]怎么变的
siny =√[(1-x)/(1+x) ]

追答

y=arcsin√[(1-x)/(1+x) ]

siny = sin{arcsin√[(1-x)/(1+x) ]} = √[(1-x)/(1+x) ]

Answer 2

y=arcsin√[(1-x)/(1+x)],
y'=1/√{1-（1-x）/(1+x)}*1/{2√[(1-x)/(1+x)]}*[-(1+x)-(1-x)]/(1+x)^2
=1/√[2x/(1+x)]*1/{2√[(1-x)/(1+x)]}*(-2)/(1+x)^2
=-1/{(1+x)√[2x(1-x)]}.

追问

错了